I

High Dimensional Euclidean Geometry

Wroted by Li Honglu

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Informed by the author of this book

June 7,2004

Brief Introduction

This book is composed of three parts:oblique axes transform,oblique axes draughting and

high-dimensional spacial analytic geometry.It has totally nine chapters：Relation and property

determined by singular linear transform；"Relation" method and preferred n-dimensional coordinate

system；Relation between figure and digit in preferred n-dimensional coordinate system；Figure

shape of preferred n-dimensional coordinate system；Figure making of preferred n-dimensional

coordinate system；Interleaving and distance between two linear figures；Included angle between two

linear figures and its linear solution；Jianshi solution of the included angle between two linear

figures；Application of High Dimensional Euclidean Geometry .This book is truly a monograph of

high-dimensional spacial analytic geometry,it is based on (but different from) euclidean space theory

in linear algebra.

This book has reference value for students and teachers of relative speciality in colleges and

universities and researchers working on operational research and graphic theory.

II

Catalog

High Dimensional Euclidean Geometry I

Brief Introduction ..................................................................................................................... I

Prefaces .....................................................................................................................................V

The First Chapter: Relation and property determined[1]

by singular linear transform .... 1

§1 Property and relations between image and image source of singular linear transform. 1

1.1 Concept of "relation"............................................................................................ 1

1.2 Relations between image and image source of singular linear transform ............ 2

1.3 Property of relations between image and image source under singular linear

transform .................................................................................................................... 2

§2 Operation rules of elements in relationσ ....................................................................... 3

2．1 Linear operation rules........................................................................................ 3

2．2 Transposition (of terms) rules ........................................................................... 3

2．3 Element exchange rules..................................................................................... 5

§3 Relations between coordinates of vectors under singular linear transform................... 7

Exercises .......................................................................................................................... 13

The Second Chapter: "Relation" method and preferred n-dimensional coordinate system

.................................................................................................................................................. 14

§1 "Relation" method and examples of preferred 4-dimensional and 5-dimensional

coordinate system............................................................................................................. 14

1．1 Examples of building preferred 4-dimensional coordinate system ................. 14

1．2 Examples of building preferred five-dimensional coordinate system ............. 16

§2 Oblique axes transform and preferred n-dimensional coordinate system ................... 17

2．1 Oblique axes transform and building of preferred n-dimensional system....... 17

2 ． 1 ． 1 Methods and steps of building preferred n-dimensional

system——oblique axes transform ..................................................................... 17

2．1．2 Structure of preferred n-dimensional system........................................ 19

2．1．3 Kinds of preferred n-dimensional system............................................. 20

§3 Property of preferred n-dimensional system ............................................................... 21

3.1 Relation between preferred n-dimensional system and oblique axes transform 21

3．2 The "preference" of preferred coordinates system .......................................... 22

§4 Punctiform figures in preferred n-dimensional system—generic point ...................... 23

4．1 Generic point、projection trace and opposite trace ........................................ 23

4．1．1 Concept of generic point ...................................................................... 23

4．1．2 Projection trace and opposite trace of generic point............................. 23

4．2 Property of generic points .......................................................................... 24

4．2．1 Shape of generic points ........................................................................ 24

4．2．2 The uniqueness of generic points relative to upright axis .................... 25

4．2．3 Uniqueness of projection trace or opposite trace.................................. 25

Exercises .......................................................................................................................... 26

The Third Chapter: Relation between figures and numbers in preferred n-dimensional

system ...................................................................................................................................... 28

§1 Translation trace of generic point—generic curved surface and generic curve........... 28

1．1 Concepts of generic curved surface and generic curve ................................... 28

III

1．2 Dimensions of generic curved surface and generic curve—dimension theorem29

1．3 Kinds of generic curved surface and generic curve......................................... 30

§2 Graphic rules of preferred n-dimensional system—three kind of graphic methods.... 30

2．1 Concepts of three graphic methods ................................................................. 30

2．2 Indirect graphic method and ordinary graphic method ................................... 31

2．2．1 Indirect graphic method........................................................................ 32

2．2．2 Common generic of points ................................................................... 33

2．2．3 Ordinary figured method...................................................................... 35

§3 Direct graphic method................................................................................................. 36

3．1 Single principal overlapping direction、oblique number and oblique coordinate

.................................................................................................................................. 36

3．2 Coordinate transform of points........................................................................ 37

3．3 The principle of direct graphic method ........................................................ 38

3.4 Drawing and distinguishing steps of direct graphic method .............................. 39

Exercises .......................................................................................................................... 42

The Fourth Chapter: Shape of figures in preferred n-dimensional system ...................... 44

§1 Shape of linear figures ................................................................................................ 44

1．1 Generic planes and their shape........................................................................ 44

1．2 Generic straight lines and their shape.............................................................. 46

1．3 Shape of many generic planes' intersection.................................................. 51

§2 Shape of non-linear figures ......................................................................................... 56

Exercises .......................................................................................................................... 58

The Fifth Chapter: Making of figures in preferred n-dimensional system....................... 59

§1 Cutting-trace method................................................................................................... 59

§2 Leading-axis method................................................................................................... 61

§3 Synthetical graphic method......................................................................................... 62

The Sixth Chapter: Interleaving and distance between two linear figures. ...................... 73

§1 Interleaving between two linear figures ...................................................................... 73

§2 Orthodromic space and normal space ......................................................................... 77

§3 Exterior sumof two linear figures................................................................................ 81

3.1 Concepts of exterior sumof two linear figures ................................................ 81

3.2 Property of outer sum...................................................................................... 82

3.3 Equation of outer sum ........................................................................................ 83

§4 Distance between two linear figures ........................................................................... 84

4.1 Distance between two parallel figures................................................................ 84

4.2 Distance between two mutually interleaving figures ...................................... 85

Exercises .......................................................................................................................... 86

The Seventh Chapter: Included angle questions and their linear solutions between two

linear figures ........................................................................................................................... 87

§1 Diversity of included angle questions bertween two linear figures of higher space ... 87

1.1 Nonuniqueness of included angle amounts ........................................................ 87

1.1.1 Two kind of projection methods................................................................. 87

1.1.2 Defination of included angle between two linear figures........................... 89

1.1.3 Common vector and uncommon vector——sameness of two linear figures'

IV

dimension ............................................................................................................ 92

1.2 Nonuniqueness of the solution of included angle questions............................... 94

§2 Linear solution of included angle questions between two linear figures .................... 95

2.1 Positive angle method ................................................................................... 95

2.2 Complementary angle method.......................................................................... 100

Exercises ........................................................................................................................ 104

The Eighth Chapter: Jianshi solution solution of included angle questions between

twolinear figures................................................................................................................... 105

§1 Orthogonal transform and principal axis questions................................................... 105

1.1 Orthogonal matrix and orthogonal transform................................................... 105

1.2 Concepts and property of exterior product among vectors—orthogonal-unitization

of vectors................................................................................................................ 106

1.3 Principal axis questions [14] ...........................................................................111

§2 Projective generic elliptic cylinder and projection of generic circle ..........................113

2.1 Projection of sufficient-order generic circle......................................................113

2.2 Projection of difficient-ordergeneric circle .......................................................115

§3 Jianshi solution solution of included angle questions between two linear figures .....118

3.1 Principle and steps of Jianshi solution solution.................................................118

3.2 Included angle questions between two planes...................................................119

3.3 Included angle questions between other linear figures..................................... 125

3.4 Other questions of Jianshi solution principle——exterior product method and

included angle ........................................................................................................ 129

Exercises ........................................................................................................................ 134

The Ninth Chapter: The application of High Dimensional Euclidean Geometry .......... 136

§1 Application of High Dimensional Euclidean Geometry in linear programming....... 136

1.1 Example one..................................................................................................... 136

1.2 Theory, steps and assumption of graphical method in preferred

n-dimensionalsystem.............................................................................................. 140

1.2.1 Theory ...................................................................................................... 140

1.2.2 Method、steps ......................................................................................... 144

1.3 An assumption.................................................................................................. 147

§2 Application of High Dimensional Euclidean Geometry in non-linear programming 150

2.1 Example 2......................................................................................................... 151

2.2 Methods and steps ............................................................................................ 153

2.2.1 Determination of search direction ............................................................ 153

2.2.2 Determination of search range ................................................................. 155

2.3 The case that objective generic curved surface is solid.................................... 158

References...................................................................................................................... 168

Postscript........................................................................................................................ 169

Solutions to part exercises.............................................................................................. 170

Indexes of special terms or symbols .............................................................................. 171

Appendix：solution process of part exercises ....................................................... 173

V

Prefaces

High Dimensional Euclidean Geometry is the author's accomplishment of seventeen years'

effort. During this time, he educated himself while he studied it ,his accomplishment was reflected in

this book.

This book is composed of three parts:"oblique axes transform",oblique axes draughting and

high-dimensional spacial and analytic geometry.

"Oblique axes transform"(the auther named it "relation method") is the theoretical basis of the

whole book. It formed another linear transform on the basis of the relation between image and image

source which is determined by singular linear transform.

Under singular linear transform, relevant concepts,properties and operation rules of relations

between image and image source are studied in depth in this book，it also overcame obstacles

appeared in "relation" method application.

"Oblique axes draughting" is the basic method in this book. On the basis of using oblique axes

transform to establish preferred n-dimensional coordinate system(an analogue figure of

n-dimensional spatial rectangular axis),this book advanced methods to study the relation between

graphics and its relevant algebraic equation in coordinate system. This includes three kind of graphic

methods(direct graphic method, indirect graphic method and general graphic method)which are the

core contents, it summarized all graphic rules of preferred n-dimensional coordinate system：graphic

principles and graphic methods of a random graph and recognition methods of objects be

graphed,etc.Besides, it also includes points having common generic theory、dimension theorem and

cutting-trace method、 leading-axis method、synthetical graphic method.All of this proceeded

beneficial study to preliminarily formed a more systematic, construction reasonable,convenient and

practical,novel graph theory system. The graphic methods are easily to learn and memerize, readers

who know engineering linear algebra can easily master it.

High dimensional spacial analytical geometry is the principal part of High Dimensional

Euclidean Geometry.Oblique axes draughting naturally extendes some relative concepts of

three-dimensional space to high space,then beginners can easily master it. But oblique axes transform

takes questions of point coincidence as a breakthrough, then large nembers of geometric problems

of high space can be readily solved.

VI

Readers who have read this book will find that although linear algebra has geometry concepts of

euclidean space,vector and coordinate etc,if you use linear algebra methods to deal with high

dimensional geometry questions,some other geometry concepts must be introduced. In this book,

the author introduced so-called orthodromic space,normal space ,orthodromic vector,normal vector,

common vector,uncommon vector when he discussed relations between figures denoted by system of

non-homogeneous linear equations and the figures denoted by its corresponding system of

homogeneous linear equations. Some above mentioned terms are defined by the author,some are

different from the popular,please pay attention to it when you read the book. To solve distances between linear figures, the author introduced concept of "exterior sum" and

discussed its property and equation. He extended some concepts and property about subspace in

linear algebra to generic linear figures. In solving included angles between linear figures, the author

put forward and solved two questions of included angle numbers and the nonuniqueness of its

solution. He used linear algebra theory to introduce Jianshi solution,and extended it to solving

included angles between two randon linear figures. He also extended the concept of "exterior

product"(another name"cross product" or "vector product")of vector of three-dimensional space to

hyperspace and put forward vector orthogonalization and another method to solve system of linear

equations. The author of this book used linear algebra methods to study graphic theory, his profitable

discussion may arouse interest and attention of some people concerned.

Shen Yidan

Beijing Institute of Technology

Hou Bingtao

Beijing Academy of Armoured Force Engenering

Pan Jianzhong

The Maths Graduate School of Chinese Academy of Sciences

July,1996

1

The First Chapter: Relation and property determined[1]

by

singular linear transform

Study of descriptive geometry and high-dimensional spacial analytic geometry often use linear

transform . This book introduced a "relation" method, viz. directly use relation between image and

image source in linear transform to solve some questions. Because elements of "relation" can

directly calculate under some certain conditions, the "relation" method has the advantage of

convenience and direct-view, then we can solve many questions not solved by ordinary linear

transform and make the method theoretical basis of this book. Under the staminal guidance and

personal participation of professor Hou Bingtao of Beijing Academy of Armoured Force

Engenering, the author studied property and operation rules of relations between image and image

source of singular linear transform in depth. Except theorem 3、6、10、11，the other seven theorems

and two inferences were strictly and scientifically proved by professor Hou himself. Of which,

theorem 4、5、8 and two inferences were newly put forward by professor Hou.

§§§§1 Property and relations between image and image source of

singular linear transform

1.1 Concept of "relation"

Many people are used to operate under equal relations.For unequal relations,many people are

used to " larger than" or "less than" relation.To top it all, people often try to change some unequal

relations into equal relations(viz. the relation between image and image source of singular

transform which will be discussed after)，so it obliterates lively difference of different things and

virtually set obstacles to subsequent operation. This almost makes some study pause for a long

time.In fact, the relation between different things is complicated and omnifarious. For

example,"parallel"、"vertical"、"homogeneous" which are known very well by people.

Let's introduce a "relation" of more extensive meanings，some "expressions" of the

"relation" can directly add and multiplicate on some particular conditions.

In discrete mathematics[2]

, a1∈A1，a2∈A2，…，an∈An are regarded as a random set. Set

which is composed of all n-component ordered classes (a1,a2,…,an) is called Cartesian product of

A1，A2，…，An，it is denoted by A1×A2×…×An，viz.

A1×A2×…×An ={(a1,a2,…,an)|a i∈A i。i =1,2,…,n}.

A random subset A1×A2×…×An of Cartesian product

is called a n-component relation of A1，A2，…，An.

For example， A1，A2，…，An all can be regarded as sets of real number field R，random

variables (a1,a2,…,an) of linear space Vn of real number field R are apparently elements of

Cartesian product，therefore，Vn is a subset of Cartesian product A1×A2×…×An ，

it is called a n-component relation of A1，A2，…，An.

An important particular case is n=2, if an ordered pair set σis a subset of

A1×A2={(a 1,a 2)| a1∈A1，a2∈A2}，

then σ is called a 2-component relation from A1 to A2 ，"relations" refered in this book belong

to this 2-component relation.

2

If σ is a 2-component relation from A1 to A2 , and

(a 1,a 2)∈σ (a 1∈A1, a 2∈A2) ，

then a 1 has relation σ with regard to a 2，marked as

a1 σ a2 .

Formulas used to express the relations between a1 and a2 are called "relational expressions"，they are called "relations" for short.

1.2 Relations between image and image source of singular linear transform

Let's assume Vn is linear space of real number field R，σ is singular linear transform of

Vn，for random α∈Vn ，σ（α）＝α′is called image of α，and α is called image source

of α. At the same time，σ(Vn)=Vn′is called image set，Vn is called image source set. Let's

assume σ is singular linear transform of linear space Vn ，α∈Vn,α′∈Vn′，if σ(α)＝α′，

then nn VV ′⊂σ ，viz. σ is a subset of Vn×Vn′，so σ is called a 2-component relation from Vn

to Vn′，and (α，α′) ∈σ ，then α has relationσwith regard to α′，marked as α σ α′，

ifσ(α)≠α′，then ( ) , , σαα ∉′ marked as α α.

1.3 Property of relations between image and image source under singular linear

transform

Theorem 1：Under singular linear transform σ，relation σ is not self reflexive.

Proof：If ∀α∈Vn ，there is ασα，then σ is an identical transform，which is in

contradiction with the singularity of σ，so relation σ doesn't has reflexivity. ▌

(note：terminal signs“ ▌”shows that the proof is finished.The same to the next).

Theorem 2：Under singular linear transformσ，relationσis not symmetrical.

Proof：Assume A is the matrix of σ to fundus e1 ，e2 ，…，en ，then A2

is the matrix

ofσ2 .

Assume relationσhas symmetry，viz.∀α,β∈Vn, because σ(α)=β, andσ(β)=α,thenσ〔σ(α)〕＝α

viz.σ2 is an identical transform，then A

2＝I ，

I is a n-moment unit matrix，which is in contradiction with that the rank r of A is less than n ，then

relationσhasn't symmetry. ▌

Theorem 3：Relation σ is not transferable under singular linear transform σ.

Proof：Assume relationσhas transferability，viz.∀α,β,γ∈Vn ，

Because σ(α)= β, σ(β)=γ, σ(α)=γ,

Then σ[σ(α)]＝σ(α )，

viz. σ(α)＝α（σmay have generalized inverse[4]），

Therefore σ[σ(α)]＝α，

but bothσandσ2 are not identical transform concluded by theorem 1 and theorem 2，so relationσhasn't transferability. ▌

3

§§§§2 Operation rules of elements in relationσσσσ

2．．．．1 Linear operation rules

Theorem 4：Ifα1 σ α1′,α2 σ α2′, then for a random real number k1，k2，it has

k1α1＋k2α2σ k1α1′＋k2α2′.

Proof： Because σ(α1)＝α1′，σ(α2)＝α2′，

then σ(k1α1＋k2α2)＝k1σ(α1)＋k2σ(α2)＝k1α1′＋k2α2′，

so k1α1＋k2α2σ k1α1′＋k2α2′

2．．．．2 Transposition (of terms) rules

Definition 1：If α σ α ，then α is called a self reflexive element of relation σ.

Assume λis a character，A is a n×n matrix，then λI-A(I is a n-moment unit matrix) is

called characteristic matrix of A. Determinant of characteristic matrix is characteristic polynomial，roots of characteristic polynomial are called characteristic root of A.But when A is a n×r or r×n

matrix whose rank is r，l is a matrix whose elements of main diagonal are 1 and all the other

element is 0，both its columns and rows are the same with A .Its characteristic matrix can be

changed into diagonal matrix by elementary transform,elements of main diagonal include first

powerλ-λ0(or polynomial aboutλ，it can be decomposed into products of same or variedλ-λ0)

about λ ，all λ=λ0 (the same λ is calculated by the times it appeared) are characteristic

roots of A.

Assumeλ0 is a characteristic root of A，α=｛x1，x2，…，xn｝ is a nonzero vector，if , 2

1

02

1

=

nn x

x

x

x

x

x

AMM

λ

then α is called a characteristic vector of A,which belongs to characteristic rootλ0.

Theorem 5：Assume A is a matrix of singular linear transform σto fundus e1，e2，…，en in Vn，

then the necessary and sufficient conditions that any element α＝x1e1＋x2e2＋…＋xnen

of Vn is a nonzero self-reflexive element of relation σ is ：transpose matrix A′of A has

characteristic rootλ0＝1，and α is a characteristic vector of A′,which belongs toλ0＝1.

Proof： Because ( ) , 2

1

21 =

n

n

e

e

e

xxx MLα and σ(α)＝α，

then ( ) ( ) ( ) ( ) ,2

1

21

2

1

21

2

1

21 ===

n

n

n

n

n

n

e

e

e

xxx

e

e

e

Axxx

e

e

e

xxx MLMLML σασ

viz. （x1 x2 …xn）A＝（x1 x2 … xn），

4

it is =′ nn x

x

x

x

x

x

A MM 2

1

2

1

(necessity after transposition)；

Whereas，because （x1 x2 …xn）A＝（x1 x2 … xn），

And α＝x1e1＋x2e2＋…＋xnen

Then ( ) ( ) ( ) ( ) ===

n

n

n

n

n

n

e

e

e

xxx

e

e

e

Axxx

e

e

e

xxx MLMLML 2

1

21

2

1

21

2

1

21 σασ ，

Viz. σ(α)＝α (sufficiency). ▌

All linear combinations of self reflexive elements are reflexive elements．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．． concluded from

theorem 4.

Theorem 6：When the matrix of singular linear transform σ is

=

−−− rrnrnrn

r

ccc

ccc

M

,2,1,

11211

1

1

1

L

LLLL

L

O

,

then among characteristic vectors α＝x1e1＋x2e2＋…＋xnen

belonging toλ0=1 of transposed matrix M ， xr+1 = xr+2 =…= xn =0，

viz. α＝x1e1＋x2e2＋…＋xr er .

Proof：Because （x1 x2 …xn）A＝（x1 x2 … xn）M＝（x1 x2 …xn），

it is

=

+++

+++

+++

=

=

′

+

−+

−+

−+

−

−

−

n

r

rnrrnrrr

nrnr

nrnr

nrrnr

rn

rn

n

x

x

x

x

x

xcxcx

xcxcx

xcxcx

x

x

x

cc

cc

cc

x

x

x

M

M

M

LLLLL

LLLLLLL

LLLLL

L

LLLLLL

L

L

M

L

LLLO

L

L

M

1

2

1

,11

2,1122

1,1111

2

1

,1

2,12

1,11

2

1

0 0

0 0

1

1

1 after transposition，so xr+1 = xr+2 =…= xn =0,

viz. α＝x1e1＋x2e2＋…＋xr er ▌

By theorem 6，when the matrix of singular linear transform σ is M，then any row vectors of

M is self reflexive element of σ.

Because transposed matrix of A may not have characteristic rootλ0 = 1，even it has

characteristic rootλ0 =1，it isn't convenient to calculate. So marix M but not A is usually directly

5

used in actual practice. Or A is used，but A=M·N on some certain conditions，so it must be

changed into M by M=A·N －1

(refer to it in example 2 of §3 of the chapter).

Theorem 7：Assume ασ α1′＋α2′，

then the necessary and sufficient conditions of α－α1′σ α2′

is ：α1′is self reflexive element. Proof：Because σ(α－α1′)＝σ(α)－σ(α1′)＝α1′＋α2′－σ(α1′)，

then，the necessary and sufficient conditions of σ(α－α1′)＝α2′，σ(α1′)＝α1′

viz. α1′is self reflexive element. ▌

Inference：Assume α1＋α2 σ β，the necessary and sufficient conditions of α2 σ β－α1

isα1 is self-reflexive element. ▌

2．．．．3 Element exchange rules

Definition 2： If α σ β，and βσα，then α andβ are symmetrical elements of

relation σ. Theorem 8：Assume A is a matrix to fudus e1，e2，…，en of singular linear transform σ in

Vn ，then the necessary and sufficient conditions that any α and β of Vn is symmetrical

element is ：the transposed matrix（A2）′of A

2 has characteristic root λ0 ＝1，both α and β are characteristic vectors belonging toλ0 ＝1 of （A

2）′.

Proof：Necessity

Because ( ) , 2

1

21 =

n

n

e

e

e

xxx MLα

Then ( ) ( ) =

n

n

e

e

e

Axxx ML 2

1

21ασ ，

Assume β＝k1e1 ＋k2e2 ＋…＋knen

Then ( ) ( )

=

n

n

e

e

e

AkkkL

L2

1

21βσ ，

From σ(α)＝β，we can get (x1 x2 … xn)A＝(k1 k2 … kn) （1）

From σ(β)＝α，we can get (k1 k2 … kn)A＝(x1 x2 … xn) （2）

Substitute formula (1) into formula (2), we can get (x1 x2 … xn)A2＝(x1 x2 … xn)，

tanspose it, then we get ( )

=

′

nn x

x

x

x

x

x

AMM

2

1

2

1

2. 〔If substitute formula (2) into formula (1),we can prove that β is also a characteristic vector

6

belonging toλ0 ＝1 of（A2）′〕.

Now let's prove the sufficiency ， because (x1 x2 … xn)A2 ＝ (x1 x2 … xn) （3）

And σ(α)＝β，

viz. ( ) ,Axxx

n

n =

e

e

eML 2

1

21 β

then when we suppose β＝k1e1 ＋k2e2 ＋…＋knen ，

it has (x1 x2 … xn)A＝(k1 k2 … kn) （4） 〔the same with formula (1)〕,substitute formula (4) into formula 3,it can get

(k1 k2 … kn)A＝(x1 x2 … xn)

so ( ) ( )

=

n

n

n

n

x

x

x

xxx

e

e

e

AkkkM

LM

L2

1

21

2

1

21 ,

viz. σ(β)＝α，viz.α，β are symmetrical elements. ▌

From theorem 8 ,theorem 4 and theorem 5，we know all linear combinations of symmetrical

elements are symmetrical elements，and self reflexive is a particular case of symmetry.

Theorem 9：At the double ends of relation σ，symmetrical elements can exchange with each

other. Proof：Assume α1 ＋α2 σ β1 +β2 ，

andα1 and β2 are symmetrical elements，because α2 σβ2 ，α1 σβ1，

and because β2 σα2 ，α1 ＋β2 σ β1 +α2 ▌

Inference：Assumeα0 andβ0 are symmetrical elements，

(1) if α ＋α0 σ β，then α σ β－β0 ；

(2) if α σ β ＋β0，then α－α0 σ β.

Theorem 10：When the matrix of relation σ is M (refered in theorem 6)，σ doesn't

contain unequal symmetrical elements.

Proof：Assume α and β are characteristic vectors belonging to λ0 ＝1 of (M2)′，but α≠β. Regard M as n×n matrix (all the elements of the following column are zero)，then

M2=M ，viz. matrix M is idempotent. From theorem 8, it has (x1 x2 … xn)(M

2)′＝(x1 x2 … xn)

and (x1 x2 … xn)M′＝(k1 k2 … kn)，

but (x1 x2 … xn)(M2)′＝(x1 x2 … xn)M′，

So it has (x1 x2 … xn) ＝(k1 k2 … kn)，

viz.α=β ，which is in contradiction with the hypothesis α≠β. So from theorem 5，it only has ασα and βσβ，

but α β and β α. ▌

At last ,let's point out by the way,because relation σ is not transferable under singular linear

transform，continuously transmit relations such as α1σα2σ…σαm (m＞2)should not be used.

7

§§§§3 Relations between coordinates of vectors under singular linear

transform

In theorem 6，we refered to a matrix whose elements of main diagonal of anterior r rows are

one and the other elements are zero，for the convenience to describe it, we call this n×r and

column nonsingular matrix M matrix. Sometimes，M is written into partitioned matrixs of upper and

lower parts.

=

C

IM ，

of which，I is r-order unit matrix，C is (n-r)×r (r＜n) matrix. Now，we choose a fundus e1 ，e2 ，…，en in linear space Vn ，choose a fundus e1′ ，e2′ ，…，er′(r＜n) of image set σ(Vn ) of Vn ，then the image of fundus e1 ，e2 ，…，en can be

represented not only by its own linearity but also by the linearity of fundus e1′ ，e2′ ，…，er′，viz. when assume

( )( )

( )

+++=

+++=

+++=

nnnnnn

nn

nn

aaa

aaa

aaa

eeee

eeee

eeee

L

LLLL

L

L

2211

22221212

12121111

σ

σ

σ

( )( )

( )

′

′

′

=

=

=

=

′++′+′=

′++′+′=

′++′+′=

rnn

nrnn

r

r

nnnn

n

n

rnrnnn

rr

rr

BA

bbb

bbb

bbb

B

aaa

aaa

aaa

A

bbb

bbb

bbb

e

e

e

e

e

e

e

e

e

eeee

eeee

eeee

MMM

L

LLLL

L

L

L

LLLL

L

L

L

LLLL

L

L

2

1

2

1

2

1

21

22221

11211

21

22221

11211

2211

22221212

12121111

hasit

, assume and

and

σ

σ

σ

σ

(A is n×n singular matrix，B is n×r and column nonsingular matrix) or can be represented as

′

′

′

=

rnrnn

r

r

nnnnn

n

n

n bbb

bbb

bbb

aaa

aaa

aaa

e

e

e

e

e

e

e

e

e

M

L

LLLL

L

L

M

L

LLLL

L

L

M

2

1

21

22221

11211

2

1

21

22221

11211

2

1

σ

Apparently, when assume

8

=

′

′

′

nr

N

e

e

e

e

e

e

MM

2

1

2

1

(Ｎ is r×n and row nonsingular matrix)，it has A＝B·Ｎ.

•= ′

′

′

= ′

′

′

= ′

′

′

=

nr

nr

nr

e

e

e

NB

e

e

e

B

e

e

e

N

e

e

e

e

e

e

A

e

e

e

B

MMMMMM

2

1

2

1

2

1

2

1

2

1

2

1

then

because

hasit condition thefrom Proof ：

viz. A＝B·Ｎ. ▌

′

′

′

=

nn e

e

e

e

e

e

MM

2

1

2

1

all when ly,Particular σ

(er+1′，er+2′，…，en′can be linearly represented by fundus e1′，e2′，…，er′)，B is a M

′

′

′

=

′

′

′

=

=

′

′

′

==

rr

nn

nr

e

e

e

e

e

e

e

e

e

e

e

e

e

e

e

N

e

e

e

C

IMB

MM

MM

MM

2

1

2

1

2

1

2

1

2

1

2

1

hasit

be so,

, because Proof

. viz.matrix,

σ

σ

：

9

Viz 　

=

nr e

e

e

N

e

e

e

MM

2

1

2

1

σ

Now，Ｎ becomes a submatrix composed of the front r rows of matrix A，viz.

=

rnrr

n

n

aaa

aaa

aaa

N

L

LLLL

L

L

21

22221

11211 ，

and in the submatrix of A which is composed of the back n-r rows of matrix A

=

+++

nnnn

nrrr

aaa

aaa

N

L

LLLL

L

21

,12,11,1

0 ，

random rows can be linearly expressed by rows of N ，so，the submatrix of A which is composed of

the back n-r rows of A can be regarded as a product of (n-r)×r matrix

=

−−− rrnrnrn

r

r

ccc

ccc

ccc

C

,2,1,

22221

11211

L

LLLL

L

L

and Ｎ： N0 = C·N ，viz. A can be changed into partitioned matrix of upper and lower parts，and

further be changed into the product of matrixes

NMNC

I

NC

NI

NC

NA ⋅=

=

⋅

⋅=

⋅

= ，

and because A＝B·N ，

( ) . matrixunit order - is then rIC

IMB == ▌

= +++

+++

nrnn

rrrr

rrrr

bbb

bbb

bbb

C

L

LLLL

L

L

21

,22,21,2

,12,11,1

Now ，

viz. C is the submatrix of B which is composed of the back n-r rows of B.

Theorem 11：Under singular linear transform σ，the relation between coordinates of vector α＝x1e1＋x2e2＋…＋xnen

and its image σ(α)＝y1e1′＋y2e2′＋…＋yrer′

10

under σ can be expressed as (y1 y2 … yr)＝(x1 x2 … xn)B.

Particularly，when σ(e i)＝e i′

( ) , ,,, fundusby drepresentelinearly becan ,,, . ,,2,1 2121 rnrr eeeeeeni ′′′′′′= ++ LLL

( ) ( ) ( )

( )

( )

( ) ( ) ( ) ( ) . then

and

because Proof

6

or

5 then

2

1

21

2

1

21

2

1

21

2

1

2

1

2

1

2

1

21

2

1

2

1

,2211

2,22211222

1,22111111

2121

•= ′

′

′

=== ′

′

′

++++=

++++=

++++=

=

−++

−++

−++

n

n

n

n

n

n

rnnn

n

nr

nrrnrrrrrr

nrnrr

nrnrr

nr

e

e

e

NBxxx

e

e

e

Axxx

e

e

e

xxx

e

e

e

B

e

e

e

A

e

e

e

e

e

e

xxx

e

e

e

N

e

e

e

xcxcxcxy

xcxcxcxy

xcxcxcxy

Mxxxyyy

MLMLMLMMMML

MM L LLLLLL LLLL

＝＝，，＝　　　　　　

，：σασ

σα

( ) ( ) ( )

( ) ( )

( ) ( )( )

( ) ( ) .

then, because , whenly,particular

. viz.

then

because and

2121

2121

2

1

21

2

1

21

2

1

21

1

21

Mxxxyyy

MBee

Bxxxyyy

e

e

e

NBxxx

e

e

e

Nyyy

e

e

e

Nyyy

e

e

e

yyy

nr

ii

nr

n

n

n

r

n

r

r

r

LL

LL

ML

ML

ML

ML

=

=′=

=

•=

=

′

′

′

=

σ

ασ

，

，

〔viz. formula (5)〕.Transpose formula（5）and write it into coordinate expressions,then we get

formula(6), this is the formula of the relation between coordinates of vector and its imageσ(α)

under σ when σ(ei )＝ei′. ▌

Under nonsingular linear transform，the relation between vector and its coordinates can be

acquired by substituting transform into the vector.But under singular linear transform,it can only be

11

acquired by matrix operation〔just like the derivation of formula (6)〕.Now by "relation" method，the formula of the relation between vector and its coordinates can be acquired by substituting

transform into the vector just like nonsingular linear transform. But for the beginners who don't

skillfully master the matrix operation technique，this is a gospel without question.

Example 1: Under the relation

21321

21

321

213

212

211

or

into changed is

vector

5

3

5

3

32

43

eYeXZeYeXe

eYeX

ZeYeXe

eee

eee

eee

′+′++

′+′

++

−−

−−

−−

σ

σ

σ

σ

Please find out the coordinate transform formula of the vector.

Solution: Substitute the above transforms into vector Xe1＋Ye2＋Ze3,

( ) ( )

−−−=′

−−−=′

−−−+ −−−

+−+−+−++

ZYXY

ZYXX

eZYXeZYX

eeZeeYeeXZeYeXe

5

334

5

323

then sign, equal theof ends double thecompare

5

334

5

323

into expressionrelation theof endright thearrange

, 5

3

5

33243 then

21

212121321 σ

But it's not the form of formula (6). If you want to acquire the result of formula (6)，the

matrix of σ must be changed into matrix M.

So，matrix operation can't be absolutely divorced from.

Example 2: Change the matrix of the example 1 into the formula of M，and find out the

relation between coordinates of the vector under new transform. Solution because when σ(e1)＝e1′, σ(e2)＝e2′, σ(e3)＝e3′

(e3′can be linearly represented by fundus e1′，e2′) A=M·N ，then M=A·N －1，

−

−=

=−−

−−=

−−

−−=

−

32

43 so

, 132

43 ,

32

43 because and

1N

NN

So

5

3

5

3

1

1

32

43

5

3

5

332

43

−

=

−

−•

−−

−−

−−

=M

and then relation σ can be expressed as

12

′−′

′

′

213

22

11

5

3

5

3

eee

ee

ee

σ

σ

σ

Assume under the latter relation，vector Xe1＋Y e2＋Ze3

is changed into X "e1′+Y＂e2′，

or Xe1＋Y e2＋Ze3 σ X "e1′+Y＂e2′，

substitute the latter transform into vector Xe1＋Y e2＋Ze3

. 5

3

5

3 getsit 2121321

′−′+′+′++ eeZeYeXZeYeXe σ

arrange and transpose the double ends of the equal sign ,it gets

,

5

35

3

then

, 5

3

5

3 so,

, 5

3

5

3 viz.

, 5

3

5

3

2121

21321

21

−=′′

+=′′

′

−+′

+=′′′+′′′

′

−+′

+++

′

−+′

+

ZYY

ZXX

eZYeZXeYeX

eZYeZXZeYeXe

eZYeZX

σ

this is completely the same as the form of formula (6).

In actual application，we usually directly wrote the matrix of transform into the form of

M(but not the form of A)，which can ensure that each relation of the transform has its reflexive

element ,and then it can has strong calculation function. Example 3: Under the relation

, or

, into changed is

vector

3

2

3

15

4

5

2

214321

21

4321

214

213

22

11

eYeXTeZeYeXe

eYeX

TeZeYeXe

eee

eee

ee

ee

′+′+++

′+′

+++

−−

−−

σ

σ

σ

σ

σ

Please find out the coordinate transform formula of the vector. Solution substitute the above

relations into vector Xe1＋Y e2＋Ze3＋Te4 ，

, 3

2

3

1

5

4

5

2 getsit 2121214321

−+

−−+++++ eeTeeZYeXeTeZeYeXe σ

13

, 3

2

5

4

3

1

5

2 so

, 3

2

5

4

3

1

5

2 viz

, 3

2

5

4

3

1

5

2 into endright thearrange

2121

214321

21

eTZYeTZXeYeX

eTZYeTZXTeZeYeXe

eTZYeTZX

−−+

−−=′+′

−−+

−−+++

−−+

−−

σ

compare the double ends of equal sign，and then it gets

,

3

2

5

43

1

5

2

−−=′

−−=′

TZYY

TZXX

this is also the same as the result of formula (6).

Exercises

1.1 Please find out the change of the vector 3e1+5e2-4e3 under the transform

+−

+−

−

213

212

211

4

2

53

eee

eee

eee

σ

σ

σ

1.2 Change the matrix of the relation of the above exercise into M ，and find out the change of

the vector under new transform.

1.3 Find out the change of vector 4e1-5e2+3e3 under the transform

+− 213

22

11

23 eee

ee

ee

σ

σ

σ

1.4 Find out the change of vector 5e1-4e2+6e3 under the transform

− 213

22

11

59

eee

ee

ee

σ

σ

σ

1.5 Find out the change of vector 12e1-7e2+8e3 under the transform

+− 213

22

11

2

1

2

2

eee

ee

ee

σ

σ

σ

14

The Second Chapter: "Relation" method and preferred

n-dimensional coordinate system

In the first chapter,we have discussed the concept and proerty of "relation" method，and

studied its operation rules in depth. In this chapter,we will begin to discuss approaches of using

"relation" method in practice，viz. how to use "relation" method to build preferred n-dimensional

coordinate system. In order to help people master the method step by step，we will begin with the

process of building preferred 4-dimensional coordinate system.

§§§§1 "Relation" method and examples of preferred 4-dimensional

and 5-dimensional coordinate system

1．．．．1 Examples of building preferred 4-dimensional coordinate system

For example assume e1 ，e2 ，e3 are unit vectors perpendicular to each other，they are fundi

of 3-dimensional euclidean space V3 ，σ1 is a singular linear transform of V3 .Regard V3 as

rectangular axis Oxyz ，fundi e1 ，e2 ，e3 are separately regarded as unit vectors of x axis, y axis

and z axis，and suppose under σ1, e1 ，e2 ，e3 and their images it has the lower relations

( )1

5

4

5

3

213

22

11

−− eee

ee

ee

σ

σ

σ

According to the above relations we can figure out an analogue image(hereinafter referred to

as model,refer to it in figure 1) of Oxyz . In the figure,directviewing figure of unit vector e3 of

vertical axis z and the vector － 3e1 － 4e2 of horizontal

coordinate plane xoy overlapped.

For the convenience，people always hope to describe things

in 3-dimensional or even higher dimensional space by using a

2-dimensional plane，it is difficult to avoid figures overlap. As

this is the case ,why still get it around but not face up to it?

Facing up to questions is solving questions. So，a vector between

two random points overlapped in figure 1 is called (of figure 1)

overlapping direction.

For example,in figure 1, point (0，0，5) of z axis and point

(－3,－4,0) of xoy plane overlapped， this shows that

overlapping direction is a vector between the two points.So，the figure 1

overlapping direction of the model can be denoted as

S＝3e1 ＋4e2 ＋5e3 .

Arrange and transpose the third relation expression of the above formula (1)，it has 3e1 ＋4e2 ＋5e3 σ1 0

or S σ1 0

This shows that in figure 1,all straight lines running parallel to overlapping direction S are

punctiform.But 0 S

15

this is because e3 isn't self reflexive element，it can't be moved to the right end of relation

expression. So，S and zero aren't symmetrical，it shows that the image of point can only be

punctiform.

Under the enlightenment of stereoscopic projection[5] ，we think transform like formula

(1)should appeared in pairs，viz. there should be relation σ2

+ 213

22

11

5

4

5

3eee

ee

ee

σ

σ

σ

(2) From formula (2)，another model (figure 2) of Oxyz can be

acquired. Z axis of figure1 and figure 2 are symmetrical relative to

origin.Because the two models represent the same 3-dimensional

rectangular coordinate system Oxyz，then we call them a couple of

preferred 3-dimensional planes. Respectively draw out a vertical

and upright ray through the origin of the preferred 3-dimensional

plane as the fourth coordinate axis t , and this forms a model of a

couple of 4-dimensional rectangular coordinate systems Oxyzt ，it

is called a couple of preferred 4-dimensional coordinate

systems(referred to as 4-dimensional system or specific 4-dimensional figure 2

figure 3 figure 4

system，refer to it in figure 3 and figure 4).

In order to distinguish it expediently，we call the 4-dimensional system of figure 3 host system

and call the 4-dimensional system of figure 4 guest system. The corresponding relations can be

expressed as

( )3

5

4

5

3

414

2113

212

111

−−

ee

eee

ee

ee

σ

σ

σ

σ

( )4

5

4

5

3

424

2123

222

121

+

ee

eee

ee

ee

σ

σ

σ

σ

16

From figure 3 and figure 4,it is observed that preferred four-dimensional system is composed

of cross axis x , vertical axis y,upright axis t and axis z with inclining direction (be called oblique

axis),which are all through the origin. Of which,the direction and unit length of oblique axis z is

symmetrical relative to the origin in host system and guest system.

In generalized case，when building preferred four-dimensional system we only write out

formula (3) and formula (4)〔viz. not from formula (1) and formula (2) to formula (3) and formula

(4). So，for figure 1 and figure 2，we usually regard them as preferred four-dimensional system of

deleting upright axis t .

1．．．．2 Examples of building preferred five-dimensional coordinate system

Based on the same method，we can also work out a couple of models of five-dimensional

rectangular coordinate systems Ox1x2x3x4x5 . For example，in Oxyzt of figure 3 and figure 4 ，order

x=x1, y=x2，z=x3, t=x4 ，

and choose a vector 3e1+4e2+6e4 of a three-dimensional subsystem Oxyt (here marked as

Ox1x2x4 ) in Oxyzt as overlapping direction. Because 3e1+4e2+6e4 σ1 0 ，

then it has ( ).436

12114 eee +−σ

but oblique axis is symmetrical relative to the origin in guest system，so it has

( ).436

12124 eee +σ

Separately substitute the two relations into formula (1) and formula (2) and make a diagram

according to it，then a couple of preferred four-dimensional planes(just call it for the moment，refer

to it in figure 5) are acquired.

figure 5

Respectively draw out a vertical and upright ray through the origin of the preferred

four-dimensional plane as upright axis x5，then a couple of preferred five-dimensional coordinate

systems are acquired(refer to it in figure 6). The corresponding relations are

( )5

6

4

6

35

4

5

3

515

2114

2113

212

111

−−

−−

ee

eee

eee

ee

ee

σ

σ

σ

σ

σ

17

( )6

6

4

6

35

4

5

3

525

2124

2123

222

121

+

+

ee

eee

eee

ee

ee

σ

σ

σ

σ

σ

figure 6

§§§§2 Oblique axes transform and preferred n-dimensional coordinate

system

According to the above ideas and methods of building preferred four-dimensional system and

five-dimensional system，we will discuss some questions of building preferred n-dimensional

system.

2．．．．1 Oblique axes transform and building of preferred n-dimensional system

2．．．．1．．．．1 Methods and steps of building preferred n-dimensional system————————oblique axes

transform

Preferred n-dimensional system actually is a kind of analogue image ( referred to as model)of

imaginary n-dimensional rectangular coordinate system Ox1x2…xn .Owing to the enlightment of

stereoscopic projection，this kind of models always appear in pairs. For the convenience of

calculation，the transform adopted are always symmetrical.

The detailed practice is ：first break up Ox1x2…xn into n-2 three-dimensional subspaces

Ox1x2x3，Ox1x2x4，…，Ox1x2xn –1 ，Ox1x2xn ，and make every subspace has common level coordinate

plane x1ox2 . Second，choose vectors

S1=a11e1+a12e2+a3e3 ，

S2=a21e1+a22e2+a4e4 ， … … … …，

Sn----3333=an--3,1e1+an－3, 2 e2+an－1en----1 ，

from the front n-3 subspaces.〔a3 ,a4 ,… , an－1 of every formula are nonzero numbers，ai1,ai2

are i numbers when group( i=1 ，2 ，… ，n-3 ) aren't zero at the same time；e1 , e2 ,…，en

are respectly unit vectors of x1 , x2 , …, xn 〕as overlapping direction. Then，under σ1 it has

S1 σ1 0 ， S2 σ1 0， … ，Sn－3 σ1 0.

Because the n-2 subspaces have x1ox2 plane aggregately，and the xn axis of the subspace

Ox1x2xn is regarded as upright axis，e1 , e2 ,…，en are determined to be self reflexive elements，

18

so the two front terms of Si (i=1 ，2 ，… ，n-3) can be moved to the right end of the relation.

Then it has ( ) ( )； 3,2,1 1

2211

2

12 −=+−+

+ nieaeaa

e ii

i

i Lσ

but underσ2 , signs of the right end of the relations are opposite，viz.

( ) ( )； 3,2,1 1

2211

2

22 −=++

+ nieaeaa

e ii

i

i Lσ

write it into complete relation：

( )

( )

( )

( )

( )

( )

( )

( )8

1

1

1

7

1

1

1

2

22,311,3

3

21

222121

4

24

212111

3

23

222

121

1

22,311,3

3

11

222121

4

14

212111

3

13

212

111

+

+

+

+−

+−

+−

−−

−

−

−−

−

−

nn

nn

n

n

nn

nn

n

n

ee

eaeaa

e

eaeaa

e

eaeaa

e

ee

ee

ee

eaeaa

e

eaeaa

e

eaeaa

e

ee

ee

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

LLLLLL

LLLLLL

〔a3 , a4 , … , an－1 of the two formulas are nonzero numbers，ai1 , ai2 are numbers of i group

(when i =1 ， 2 ， … ， n－3 ) aren't zero at the same time〕. Make a diagram according to

formula (7) and formula (8)，a couple of analogue images( refer to it in figure7，in the figure， ai1=

a 1 ，ai2= a 2 ；a1 , a2 , … , an－1＞0 ) of n-dimensional rectangular coordinate system Ox1x2…xn

can be acquired ，we call the couple of analogue images a couple of preferred n-dimensional

coordinate system(referred to as n-dimensional system or specific n-dimensional system). Of

which， the n-dimensional system generated from formula (7) is called host system， the

n-dimensional system generated from formula (8) is called guest system.Accordingly,formula is

called positive transform or main transform，formula (8) is called subsidiary transform or assistant

transform.Preferred n-dimensional systems always appear in pairs，usually host system is on the top

or on the left，guest system is below or on the right. Sometimes,it appears alone. When it appears

alone, it is always host system if there is no special version .

19

figure 7

In formula (7) and formula (8)，the first ,the second and the n-th couple of relations determine

the existence of rectangular coordinate system Ox1x2xn.If people assume Ox1x2xn is existent through

common practice，at the same time we think using letter as sign and using figure as sign have the ．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．same significance．．．．．．．．．．．．．．．． ，so respectively substitute．．．．．．．．．．．．．．．．．．．．．．．． “ □ ” and “ ○ ” for “σ1” and “σ2”,

then formula (7) and formula (8) can be respectively written into

( )

( )10

9

22,311,311

22212144

21211133

22,311,311

22212144

21211133

+

+

+

−−

−−

−−

−−−−

−−−−

eaeaea

eaeaea

eaeaea

eaeaea

eaeaea

eaeaea

nnnn

nnnn

○○○□□□

LLLL

LLLL

for short. Here，we may as well call “ □ ” of the two formulas sign “image relation sign”，and call

“ ○ ” sign of“shadow relation”(of which，“α□β ”can be read “the image of α is β ”,“γ ○ δ ”can be read “the shadow of γ is δ” ).

The introduction of sign “ □ ” and “○” enhances visualization and visualizability of

the transform，and it also lessens much descriptive trouble in the following application.

Figures of axis Ox1x2…xn－1 in preferred n-dimensional system are always oblique.So，we call

them oblique axes，and accordingly,formula (7) and formula (8)〔or formula (9) and formula (10)〕can also be called oblique axes transform．．．．．．．．．．．．．．．．．．．． .

2．．．．1．．．．2 Structure of preferred n-dimensional system

Preferred n-dimensional system is composed of cross axis x1 ,vertical axis x2 ,oblique axis

Ox3x4…xn－1 and upright axis xn which are all through the origin(under particular circumstances,

x1, x2 can be selected for upright axis or oblique axis，and xn can be select for cross axis,vertical axis

or oblique axis，etc. But without special version，x1 , x2 are still served as cross axis and vertical

axis，xn is still served as upright axis)，we can also consider that it is composed of models of a

couple of n-1-dimensional rectangular coordinate systems and upright axes. But n-1-dimensional

model represents a n-1-dimensional subspace of Vn ,so we call them a couple of preferred

n-1-dimensional planes for the moment，they are formed of overlapping of the models of n-3

three-dimensional subspace Ox1x2x3，Ox1x2x4，…，Ox1x2xn－1，of Vn . They have common plane

x1ox2 ，but figures of each oblique axis sometimes are overlapped , it is difficult to differentiate and

not convenient to be identified.So separate the overlapping oblique axes when we make a diagram.

They are listed parallel to a straight line(the straight line should be perpendicular to every oblique

axis in essence)which is through the origin，the straight line represents the origin of every oblique

axis.

20

For upright axis，because its actual position is through the origin and perpendicularily goes

away from the plane,it is difficult to be drawn on figures.According to the draughting of oblique

axis，we draw it in a random direction through the origin.But in order to differentiate it from

oblique axes，figure direction and unit length of upright axis is identical in host system and guest

system,but direction of the same oblique axis is opposite in host system and host system.

Through choosing reasonable direction and unit length,the upright axis of specific system can

not only offer calculation convenience but also lead some elements (such as point,line,plane,etc)of

geometric figure drawn in preferred n-dimensional system to suitable position，thereby it make

the whole geometric figure occupancy small chart size and has satisfactory and goodly shape.

Speaking from this meaning ，the upright axis of preferred n-dimensional can be called leading

axis.

2．．．．1．．．．3 Kinds of preferred n-dimensional system

According to the permutation pattern of oblique axis，preferred n-dimensional system can be

divided into convergent oblique pattern and emanative oblique pattern.In formula (7) and formula

(8)〔or formula (9) and formula (10)〕, when λ1a11 =λ2a21 = … =λn－3 an－3,1 = a1， λ1a12 =λ2a22 = … =λn－3 an－3,2 =a2

(λ1，λ2，…，λn－3 are nonzero numbers)， directions of every oblique axis are parallel.

So their figures are overlapped，we think this kind of preferred n-dimensional system is

convergent oblique pattern(but when concretely making a diagram ，we should avoid figures

overlapping of oblique axes.So we should differentiate them one from another and parallelly

arrange them on a straight line which is through the origin, use the straight line as their origin for

the moment). Figure 8 is a couple of emanative oblique pattern examples when n is even

number(accordingly，its transform is

( ) ( )

−−

−−

+

−−

−− 221111

2211

221144

221133

11

eee

eee

eee

eee

aaa

aaa

aaa

aaa

nn

ii

ii □□□□

LLL

LLL

i=3 ，4 ， … ，n－1，readers are expected to write out its subsidiary transform)，and figure 9

convergent oblique pattern examples when n is odd number(the corresponding transform is

( ) ( )

−−

−−

+

−−

−−

−−

++

221111

22

1

11

1

221155

221144

221133

11

2

2

eee

eee

eee

eee

eee

aaa

aaa

aaa

aaa

aaa

nn

ii

ii □□□□□

LLL

LLL

i=5 ，6 ，… ，n－1 ， the readers are expected to write out its subsidiary transform).

21

figure 8

If it dissatisfies the above terms，viz. in formula (7) and formula (8)〔or formula (9)，formula

(10)〕，if and only if λ1，λ2，…，λn－3 are zero at the same time, λ1a11 =λ2a21 = … =λn－3 an－3,1 ， λ1a12 =λ2a22 = … =λn－3 an－3,2

are founded，figure directions of every oblique axis are emanative，we call this kind of preferred

n-dimensional system emanative oblique pattern.Figure 10 is a couple of preferred six-dimensional

systems of emanative oblique pattern.

figure 9

§§§§3 Property of preferred n-dimensional system

3.1 Relation between preferred n-dimensional system and oblique axes transform

The above discussion has shown：given a transform ，we can immediately make out the

corresponding preferred n-dimensional system. Vice versa，given a preferred n-dimensional

system，can we immediately make out the corresponding oblique axes transform? The answer is

affirmative.

For example，in the host system of figure10，points

(0，0，2，0，0，0)，(0，0，0，3，0，0)，(0，0，0，0，5，0)

figure 10

22

of every oblique axis respectively coincide with points

(-2,-1，0，0，0，0)，(1，-1，0，0，0，0)，(-3，-1，0，0，0，0)

of x1ox2 plane，it shows the overlapping directions of Ox1x2x3 ，Ox1x2x4 ，Ox1x2x5 respectively is

2e1 + e2 + 2e3, -e1 + e2 + 3e4 , 3e1 + e2 + 5e5 .

Because they are all punctiform，viz. under singular linear transform σ1，their images are

all zero，viz. 2e1 + e2 + 2e3 □ 0, -e1 + e2 + 3e4 □ 0, 3e1 + e2 + 5e5 □ 0.

From figure 10，the cross axis、the vertical axis and the upright axis are respectively x1 , x2 , x6 ，so，e1 , e2 are self reflexive elements. Transpose the above three relations then the abbreviated

formula of positive transform is acquired.

−−

−

−−

215

214

213

35

3

22

eee

eee

eee □□□

Because the oblique axes directions in guest system are opposite to those of host system，

then the assistant transform is

+

+−

+

215

214

213

35

3

22

eee

eee

eee ○○○

In fact，according to that vectors 2e3 , 3e4 , 5e5 of oblique axes x3，x4，x5 in host system

respectively coincide with vectors -2e1-e2 , e1- e2 , -3e1-e2

of plane x1ox2 ，we can immediately determine it has relation σ1 between them, so the

corresponding oblique axes transform can be directly written out.

3．．．．2 The "preference" of preferred coordinates system

Defination 1：The overlapping direction of host system in preferred n-dimensional system

can also be called principal overlapping direction of preferred n-dimensional system，referred to

as principal overlapping direction (rampart：it is the rampart that be referred as ramparting walls，building by piling bricks can also be called ramparting walls in Beijing dialect. Here,“rampart”can

be extended as overlap or coincidence).

Defination 2 ： All linear combinations of principal overlapping direction form a

n-3-dimensional subspace ，it is called principal overlapping space .

Here，we can catch a glimpse of the advantage of abbreviated formula (9) and abbreviated

formula (10) .

Order S1=a11e1+a12e2+a3e3 ，

S2=a21e1+a22e2+a4e4 ， … … … …，

Sn----3333=an--3,1e1+an - 3, 2 e2+an-1en----1 ，

From the linear independence among e3, e4,…, en－1 ，we can extrapolate that S1，S2，…，Sn----3333 are linearly independent.Transpose formula(9)，it has

S1 □ 0, S2 □ 0, …, Sn----3333 □ 0 ，

so ( ) ( )11 numbers are 03

1

i

n

i

ii kSk □∑−=

As will be readily seen，S1，S2，…，Sn----3333 are respectively overlapping directions of n-3

23

three-dimensional models of preferred n-1 dimensional plane in host system these overlapping

directions are called principal overlapping directions of preferred n-dimensional system. All their

linear combinations filled a n-3-dimensional subspace，viz. principal overlapping space.When we

take all numerical values of ki ， formula (11) represents that principal overlapping space

accumulates into punctiform. So，the "preference" of preferred n-dimensional system is ：it suits a

known﹑preferred principal overlapping space and make it accumulate into punctiform.

In our realistic﹑ three-dimensional world ，people can't make out unified models of

n-dimensional space，but only make out "preferred" n-dimensional system satisfying different

specific conditions according to different needs.

§§§§4 Punctiform figures in preferred n-dimensional system————generic

point

4．．．．1 Generic point、、、、projection trace and opposite trace

4．．．．1．．．．1 Concept of generic point

In descriptive geometry，use a bundle of parallel light to irradiate a geometrical body，then the

geometrical body leaves shadow on the image plane. Figures on the image plane(also referred to as

projection plane) and completely coincide with the shadow are called projection of the geometrical

body. The direction of light travelling is called projection direction，this kind of projection is

called parallel projection[6]

.

Preferred n-1-dimensional plane is equivalent to be formed of parallel projection，so，random

punctiform figures of n-1-dimensional plane in host system are equivalent to principal overlapping

space. And because preferred n-dimensional system is composed of n-1-dimensional plane and

upright axis，we can imagine the whole n-dimensional space is formed of the translational trace of

preferred n-1-dimensional plane along the direction of the upright axis. So，random punctiform

figures in preferred n-dimensional system are equivalent to punctiform figures of preferred

n-1-dimensional plane. Or，random punctiform figures of preferred n-dimensional system(referred

to as host system)can represent figures parallel or congruent with principal overlapping space.

Defination 3: Random punctiform figures in preferred n-dimensional syetem are called

generic points(the meaning is “generally referred to as "all punctiform figures")：

(1)When a punctiform figure represents a figure equivalent to (or parallel and congruent )

principal overlapping space，it actually represents the dimension is n-3(the directviewing dimension

is zero)，it is called sufficient-rank generic point，or directly call it n-3-rank generic point，the

following generic points referred to are of this kind if without special version；

(2)when it is ordered only to represent a p-dimensional figure(P＜n-3) ,we call it deficient

-rank generic point or directly call it p-rank generic point；

(3)For a point of common meaning，because it represents the dimension P=0，sometimes it is

also called zero-rank generic point (but it use the original name under most circumstances.Since

then, "points " has no "generic" before them are referred to be this kind)，it is regarded as a

particular case of generic point.

4．．．．1．．．．2 Projection trace and opposite trace of generic point

Parallel projection can be divided into two kinds：

1.The included angle between projection direction and image plane is less than 90°,it is

called oblique projection[6] ；

24

2.The projection direction is perpendicular to image plane， it is called orthographic

projection[6]

.

The following projections referred to are all orthographic projections if without special

version.

The intersection point of random geometrical figure (referred to as their original shape,but not

their projection) and the projection plane is called trace of the geometrical figure[6]

.

Defination 4: For the projection A′of a generic point A on preferred n-1-dimensional

plane, its trace on horizontal projection plane x1ox2 is called projection trace of generic point，marked as Ja′.

Defination 5: The point of generic point A which corresponds with its projection trace is

called opposite projection trace of A，it is called opposite trace for short，marked as Ja.

The concept of opposite trace can be explained as：translate n-1-dimensional plane along

upright axis and make the projection A′of A coincide with A，then the projection trace Ja′

become opposite trace Ja′.So，a random generic point on preferred n-1-dimensional plane ，its

projection trace and opposite trace are the same point (the point can be called "trace" or "trace

point' of the generic point).

From defination 4 and defination 5,we can extrapolate：

1)If a generic point has projection trace, then it must have opposite trace. Except

sufficient-rank generic points，it isn't all generic points have projection trace and opposite trace.

Such as some (not all) difficient-rank generic points ；

2)Projection trace or opposite trace of generic points are points like these：their coordinates

relative to oblique axes x3, x4,…, xn－1 are simultaneously zero (of which，the coordinate of

projection trace relative to upright axis is zero too；but the coordinate of opposite trace relative to

upright axis is nonzero uner generalized case).

Among all points represented by a random generic point in preferred n-dimensional system，only the positions of projection trace and opposite trace are easy to be found out, but if the the

positions of projection trace and opposite trace are determined，the position of generic point is

determined naturally.

People are used to study in three-dimensional space，but preferred n-dimensional system

often choose subsystem Ox1x2…xn to simulate spatial three-dimensional rectangular coordinate

system，projection trace or opposite trace nicely give the directviewing position of the generic point

in three-dimensional coordinate system. If you master concerned knowledge of projection trace or

opposite trace，then you catch hold of the trace of generic points.

4．．．．2 Property of generic points

4．．．．2．．．．1 Shape of generic points

Theorem 1：Random points are in the shape of punctiform in host system，and they are in the

shape of line(in convergent oblique pattern) or plane (in emanative oblique pattern)in guest system .

Proof：Because generic points are parallel or equivalent to principal overlapping space，from

formula (11), we can immediately prove that generic points are in the shape of punctiform in host

system.

Simultaneously add ai1e1+ai2e2

(i=1 ， 2 ， … ， n-3) to the two ends of formula(10)，it has

S1 ○ 2(a11e1+a12e2 )，

S1 ○ 2(a11e1+a12e2 )，

25

… … … … …，

Sn－3 ○ 2(an－3,1e1+an－3,2e2 )，

So ( ) ( )12 2 2211

3

1

3

1

eaeakSk ii

n

i

i

n

i

ii +∑∑ −

=

−

=

○

(ki are numbers)，in convergent oblique pattern λ1a1j =λ2a2j =…=λn－3an－3,j (j=1 ，2)

( )

( ) ( ) ( )13 22 so

21321 so

3

1

3

1

22112211∑ ∑−

=

−

=

+=+

=−==

n

i

n

i i

iiii

i

j

ij

eaeak

eaeak

jnia

a

λ

λ，；，，，L

When we take all numbers of ki ，the right end of formula (13) represents that all linear

combinations of a1e1+a2e2 fill a straight line. viz. generic points are in the shape of line in host

system of convergent oblique pattern.

In emanative oblique pattern，it always has two vectors which are linearly indeqendent among a11e1+a12e2 ,

A21e1+a22e2 , … … ,

an－3,1e1+an－3,2e2 ,

so,when take all numbers of ki ,all their linear combinations fill a two-dimensional plane, viz.

generic points are in the shape of plane in host system of convergent oblique pattern. ▌

4．．．．2．．．．2 The uniqueness of generic points relative to upright axis

Theorem 2：Coordinates of a random generic point relative to upright axis are unique .

Proof：First suppose coordinates of a generic point relative to upright axis are not unique，viz. besides xn =xn1 ,

coordinates of a generic point relative to upright axis still have

xn =xn2，

So, the generic point includes the vector (xn－1xn2 )en .

And because figures represented by generic points are parallel or equivalent to principal

overlapping space，and principal overlapping space is all linear combinations of principal

overlapping direction，then (xn－1xn2 )en .

is a linear combination of principal overlapping direction，from formua (11) we know

(xn－1xn2 )en□ 0 .

but from formula (7)，formula (8)，en is reflective element，it should has

(xn－1xn2 )en□ (xn－1xn2 )en

and viz. xn1 = xn2 . ▌

4．．．．2．．．．3 Uniqueness of projection trace or opposite trace

Theorem 3：Projection trace of a random generic point is unique，the opposite trace is

unique ,too.

Proof：If we can prove that the trace of a random generic point of n-1-dimensional plane is

unique，the theorem can be immediately proved.still use disproof ：

Suppose a random generic point A of n-1-dimensional plane has at least two traces on x1ox2

26

，　

　　：　　　　　

) 0,,0,,( and

) 0,,0,,( plane

2

20100

2

20100

876

L

876

L

−

−

n

n

bbB

aaA

and α1=b10-a10 , α2=b20-a20 ，

then principal overlapping space must include the vector

221100 eeBA αα += ，

From formula(11) we know α1e1+α2e2 □ 0 ,

but e1，e2 are all reflective elements，it should has α1e1+α2e2 □ α1e1+α2e2 ，

viz. α1e1+α2e2=0 ,

so， α1=α2 = 0 ,

viz. a10 = b10 , a20 = b20 .

then A0，B0 are the same pont. ▌

Equations of generic point (referred to in theorem 3 of the four chapter) can be discussed after

solving the graphic rules of preferred n-dimensional system.

Exercises

2.1 What's principal overlapping direction? what's principal overlapping space? what's the

"preference' of preferred n-dimensional system?

2.2 what's generic point ? what kinds can generic point be divided into?why?

2.3 What's the projection trace and opposite trace of generic point ?what's its principal

character(note：its coordinates relative to some coordinate axes)?

2.4 Why say“the projection trace and opposite trace of a random generic point on preferred

n-1-dimensional plane are the same point”?

2.5 Please give out three characters of generic point .

2.6 According to the given transformation

−

−

+−

+−

525

2124

2123

222

121

515

2114

2113

212

111

3

2

6

53

1

3

2

3

2

6

53

1

3

2

ee

eee

eee

ee

ee

ee

eee

eee

ee

ee

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

make out a couple of preferred five-dimensional systems.

2.7 According to formula (9)、formula(10)，change the transformation formula of the above

exercise(exercise 2.6) into abbreviated formula.

2.8 Suppose the given transform formulas are

27

+

+

+

−−

−−

−−

215

214

213

215

214

213

643

964

325

643

964

325

eee

eee

eee

eee

eee

eee ○○○□□□

then please make out a couple of preferred six-dimensional systems.

2.9 Assume the given transform formulas are

−−

−

+−

+

+

+−

−

−−

216

215

214

213

216

215

214

213

435

65 3

734

35 5

43 5

653

73 4

355

eee

eee

eee

eee

eee

eee

eee

eee ○○○○□□□□

then please make out a couple of preferred seven-dimensional systems.

2.10 Among the above preferred n-dimensional systems of exercise 2.6、exercise 2.7、exercise

2.8 and exercise 2.9，which are emanative oblique pattern? which are convergent oblique pattern?

2.11 Please write out the oblique axes transform formulas of the following preferred

six-dimensional system：

2.12 Please write out the corresponding principal transform and subsidiary transform formula

according to the following host system of preferred seven-dimensional system，and make out the

guest system of the preferred seven-dimensional system.

exercise 2.12 exercise 2.13

2.13 Please write out the corresponding principal transform and subsidiary transform formula

according to the following host system of preferred nine-dimensional system，and make out the

guest system of the preferred nine-dimensional system.

28

The Third Chapter: Relation between figures and numbers

in preferred n-dimensional system

In the front two chapters we have discussed concept of oblique axes transform ,preferred

n-dimensional system、generic point and their property. On the basis，this chapter will discuss

relation between figures and their corresponding algebraic equations in preferred n-dimensional

system. This inclues two aspects of contents：(1) Given figures of a preferred n-dimensional

system，we can immediately make out the corresponding algebraic equations；(2) Given a algebraic

equation，we can also immediately make out the corresponding figure of preferred n-dimensional

system.

§§§§1 Translation trace of generic point————generic curved surface and

generic curve

1．．．．1 Concepts of generic curved surface and generic

curve

A

ssume during the course of a generic point of n-dimensional

system accompanied with n-1dimensional plane continuously

transposing along the upright axis，A is also continuously

transposing on the n-1-dimensional plane，then the trace of A is

in the shape of curve(figure 11).we call the curve-shape figure

generic curve .When A sufficient-rank，the curve-shape figure is

also called sufficient-rank generic curve ； When A is

difficient-rank ，the generic curve is also called difficient-rank figure11

generic curve.

And suppose there is a generic curve on n-1-dimensional

plane，it is regarded as the trace of generic point A. When

n-1-dimensional plane is continuously transposing along upright

axis ， the generic curve is also transposing on the

n-1-dimensional plane. and its shape either continuously

changes or has no change (viz. the movement of the generic

curve on the n-1-dimensional planemay not be transposition)，then the trace of the generic curve must be in the shape of

curved surface(figure12)，the figure is apparently regarded as the

trace of A . So，we call the figure which is in the shape of curved

surface generic curved surface. When A is sufficient-rank，the

generic curved surface is also called sufficient-rank generic

curved surface；When A is difficient-rank ，the generic curved figure12

surface is also called difficient-rank generic curved surface.

Generic curved surface and generic curve are usually referred to as hypersurface and

hypercurve. Here，we call them generic curved surface and generic curve，first, in order to

29

accommodate to the concept of trace of generic point in preferred n-dimensional system；second，it has Chinese feature in language. It is believed to be more easily accepted.

A particular case of generic curved surface and generic curve is curved surface and curve of

ordinary meaning ，they are regarded as the trace of zero-order generic point，so sometimes they

are called zero-order generic curved surface and zero-zero generic curve. Since then，all curved

surface and curve without "generic" before them are to be this kind.

1．．．．2 Dimensions of generic curved surface and generic curve————dimension theorem

Dimensions of generic curved surface or generic curve can be divided into actual dimension

(marked as W ) and directviewing dimension(marked as Z ). Here it is mainly referred to be actual

dimension.

For generic curved surface and generic curve (include generic point) determined by system of

linear equations，they can be solved by linear algebra theory. Such as the actual dimension of

figures determined by system of homogeneous linear equations

( )∑=

==n

i

iij jxa1

2 1 0 L，，

is W=n－r，of which，r is the order of coefficients matrix of the equation set. But for the figures

determined by system of non linear equations，this method is inconvenient . Here，we introduce

another method.

Figures of preferred n-dimensional system can all be regarded as the trace of generic point.

From the analysis of the second chapter we know，the actual dimensions W of generic point is equal

to its orders (marked as J ). And the directviewing dimension of generic point is zero，so，the

dimensions of generic point can be expressed as Ｗ=Z+J=0+J=J ;

but generic curve is the trace of generic point transposing along certain paths，on the moving

paths ,it has directviewing dimension Z=1. So，the actual dimension of generic curve is

W=Z+J=1+J,

viz. its actual dimension is one dimension more than its above dimension of generic point; Generic

curved surface can also be regarded as the trace of generic curve transposing along certain paths，apparently，the moving path has one directviewing dimension. So，the directviewing dimension of

generic curved surface is Z=2，one dimension more than that of generic curve，two dimensions

more than that of generic point，viz. the actual dimension of generic curved surface

W=Z+J=2+J.

In addition，there are some curved surfaces of directviewing dimension Z=3 and somen

curves of directviewing dimension Z=2，they are all transposition traces of difficient-order generic

point and separately be called difficient-order generic solid and difficient-order generic curved

surface. The dimension of difficient-order generic curved surface has suited W=Z+J ；Difficient-ordergeneric solid is the transposition trace of difficient-order generic curvd surface，it is

three dimensions more than the directviewing dimensions of above difficient-order generic point，viz. W=Z+J=3+J.

Generalize the above description，we have the following dimension theorem:

Theorem 1：The relation between the actual dimension W of random figures in preferred

n-dimensional system and their directviewing dimension Z and order J is

W=Z+J. ▌

30

1．．．．3 Kinds of generic curved surface and generic curve

Generic curved surface and generic curve can be divided into the sufficient-order and the

difficient-order，they can also be divided into the linear and the nonlinear according to equation

property. Of which，the generic curved surface and generic curve of linear equation are respectively

called generic plane and generic straight line，they can be regarded as a particular case of generic

curved surface and generic curve. Preferred n-1-dimensional plane can be regarded as a particular

case of generic plane，it is called generic coordinate plan．．．．．．．．．．．．．．．．．．．．．e ．.

Generic plane、generic straight line and generic point are generally called linear fiqures，and

generic curved surface and generic curve of nonlinear equation are generally called non-linear

fiqures.

Equations of generic curved surface and generic curve will be discussed wehn solving the

figured rules of preferred n-dimensional system.

§§§§2 Graphic rules of preferred n-dimensional system————three kind of

graphic methods

2．．．．1 Concepts of three graphic methods

Graphic rules of preferred n-dimensional system is how to identify the figured principle of

figures,graphic methods and the objects be figured，these contents form the basic principle and

methods of oblique axes draughting and high-dimensional spacial and analytic geometry.

Because figures of preferred n-dimensional system are regarded as the trace of generic point，and in the final analysis，generic point is regarded as the trace of points(moving along principal

overlapping direction and directions parallel to all linear combinations). In this meaning，we can

build the relation between the directviewing figures of generic curved surface or generic curve and

their corresponding algebraic equations：all points coordinates on the generic curved surface or

generic curve satisfy the equations of the generic curved surface or generic curve；All points

coordinates that aren't on the generic curved surface or generic curve don't satisfy their equations.

Contrary，all points that satisfy equations of generic curved surface or generic curve must be on the

generic curved surface or generic curve ；all points that don't satisfy equations of generic curved

surface or generic curve must not be on the generic curved surface or generic curve.

Judge a point to be on the generic curved surface(or generic curve) or not，it is not onoy

determined by direct-view(because for difficient-order generic curved surface and difficient-order

generic curve，some points are on the figures by direct-view，but they are not on them by the

analytic property of equations)，but also be determined by figured rules(viz. three figured methods).

Defination 1: Coordinates of points be figured can be directly read from figures without

explanations ，this kind of figured method is called direct graphic method..

Defination 2: Coordinates of points be figured can't be directly read from figures, they need

explanations(include the analytic property of equations or proper auxiliary lines ) to give

coordinates of the point or point out some of their geometric properties，this kind of figured method

is called indirect graphic method..

Defination 3: Between the above two methods，method that using a sufficient-order generic

point and its translation trace to express all points belonging to them, it is called an ordinary figured

method.

Direct graphic method.s uses couples of preferred n-dimensional system，they represent the

31

coordinates of ordinary points(viz. zero-order generic point)；both indirect graphic method.s and

ordinay figured methods use single n-dimensional system(referred to host system). Of which，indirect graphic method.s are not only used for ordinary points but also be used for difficient-order

and sufficient-order generic points，but ordinary figured methods are only used for sufficient-order

generic points and their translation

trace.

Before introducing three kind of

graphic methods，let's introduce

along-axis measuring method by a

visualized example[6]，this is the

necessary basis of constituting figured

rules of preferred n-dimensional system——because the position of a random

point of n-dimensional euclidean space

in a random preferred n-dimensional

system is determined by this method.

Example 1: The preferred

five-dimensional system of figured 13 figure13

is generated by transform

−−

−−

214

213

233

232

eee

eee □□ ，

let's introduce the method of determining the position of point B(2，1，1，-3，2) in the system：

Accoring to coordinates of B relative to each axis

x1 = 2 , x2 = 1 , x3 = 1 , x4=－3 , x5 = 2

and take the following steps first：begin with the origin and make

11 2eOB = ；

secondly：begin with point B1 and make 221 eBB = ；

thirdly：begin with point B2 and make 332 eBB =

or 2132 51 ee.BB −−□ ；

fourthly：begin with point B3 and make 443 3eBB −=

or 2143 23 eeBB +□ ；

last(viz. fifthly)，begin with point B4 and make 54 2e=　　BB .

then，the position of point B(2,1,1,-3,2) is determined (referred to in figure13).

2．．．．2 Indirect graphic method and ordinary graphic method

Contents of direct graphic method is much，we will discuss it in §3 of this chapter. Here we

will first discuss indirect graphic method. and ordinary figured method.

32

2．．．．2．．．．1 Indirect graphic method

When coordinates of a point relative to each axis are known，first we find the proper position

of the point in preferred n-dimensional

system by along-axis measuring method，secondly label its coordinates relative to

each axis beside it，or point out it among

corresponding words or equations. Figure 14

is an example of removing auxiliary point

and auxiliary line of figure13 and label its

coordinates under point B to represent the

position of point B(2,1,1,-3,2) in the

preferred five-dimensional system.

We often take indirect graphic method

for points that can't be directly figured，sometimes we take indirect graphic method figure 14

for points that can be directly figured. The following are examples of building equations of generic

curved surface by indirect graphic method.

Example 2: figure 15 is a generic curved

surface whose centre is on the origin and the

radius is R of four-dimensional space，please find

out the equation.

Solution：The exercise give that the distance

between a random point of the generic curved

surface and the orgin is R，suppose the random point

is M( x1 , x2 , x3 , x4 ) ，

then it has |OM|=R，

viz. Rxxxx =+++ 2

4

2

3

2

2

2

1 ，

square the two ends of it,it has figure 15

22

4

3

3

2

2

2

1 Rxxxx =+++ .

this is the equation of the generic curved surfaceto be solved.

The generic curved surface is called generic sphere.

Example 3: It is known that a generic plane of preferred four-dimensional system

perpendicularily halve the line segment between point A(3,5,2,6) and point B(8,7,4,3)(refer to

it in figure16)，please find out its equation.

Solution：from the exercise we know the generic curved surface to be solved is trace of points

that are equidistant from point A and point B，a random point on the generic plane

M(x1 , x2 , x3 , x4 )

has relation |AM|=|BM| ，viz.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2

4

2

3

2

2

2

1

2

4

2

3

2

2

2

1 34786253 −+−+−+−=−+−+−+− xxxxxxxx

Square the two ends of equal mark and simplify it ,it has

5x1+2x2+2x3-3x4-32=0,

33

This is the equation of generic plane to be solved.

As will be readily seen，the figures and methods used in "analytical geometry of three

dimensions" of High Mathematics [7]，are mostly equivalent to the indirect graphic method. to be

referred to.

2．．．．2．．．．2 Common generic of points

The principle of ordinary graphic method is

based on points having common generic theory，and points having common generic theory

condense oblique axes transform to a kind of

simple sign and make the description more

shortcut and convenient.

Defination 4: Suppose the principal

overlapping direction of a preferred

n-dimensional system is

S1=r11e1+r12e2+r3e3,

S2=r21e1+r22e2+r3e3 , … … … ， figure 16

Sn－3=rn－3,1e1+rn－3,2e2+rn－1en－1 ,

(in each formula，ri1 , ri2 aren't zero at the same time，ri+2 ≠0；i=1 ，2 ，… ，n-3 )，then its

coefficient matrix.

=

−−− 12,31,3

42221

31211

nnn rrr

rrr

rrr

RLLL

is called principal overlapping matrix.

Defination 5: In preferred n-dimensional system， if more than two points A1，A2，…，Am

(m≥2) belong to the same generic pointA (or explain that they have common generic A )，we call

these points common-generic，marked as

A1A2…Am or A~ A1A2…Am

Theorem 2：Suppose more than n-2 points A1A2…Am (m≥n-2)，the vector of a random point

Ai(1≤i≤m) to other points is

{ } ( )jimiijAAaAA jnjjji ≠+−== ；　　　 ,,1,1,,1,,, 21 LLL ，

then it make the necessary and sufficient conditions of

A1A2…Am

is ，matrixes

=+++

−−−

mnmm

niii

niii

n

aaa

aaa

aaa

aaa

A

L

LLLL

L

L

LLLL

L

21

,12,11,1

,12,11,1

11211

composed of coordinates of these vectors can be changed into principal overlapping matrix R by

34

elementary transform.

Proof ：For the convenience of description，we may suppose i=m and j=1 ，2 ，… ，m-1

of ji AA .If A can be changed into principal overlapping matrix R，then vectors 1AAm ，2AAm ，…， 1−mm AA are respectively a linear combination of principal overlapping direction S1，

S2，…，Sn－3 ，so when order

( )numbers are 0 because

3

1

22311311

22212144

21211133

i

n

i

i

,n,nnn

kSk

rrr

rrr

rrr

∑−=

−−−−

−−

−−

−−

□ □□□ eee

eee

eee

LLL ，

〔refer to it in formula (11) of the second chapter〕，so it also has

( )numbers are 0 3

1

i

n

i

jm kAA∑−=

□

then ，in the preferred n-dimensional system taking S1，S2，…，Sn-3 as principal overlapping

direction .， A1A2…Am

this is sufficiency.

If A can't be changed into principal overlapping matrix R，then despite a random group of S1，S2，…，Sn-3 are choosed as principal overlapping direction ，among A1，A2，…，Am there are always

such points，the vectors of them to other points can't be linearly represented by the group S1，S2，…，Sn-3 ，so，in preferred n-dimensional system taking a random group S1，S2，…，Sn-3 as

principal overlapping direction ，they all can't make

( )numbers are 0 3

1

i

n

i

jm kAA∑−=

□ ,

viz. they can't make A1A2…Am

this is necessity. ▌

From theorem 2 of the second chapter，the coordinates of generic points to upright axis are

uniquely determined，so the coordinates of A1，A2，…，Am (m≥n-2) to xn are the same. But when

the coordinates of A1，A2，…，Am (m≥n-2) to xn are not identical but identical to another axis

xk (1≤k≤n-1)，we can take axis xk as upright axis and change xn into another axis(accordingly，the nonzero column suffix of the k column in principal overlapping matrix are all changed into n)，we can still order A1A2…Am

In actual application，we usually first take the opposite trace of generic points. But for the

other points(altogether n-3), we take the vector terminal which takes the opposite trace as starting

point and parallel to every principal overlapping direction.

35

2．．．．2．．．．3 Ordinary figured method

Suppose there a group of points A1，A2，…，Am (m≥n-2) ，they satisfy the conditions of

theorem，find out the principal overlapping matrix R according to the coordinates of these points，then write out the corresponding oblique axis transformation，in order to make out preferred

n-dimensional system which orders

A～ A1A2…Am

(m≥n-2) ，in the n-dimensional system，first find out the projection trace of generic point A and

mark out by Ja′，then make out nnaa eaJJ □′

(point is marked as Ja A，Ja is the opposite trace of generic point A ) according to the coordinates

of these points to the coordinate xn = an of upright axis，rub out the connecting line between Ja′

and Ja (is marked as A ) then the ordinary figure to generic point is finished. The method can be

generalized into three steps：

(1) choose preferred n-dimensional system that can make

A～ A1A2…Am

(m≥n-2) ；

(2) determine the position of projection trace

Ja′of generic point A ；

(3) determine the opposite trace Ja viz. the

position of generic point A (so we mark it as A

not Ja ).

In ordinary figured method we should pay

attention to：while we figure a generic point,we

must mark out its position of projection trace. If

we don't mark out the position of projection trace,

then it indicates that the generic point is on figure 17

generic coordinate plane and it coincides with the trace point.

Example 4: Figure 17 shows a generic plane of preferred six-dimensional system，please find

out its equation.

Solution： The preferred six-dimensional system that the generic plane is in is generated by

transform

5213

4212

3211

115

214

213

5 3 3

5

is system theofdirection ramparting principal theindicatesit

335

5

eeeS

eeeS

eeeS

eee

eee

eee

+−−=

+−−=

+−−=

+

+

+□□□

So， random generic points on the generic plane are all correspondent with the vectorS1，S2，S3 .

In figure17，generic points C and D are on the generic plane and they concide wit their trace，then we can easily know the two points are

36

Jc (-4,3,0,0,0,0)，Jd (-2,0,0,0,0,0)，

but the projection trace of generic point A is

Ja′(-4,0,0,0,0,0)，

so its oppposite trace is Ja (-4,0,0,0,0,6).

We suppose the generic plane is composed of the translation trace of generic point C along

. 3

1

32 suppos and

and

625

214

eeJJ

eeJJ

JJJJ

ac

dc

acdc

−=−=

−==

S

S

　　　　　　　　　　　

，

Respectively regard the coordinates of S1，S2，S3 ，S4，S5 as elements of the following

determinants from the second row to the sixth row，then the normal direction vector of the generic

plane { }1 5 3 1 2 3

200010

000032

010011

005033

000511

50

1

654321

,,,,，=

−

−

−

−

−

=

eeeeee

H ，

is acquired.Choose a known point Jd and a random point

X (x1 , x2 , x3 , x4 , x5 , x6 ) ，

and use dot-method formula [7]

{ } ( ){ } .xxxxxxXJ d 　　，，，，，，，，，， 021 5 3 1 2 3 654321 =−−⋅=⋅H

then we get the general equation of the generic plane

065323 654321 =++++++ xxxxxx .

§§§§3 Direct graphic method

3．．．． 1 Single principal overlapping direction、、、、 oblique number and oblique

coordinate

Defination 6：Suppose the principal overlapping direction of a preferred n-dimensional system

is ( )3 2 1 222211 −=++= ++ nieaeaea iiiii ，，， LS ,

of which， ai1 and ai2 aren't zero at the same time，ai+2 ≠0 , and at least two numbers of

each group of ai1 ，ai2 ，ai+2 are relatively prime ．．．．．．．．．．．．．．．．．．. And suppose

{ }0 call then we

1321321

22,32212

11,32111

，，，，，

，

，

−−

−

−

=+++

=+++

=+++

nn

n

n

aaaa

aaaa

aaaa

LL

L

L

SSS

is single linear combination of principal overlapping direction，it is also called single principal

overlapping direction of the preferred n-dimensional system.

Defination 7 : From the third coordinate to the (n-1) -th coordinate a3 ，a4 ，…，an－1 of

single principal overlapping direction { }0 1321 ，，，，， −= naaaa LS

37

are called oblique axes transform number，they are called oblique number for short，we use a vector

S(xs) to denote them，viz. S(xs) ={0，0，a3，a4，…，an－1，0} .

Defination 8: The coordinates x3 , x4 , … ，xn－1 of a random point A（x1 ， x2 ，… ，xn ） of

preferred n-dimensional to each oblique axis are called oblique coordinates of A，we also use a

vector A(xb) = {0，0，x3，x4，…，xn－1，0}

to denote it.

3．．．．2 Coordinate transform of points

Supppose images of a point A（x1 ， x2 ，… ，xn ）

under singular linear transform σ1 and σ2 are

A01 ( x1′, x2′，0 ，… ，0 , xn′)，A02 (x1＂, x2＂， 0 ，… ，0 ，xn＂) 〔transform formula like formula (9) and formula (10) in the second chapter〕，regard A as the

terminal point of vectorOA (O is the origin)，then the transform of A coordinate can be regarded

as the transform of OA coordinate. Successively substitute formula(9) and formula(10) of the

second chapter into OA , then it has

，，○

□

OA

21

1

2,3

4

4

223

3

122

11

1

1,3

4

4

213

3

111

21

1

2,3

4

4

223

3

122

11

1

1,3

4

4

213

3

111

nnn

n

n

n

n

n

nnn

n

n

n

n

n

exexa

ax

a

ax

a

ax

exa

ax

a

ax

a

ax

exexa

ax

a

ax

a

ax

exa

ax

a

ax

a

axOA

+ +++++

+ ++++

+ −−−−+

+ −−−−

−

−

−

−

−

−

−

−

−

−

−

−

L LL L

but it also has the terms ，　　　 ○□ , 0201 OAOAOAOA

.

viz.

2211

2211

nn

nn

exexexOA

exexexOA

′′+′′+′′

′+′+′○□　　　　　　　　　　　 ，　　　　　　　　　　

( )

( )2

1 hasit so

1

1

2,3

4

4

223

3

1222

1

1

1,3

4

4

213

3

1111

1

1

2,3

4

4

223

3

1222

1

1

1,3

4

4

213

3

1111

=′′

++++=′′

++++=′′

=′

−−−−=′

−−−−=′

−

−

−

−

−

−

−

−

−

−

−

−

nn

n

n

n

n

n

n

nn

n

n

n

n

n

n

xx

xa

ax

a

ax

a

axx

xa

ax

a

ax

a

axx

xx

xa

ax

a

ax

a

axx

xa

ax

a

ax

a

axx

L

L

L

L

The above formula(1) and formul(2) can be derived from formula(7)、formula(8) of the second

38

chapter and formula(6) of the first chapter.

Suppose the relation between the oblique coordinates A(xb) and the oblique numbers S(xs) of

point A is A(xb) =λS(xs) ，

(it is read as “the oblique coordinates of A are equal to the oblique numbers of λS ”) ，viz.

λ====−

−

1

1

4

4

3

3

n

n

a

x

a

x

a

xL

(λ is a number)，because a11 + a21 + … + an－3,1 = a1 ，

a12 + a22 + … + an－3,2 = a2 ，

then，the above two formulas are changed into

( )

( )4

3

222

111

222

111

=′′

+=′′

+=′′

=′

−=′

−=′

nn

nn

xx

axx

axx

xx

axx

axx

λ

λ

λ

λ

because A(xb) =λS(xs) ，

it has x3 =λa3 , x4 =λa4 , … , xn－1 =λan－1 ，

add up formula (3) and formula(4) ，it has

( )

( )

( )

( )5

1,,4,3

2

1

2

1

11

33

222

111

′′=′=

=

−==

=

′′+′=

′′+′=

−−

nnn

nn

ii

xxx

ax

niax

ax

xxx

xxx

λ

λ

λ

LL

L

LL

For direct graphic method, point coordinate transform under oblique axes transform has great

significance. In fact，formula(3) and formula(4) are basis of the figured methods when point A of

preferred n-dimensional satisfies A(xb) =λS(xs) ，and formula(5) is the basis of reading the

coordinates of the points be directly figured.

3．．．．3 The principle of direct graphic method

Theorem 3：The necessary and sufficient conditions of a random point A of n-dimensional

space can be directly figured in a couple of preferred n-dimensional system is

A(xb) =λS(xs)

(λ is a number).

39

Proof：From formula (3)，formula(4)，formula(5)，if coordinates of A satisfy

A(xb) =λS(xs) ，

then it can be directly figured in preferred n-dimensional system which takes S as single principal

overlapping direction，this is sufficiency.

If coordinates of A don't satisfy A(xb) =λS(xs) ，

What about the result? Please look at the following example：in figure 18，

S1 =3e1 + 2e2 +2e3 ，S2 =3e1 + 2e2 +3e4 ，

then S=｛6，4，2，3，0｝， S(xs) =｛0，0，2，3，0｝ ，

coordinates of point A(0.9，-0.6,0.5,0.75,6) satisfy

A(xb) = 1 4 S(xs)，

Figure 18 indicates that it can be directly figured in the couple of preferred five-dimensional

system. but it is also observed that，point (0.9，-0.6,1,0,6) ,(0.9，-0.6,1.5,-0.75,6) and

point A belong to the same generic point，their position in host system and guest system concide

with A

figure 18

completely.

If we don't regulate the relevance between the oblique coordinates and oblique numbers of

points，it will arouse chaos，The direct diagrammatic representation of points become yeasty

talk.This is necessity. ▌

From theorem 3，the following two inferences can be immediately acquired.

Inference 1：The necessary and sufficient conditions that more than

two points A1，A2，…，Am (m≥2)

of n-dimensional space can be directly figured in the same couple of preferred n-dimensional

system is λ1A1(xb) =λ2A2(xb) = … =λmAm(xb) = S(xs)

(λ1，λ2，…，λm are numbers). ▌

Inference 2: More than two points of four-dimensional space can be directly figured in the

same couple of four-dimensional system . ▌

3.4 Drawing and distinguishing steps of direct graphic method

For a random point A（x10 , x20 , … ，xn0 ）

of n-dimensional space，we can directly figure it by the following steps：

First，choose a group of nonzero oblique axes transform numbers according to the oblique

coordinates of A a3 ，a4 ，… ，an－1

and represent it by vector S(xs) ，in order to make

A(xb) =λS(xs) .

Second，determine a group of principal overlapping directions according to S(xs)

40

Si = ai1e1 + ai2e2 + ai+2ei+2

(i= 1 ，2 ，… ，n-3 ) notice：ai+2 ≠ 0 ，ai1 and ai2 aren't zero at the same time，and at least two

numbers of ai1 ，ai2 ，ai+2 are relatively prime ．．．．．．．．．．．．．．．．．．. Then make out a couple of preferred n-dimensional

system according to the group of principal overlapping directions .

Third，respectively mark out the projection A′(x10 , x20 , … , xn－1,0 , 0 )

of A on generic coordinate plane by along-axis measuring method in host system and guest system，last respectively mark out the position of A in host system and guest system(if the projection A′of

A is not marked out，it indicates that A is on generic coordinate plane)，then the direct figure of A is

to be finished.(figure18).

The distinguish (viz. the reading of

coordinates of A and other ppints)process is

finished in guest system. Take the guest system

of figure 18(figure19)as example：first find out

the projection trace position of generic point A

that A is on and mark it out by Ja′，connect

Ja′and A′，from the front two equalities of

formula(5) we know，the projection of A on

plane x1ox2 is at the midpoint A＂of line

segment Ja′A′，

respectively figure out vertical lines to axes figure 19

x1 , x2 through point A＂，then its coordinates to axes x1 , x2 are x1 = x10 ，x2 = x20；and

( ) ( )xsxb SAAA λ==′′′ ,

then coordinates of A to each oblique axis are

x3 =λa3 = x30 ，

x4 =λa4 = x40 ， … … … … ，

xn－1 =λan－1 = xn－1,0 ，

this is identical to formula (5). From directly measurement,it has nn exAA 0=′ ,

so the coordinates of A to upright axis are xn = xn0 (refer to it in figure 19).

figure 20

Besides，the figured objects of direct graphic method also includes such points as projection

traces or opposite traces：In A(xb) =λS(xs) when λ=0，the coordinates of A to each oblique axis

are all zero. So，in host system and guest system，their (include their projection on the generic

41

coordinate plane) positions to axis x1 , x2 are identical.

Figure 20 is an example of using theorem 3 and reference 1(in the figure，upright axis xn is

omitted).In the preferred n-dimensional system，point A1 and point A2 are directly figured at the

same time. But figure 21 gives the reading results of the coordinates of the two points：

A1 (a11 , a12 , a3 , a4 , … , an－1 , 0 )，

A2 (a21 , a22 , 2a3 , 2a4 , … , 2an－1 , 0 ).

The following is an example of building

generic curved surface equation using theorem 3

and reference2：

Example 5: In preferred four-dimensional

system of figure 22,there is a curved surface in the

shape of cylindroid (in host system，upright axis x4

is omitted)，please find out its equation according to

the conditions given by the figure.

Solution：The cut trace of the cylindroid-shape

surface cut by generic coordinate plane is a piece of

elliptical continuum，after being read，the four

points of the elliptical continuum separately are: figure 21

( ) , 0,2

,2

,2

, 0 , , 0 , 0 , 0,2

,2

,2

, 0,0,2

,2

−−

−

−

RRRDRC

RRRB

RRA

figure 22

apparently，they aren't coplanar，and their distance to the origin is:

|OA|=|OB|=|OC|=|OD|=R ，

then we can extrapolate that the distance of a random point M (x1 , x2 , x3 , 0 ) of the elliptical

continuum to the origin is |OM|=R ，viz.

Rxxx =++ 2

3

2

2

2

1

square the two endas,it has 22

3

2

2

2

1 Rxxx =++ .

Now look at the elliptical continuum where x4 =x40 ，it is identically equal to the elliptical

continuum of generic coordinate surface，this shows that when the elliptical continuum with

generic coordinate surface translates along upright axis，its position and shape with respect of

upright axis are changeless，so，the equation of the cylindroid-shape surface has no relation with

42

x4.then，equation 22

3

2

2

2

1 Rxxx =++

without variable x4 is the equation of the cylindroid-shape surface，we call the cylindroid-shape

surface generic cylindrical surface.

The elliptical continuum of generic coordinate surface can be called generic circular curve or

generic circle for short，it is generic intersecting line of generic cylindrical surface and generic

coordinate surface，that is to say ，points of the generic circle is not only on the generic cylindrical

surface but also on the generic coordinate surface. then，a random point of the generic circle

simultaneously satisfies equations of generic cylindrical surface and generic coordinate surface.

So，equations of the generic circle are simultaneous system of equations of generic cylindrical

surface and generic coordinate surface

.04

22

3

2

2

2

1 　　

=

=++

x

Rxxx

In fact，the generic circle is a sphere of three-dimensional space . Its centre is on the origin

and its radius is R.

From the above analysis, it is observed that a generic curved surface of preferred

n-dimensional system can be expressed by an equation

f ( x1 , x2 , … , xn ) = 0

among x1 , x2 , … , xn ，contrarily，an equation

f ( x1 , x2 , … , xn ) = 0

among x1 , x2 , … , xn can express a generic curved surface of n-dimensional system. and generic

curve serves as intersecting line of more than two generic curved surface

f1 ( x1 , x2 , … , xn ) = 0

f2 ( x1 , x2 , … , xn ) = 0 … … … … …，

fm ( x1 , x2 , … , xn ) = 0

(m≥2)，its equations are simultaneous system of equations of esquations of these generic curved

surface

( )( )

( )

=

=

=

0

0

0

21

212

211

nm

n

n

x,,x,xf

x,,x,xf

x,,x,xf

L

LLL

L

L

(m≥2).

Exercises

3．1 How to determine the dimension of generic curved surface and generic curve?

3．2 What are linear figures? What are non-linear figures?

3．3 What are figured rules of preferred n-dimensional system?

3．4 What are three kind of graphic methods? Why there are neither more nor less than three

kind of graphic methods?

3．5 In the following preferred six-dimensional system，please mark out the position of point (3,5,2,7,-6,4) according to along-axes measuring method.

43

3．6 Suppose the principal ramparting direction of a preferred five-dimensional system is

2e1 +3e2 + 5e3 , 4e1 + 6e2 + 3e4 ，

the projection trace of a generic point is Ja′(2,-3,0,0,0) ,

its opposite trace is Ja (2,-3,0,0,4) ,

please picture A.

3．7 Suppose there are five points

A0 (2,-1,0,0,0,0,5), A1 (5,3,4,0,0,0,5),

A2 (5,3,0,3,0,0,5), A3 (5,3,0,0,5,0,5) , A4 (5,3,0,0,0,4,5)， please make out a preferred seve n-dimensional system and make

A0A1A2A3A4 .

3．8 Please write out equations of the following generic plane according to the methods of

example 4.

3．9 Please picture point A (2,5,4,6,5,6)

by direct graphic method.

3．10 Please determine coordinates of point A according to conditions offerred by the

following figures.

3.11 Please judge which points of the following can be directly pictured in the same couple of

preferred n-dimensional systems，which can't be directly pictured：

A1 (3,2,5,3,3,5), A2 (5,4,7,1,5,6)，A3 (2,5,3,6,4,3), A4 (4,-3,10,6,6,7), and picture the points that can be directly pictured in the same couple of n-dimensional system.

44

The Fourth Chapter: Shape of figures in preferred

n-dimensional system

Figures in preferred n-dimensional system can be divided into linear and non-linear according

to their property.Their shape is referred to whether they are in the shape of directviewing

plane,line or point or not in a preferred n-dimensional system. The following respectively narrate

them.

§§§§1 Shape of linear figures

1．．．．1 Generic planes and their shape

When the equation of generic curved surface F

f (x1 ，x2 ，… ，xn ) =0

is a first-order equation about x1 ，x2 ，… ，xn ，F represents a generic plane(refer to it in example

3 and example 4 of the third chapter)，it can be regarded as a particular case of generic curved

surface，and generic coordinate plane can be regarded as a particular case of generic plane，its

equation is always xn = 0 .

The same generic plane isn't always in the shape of plane in different preferred n-dimensional

system，different generic planes are not always in the shape of plane in the same preferred

n-dimensional system. In order to simplify it, we are always ready to change generic planes we

studied into planes as many as possible. Our desire can at least be satisfied for each separate

generic plane.

Example 1: Suppose there is a generic plane of six-dimensional space

F： x1 +3x2 +4x3 +2x4 +x5 +3x6 + 5 = 0

please change it into a figure which is in the shape of plane.

solution：：：：choose four coordinates on F, their x 6 are all zero at the same time

A0（－5，0，0，0，0，0），A1（1，2，－3，0，0，0），

A2（1，2，0，－6，0，0），A3（1，2，0，0，－12，0），

then it gets a group of vectors

. 1226

626

326

521303

421202

321101

eeeAA

eeeAA

eeeAA

−+==

−+==

−+==

S

S

S

，

，

order A~ A0A1A2A3

then it gets a preferred six-dimensional system which take S1 ，S2 ，S3 as principal overlapping

direction，A is a sufficient-order generic point of the six-dimensional system andA is on F. So，F

can be regarded as the trace of A，viz. the orders of F. J = 6-3 = 3 . From linear algebra theory，we

know the actual dimensions of F are W = 6-1 = 5 ，then according to dimension theorem of the

third chapter,the directviewing dimensions of F are

Z = W-J = 5-3 = 2 ，

viz. F is changed into a figure in the shape of plane. (refer to it in figure 23).

Example 2: Suppose an equation of a generic plane in five-dimensional space is

F：3x3 +2x4 -x5 –3 = 0

45

please change it into plane .

solution：Because coefficients of the front two terms are zero， points that whose coordinates

relative to x5 are identical and satisfy other conditions of theorem 2 in the third chapter are

ompletely

figure 23

exist. So，choose x5 asvertical axis，and change x2 into upright axis. So，choose three points on F

A0 (2,4,0,0,-3), A1 (5,4,2,0,3), A2 (5,4,0,3,3).

then it gets . 633 623 5412053110 eeeAAeeeAA ++=++= ，

order A~ A0A1A2 ，

viz. use transform

−−

−−

514

513

2

632

eee

eee □□

then we can change F into plane.(figure 24).

Normally，for a random generic plane whose

a1 and a2 aren't zero at the same time(or a1 and

a2 are zero at the same time，but a3 , a4 , … , an－1 are also zero at the same time)

F：a1 x1 + a2 x2 + … + an xn + a = 0

On it, we can all find n-2 points A0 ，A1 ，… ，An－3 figure 24

with coordinates xn are identical and make

302010 −nAAAAAA ，　，　，　 L

linearly indeqendent and satisfy other conditions of theorem 2 in the third chapter，order

A~ A0A1A2…An-3 ，

then it gets a preferred n-dimensional system(sometimes， 302010 −nAAAAAA ，　，　，　 L can

become its principal overlapping direction)，A is a sufficient-order generic point of the

n-dimensional system and A is on F，viz. F is regarded as the trace of A .

46

So，the order of F is J = n – 3 ，its actual dimension is W = n –1 ，So from dimension

theorem W=Z+J ，then the directviewing dimension of F become

Z = W – J = ( n - 1) - ( n – 3 ) = 2 ，

viz.F become in the shape of plane.

But，like generic plane of example 2, coefficients a1 and a2 of the front two terms of their

equations are zero at the same time，but their coefficients from the third term to the n-1-th term are

not zero at the same time. If we still order x1 ，x2 ，x5 separately as cross axis ,vertical axis and

upright axis，because the points we choosed can't satisfy theorem 2 of the second chapter (the

coordinates of generic points relative to upright axis are uniquely determined),then they have no

common generic.〔for example，order x3 ，x4 ，…，xn－1 are zero at the same time，it has

n

na

ax −=

(when determine the opposite trace of generic points ,we should take values like this)，if order one

of x3 ，x4 ，…，xn－1 isn't zero，the others are still zero at the same time (when determine the

terminal points of vectors parallel to every principal overlapping direction and take the opposite

trace of generic points as starting points,we should take values like this)，but it has

n

na

ax −≠ 〕，

so it is impossible to change the generic plane into plane .But if we change one of x1、x2 into upright

axis，and change xn into cross axis(or vertical axis)，then the generic plane can be changed into

plane.

It should be specialized that，when the coefficients of variable x3 ，x4 ，…，xn－1 of equations

of a generic plane are zero at the same time，no matter how take values to x3 ，x4 ，…，xn－1, they

all can make the equation be ture. So，the generic plane can still be changed into plane.

As the above description，we has：

Theorem 1: For a random generic plane of n-dimensional space

∑=

=+n

i

ii axaF1

0：　

we can always change it into the shape of plane in a preferred n-dimensional system by proper

oblique axes transformation. ▌

When a generic plane isn't in the shape of plane in a preferred n-dimensional system，the

generic plane can be regarded as trace of difficient-order generic points，so it is also called

difficient-order (or n - 4-order ) generic solid.〔Proof：From linear algebra theories，the actual

dimension of generic plane is W = n – 1，and from dimension theorem W = Z + J ，and the

maximum directviewing dimension of a random figure Z≤3，so，when generic plane isn't in the

shape of plane，its order J = W - Z = ( n - 1 ) - 3 = n - 4〕.

1．．．．2 Generic straight lines and their shape

In three-dimensional space,there is a concept that the intersection of two planes determines a

straight line，extend the concept to n-dimensional space，then it is the intersection of two generic

planes determines a generic straight line.

Suppose two generic planes

F1 ：a11 x1 + a12 x2 + … + a1n xn + a1 = 0

F2 ：a21 x1 + a22 x2 + … + a2n xn + a2 = 0

intersect on generic straight line L，then the equation of L is compatible system of equations

=++++

=++++

0

0

22222121

11212111

axaxaxa

axaxaxaL

nn

nn

L

L

：　

47

( ) 2 , 1 0 or 1

==+∑=

jaxaLn

i

jiji：　

A same generic straight line isn't always in the shape of line in different preferred

n-dimensional systems，different generic straight lines aren't always in the shape of line in a same

preferred n-diemensional system. But in order to simplify it，we are always ready to change generic

straight lines we studied into the shape of line. Our desire can at least be separately satisfied for

every single generic straight line.

Example 3: Suppose there is a generic straight line in six-dimensional space

, 021533

0145332

654321

654321

=++++++

=++++++

xxxxxx

xxxxxxL：　

Please change it into the shape of line.

Solution：first find out four points on L

A0（－7，0，0，0，0，0）， A1（－5，1，－7，0，0，0）， ，　， 0 , 5

7 , 0 , 0 , 1 , 5 0 , 0 , 0 ,

3

7 , 1 , 5 32 −− −− AA

and order A0A1A2A3

then a preferred six-dimensional system is aquired. Because vector 302010 AAAAAA ，　，　 can be

regarded as the principal overlapping direction of the preferred six-dimensional system，and all of

its linear combinations are a sufficient-order generic point，so L is regarded as the trace of

sufficient-order generic points，viz. now its order is J = 6 – 3 = 3. From linear algebra theory，its

actual dimension should be W = 6 - 2 = 4，then from dimension theorem its directviewing

dimension is Z = 4 - 3 = 1，viz. L is changed into the shape of line(figure 25).

Example 4: Suppose there is a generic straight line in five-dimensional space

Figure25

，：　 0223

074232

54321

54321

=++++

=+++++

xxxxx

xxxxxL

please change it into the shape of line.

Solution： First find out three points on L

A0（1，－3，0，0，0）， A1（0，－7，7，0，0）， A2（－4，－2，0，7，0），

48

and order A0A1A2

then a preferred five-dimensional system of emanative

oblique pattern，in the five-dimensional system，L is in the

shape of line(figure 26).

Example 5: Suppose

=+−+

=−−+

013

04322

421

321

xxx

xxxL：　

is a plane of four-dimensional space，please change it into the

shape of line.

Solution：because the rank of the coefficients matrix of its figure 26

front two terms' coefficients is 1， a vector between two random points α11e1 +α12e2 +α3e3

can't makeα11 ，α12 zero at the same time andα3≠0. Then，change x4 into vertical axis and

change x2 into upright axis，choose point

( )( )， 2,2,0,5 and

1,0,0,2

1

0

A

A

on L, then a vector between the two points is acquired

3e1 + e4 +2e3 ，

Order A0A1 ，

viz. using the transform 2e3 □ -3e1 - e4 ，

can change L into the shape of line.(figure 27).

Let's analyze the front examples：in example 3，the rank of coefficients matrix

( )

=

× 153113

153132

2 njia

is r = 2 ， of which the rank of coefficients

matrix ( it is a submatrix of ( a j i )2 × n ) of

variables x3 ，x4 ，x5

( )

=

−×+531

531

)3(22, nija

is r′= 1 . But in example 4, both the ranks of coefficients

matrix (aji)2×n and skew matrix (aj,i+2 )2×(n－3) are 2. So，the

generic straight lines of example 3 can be in the shape of line in figure 27

a convergent oblique pattern，but the generic straight lines of example 4 can only be in the shape of

line in a emanative oblique pattern. But，when we exchange variable x4 and x5 of the equation of

example 4 and newly arrange them，it will has

=++++

=+++++′

0223

074232

54321

54321

xxxxx

xxxxxL ：　

49

because the rank of the skew matrix becomes 1，then L′can be in the shape of line in a preferred

five-dimensional system of convergent oblique pattern.(figure 28).

Normally，for a random generic straight line of n-dimensional space

( )2 , 1 01

==+∑=

jaxaLn

i

jiji：　

when the rank of coefficients matrix of its front columns is 2，we can find out n-2 points

A0 ，A1 ，A2 ，…，An–3 whose coordinates relative to xn are identical，Of which，A0 is the opposite

trace of a sufficient-order generic point on generic straight line L，A0 ，A1 ，A2 ，…，An–3 separately

is terminal point of every principal overlapping direction which takes A0 as starting point. The

following is our detailed proof.

First suppose xn =αn ，and suppose

2221

1211

aa

aaD =

(because the rank of coefficients matrix formed of the front two terms' coefficients of the

equation of L is 2，so D ≠ 0 )，and

， , 221

111

2

222

121

1ta

taD

at

atD ==

of which， t1 = - a1nαn - a1 ， t2 = - a2nαn - a2 ，

when suppose x3 = x4 = … = xn－3 = 0 ，and D ≠ 0 ，according to Cram's rule，it has

，，　　　 222

111

D

Dx

D

Dx ==== αα

Then point A0（α1 ，α2 ，0，…，0，αn ）

is first choosen.

And suppose the other n–3 points are expressed by

Ak ( xk1 ，xk2 ，0 ，…，0 ，xk+2 ，0 ，…，0，αn )

( k=1 ， 2 ， … ，n -3 ) ，for the convenience ，we can suppose

( )

. which of

suppose and number realany can take , 0hen 0

21

22

122222222

111 1 12211

22121

21

22

+

+

++

++

+++

+

++

=

−=−−−=−−−=

−=−−−=−−−=

=≠

==

k,

k,

k

knnnnkk,k

nnnnkk,k

kk,k,

k,

kk

a

a

DtaaDkaaat

DtaaDaaat

xawa

a

Dx

λ

λααα

ααα

α

，　

，

，　，

because we have already suppose D ≠ 0 ， and when we suppose

， , 221

111

2

222

121

1

k

k

k

k

k

kta

taD

at

atD ==

according to Cram's rules，it has

50

，，　　　 2

22

1

11D

Dx

D

Dx k

kk

k

kk ==== αα

viz. point A1（α11 ，α12 ，α3 ，0，…，0，αn）,

A2（α21 ，α22 ，0，α4 ，0，…，0，αn）, … … … … … … … ，

An-3（αn-3,1 ，αn-3,2 ，0，…，0，αn-1 , αn）

is also choosen.Order A0A1A2…An－3

a preferred n-dimensional system taking

302010 −nAAAAAA ，　，　，　 L as principal overlapping

direction is acquired. Now，A0 ，A1 ，A2 ，…，An–3 have a

common sufficient-order generic point，and the generic point is

on L . So ，L can be regarded as trace of the generic point，viz.its

order is J = n- 3，from linear algebra theory，the actual

dimension of L is W = n – 2 ，then from dimension theorem

we know the directviewing dimension of L is Ｚ = W – J = (n – 2 ) - (n – 3 ) = 1

viz. L become in the shape of line.

It is easy to know the rank of coefficients matrix figure 28

( )

=×

n

n

njiaaa

aaaa

22221

11211

2L

L

of L is r = 2 , when the rank of its skew matrix

( )

=−

−

−×1,22423

1,11413

)3(2n

n

njiaaa

aaaa

L

L

r′≤1(viz. r′=0 or r′= 1 )，it has λ1 =λ2 = … =λn－3 ，

then it has t12 = t22 = … = tn－3,2 .

so α11 =α21 = … =αn－3，1 , α12 =α22 = … =αn－3,2 ，

viz. the preferred n-dimensional system is convergent oblique pattern. When r′=2 ，only when λ1 =λ2 = … =λn－3 = 0

t12 = t22 = … = tn－3,2

is true，but because 02,1

2,2==

+

+

k

k

ka

aλ ，

it must has a23 = a24 = … = a2,n－1 = 0 ，

then ，the rank of skew matrix become r′=1 ， it contradicts with the assumption r′=2. So the

n-dimensional system can only be emanative oblique pattern.

When the rank of the front two terms' coefficients matrix of L is 1，then the above point

A0 ，A1 ，A2 ，…，An–3 which simultaneously satisfy theorem 2 and theorem 3 of the second

51

chapter can't be simultaneously found out，so the generic straight line can't be directly changed into

the shape of line，now，we can choose another axis as cross axis (or vertical axis) instead of x1（or x2 ），and change x1 (or x2 ) into other axis，in order to change the rank of the two terms'

coefficients matrix of L corresponding to new cross axis and new vertical axis into 2，viz. change L

into the shape of line.

As the above description，we has：

Theorem 2：：：：For a random generic straight line of n-dimensional space

( )2 , 1 01

==+∑=

jaxaLn

i

jiji：　

we can always make it in the shape of line in a preferred n-dimensional system by proper oblique

axes transform. Of which，when the rank of coefficients matrix (aji )2×n r=2，the rank of the front

two columns of submatrixs (ajk )2×2 is 2 ，when the rank of skew matrix (aj,i+2 )2×(n－3) r′≤1，the preferred n-diemensional system can be a convergent oblique pattern；But when r′= r = 2 ，the n-dimensional system can only be a emanative oblique pattern. ▌

When a generic straight line isn't in the shape of line but plane and solid in a certain preferred

n-dimensional system，it can also be regarded as trace of difficient-order generic point，so it can

also be called difficient-order generic plane or difficient-order generic solid.

1．．．．3 Shape of many generic planes' intersection

Now let's discuss cases when more than three generic planes

0

0

0

11

221212

111111

=+++

=+++

=+++

mnmnmm

nn

nn

axaxaF

axaxaF

axaxaF

L

LLLLLL

L

L

：

　

：

：

( m≥3 ) intersect .Assume

( )　；：　 3 , , 2 , 1 01

≥==+∑=

mmjaxaMn

i

jiji L

is a figure determined by m generic planes' intersection，and assume the rank of coefficients matrix

of M ( aji )m×n , r = m ≤n ，and the system of equations is compatible，then M can be regarded as n –m -3-order generic solid (when n≥6，m≤n-3 )or n–m -2-order generic plane (n≥5,m≤n-2 )

or n-m-1-order generic straight line( n≥4, m≤n-1). Apparently，when m=n-2，M can be regarded

as a zero-order generic plane(viz. ordinary plane)or one-order generic straight line；When m=n-1，M can only be regarded as a straight line or an one-order generic point. When m=n ，M represents a

point. It is necessary to be pointed out, when m=3 ，sometimes M represents a sufficient-order

generic point.

Example 6: Assume a linear figure of five-dimensional space

，：　

07631210

0434243

042244

54321

54321

54321

=−−++

=+−−++

=−−+−+

xxxxx

xxxxx

xxxxx

M

52

please change it into the shape of point.

Solution：Make elementary transformation for the augmented matrix of M

−−

−−−

−−

07631210

434243

422144

(the third row minus twice of the second row ,then minus the first row),it gets

，

410000

434243

422144

−−−

−−

viz. the equation of M becomes

，：　

04

0434243

042244

5

54321

54321

=−

=+−−++

=−−+−+

x

xxxxx

xxxxx

M

Now，M can be regarded as intersection of a generic straight line

=+−−++

=−−+−+

0434243

042244

54321

54321

xxxxx

xxxxxL：　

and a generic coordinate plane x5 = 4 .

In three-dimensional space，there is a concept that a plane and a straight line intersect to a

point. Extend the concept to higher space，then the intersection of a generic straight line and a

generic plane determines a sufficient-order generic point. Then，first change L into the shape of line

and then make it intersect with generic coordinate plane according to theorem 2, thus problems can

be solved.

In order to make people have deep impression on theorem 2，we will give a detailed solution

process of n-2 = 5-2 = 3 points A0 ，A1 ，A2 whose coordinates are all four with relation to x5

according to the proof methods of theorem 2.

. 483

124 16

48

412 then

8443 12442

443

44 Because

21

21

−====

=−×==+×=

==

DD

tt

D

，　，　　　　　

，　，　　　　　　　　　

　　　　　　

so when order x3 = x4 = 0，it has

，，　　 1 4 22

11 −====

D

Dx

D

Dx

viz. point A0 (4,-1,0,0,4) is first choosen.

For the convenience，we can assume

( )

( )，

，

2 22

4

1 41

4

2,1

24

2,1

23

=====

=−=−

===

+

+

+

+

ka

Dxx

ka

Dxx

k

k

k

k

53

in turn，assume tk1 = t1 - D, tk2 = t2 –λk D

simultaneously, (k=1,2), of which λ1 =λ2 = - 2

(the rank of skew matrix is 1)，so tk1 = 8 ， tk2 = 16 ，

when assume 221

111

2

222

121

1

k

k

k

k

k

kta

taD

at

atD == 　，

，，　

，，

104

40 8

4

32 so

40163

84 32

416

48 hasit

22212111

22122111

===−=−

==

===−===

xxxx

DDDD

viz. point

A1 (-8,10,-4,0,4)，A2 (-8,10, 0, 2,4)，

are also choosen. Now

,21112

,41112

42102

32101

eeeAA

eeeAA

−−=

+−=

order M～ A0A1A2

then a preferred five-dimensional system taking

2010 , AAAA

as principal ramparting direction is acquired (because λ1 =λ2 = - 2 ，then the preferred

five-dimensional system is convergent oblique pattern)，in the preferred five-dimensional system，M is in the shape of point (figure 29).

Theorem 3: Assume the rank of coefficients matrix (aji )3×n of consistent system of linear

equations ( )3 , 2 , 1 01

==+∑=

jaxan

i

jiji

is r =3；The rank of its front two columns of submatrixes

(ajk )3× 2 is r′= 2， the rank of front n-1 columns of

submatrixes ( ajl )3×(n－1) is r＂= 2；and the rank of skew

matrix (from the third column to the n-1-th column of

submatrix) ( aj,i+2 )3×(n－3) is r = 2 or r≤1，then under proper

oblique axes transform，M represents a sufficient-order

generic point. Of which ,when r≤ 1，M can suit a

convergent oblique pattern，when r = 2, M only suits

emanative oblique pattern. figure 29

Proof ： From theorem1 and theorem 2 of the second chapter，the coordinates of generic point

with relation to upright axis are uniquely determined，its opposite trace is also unique. So，the

solution of equation M with relation to xn should be unique，at the same time，when assume the

solutions of M with relation to x3，x4，…，xn－1 are all zero，its solutions with relation to x1，x2 should be unique.

Because r″= 2，then there must be numbersλ1 ，λ2，λ3 which aren't zero at the same

54

time ,they make ∑∑=

−

=

=3

1

1

1

0j

n

i

ijij xaλ

But from the equation of M,it apparently has

∑∑= =

=+3

1 1

0)(j

n

i

jijij axaλ

then ( ) ∑∑∑∑=

−

== =

=−+3

1

1

1

3

1 1

0j

n

i

ijij

j

n

i

jijij xaaxa λλ

viz. (λ1a1n +λ2a2n +λ3 a3n )xn= -λ1a1 +λ2a2 +λ3 a3

and because r =3，so λ1a1n +λ2a2n +λ3 a3n≠0 ,

then ， 332211

332211 αλλλ

λλλ=

++

++−=

nnn

naaa

aaax

viz. the solution of M with relation to xn is unique.

In fact，because r = 3 ，r″= 2，then in the equations of three generic planes of M，at least

equations of one generic plane can be directly changed into xn =α by augmented matrix operation

of M，then the remainder two equations are equations of a generic straight line，then according to

the methods of theorem 2,we can find out n - 2 points

A0（α1，α2，0，…，0，α），

A1（α11，α12，α3，0，…，0，α）, … … … … …,

Aj（αj1，αj2,0，…，0，αj+2，0，…，0，α）, … … … … …，

An－3（αn－3,1，αn－3,2，0，…，0，αn－1，α），

on the generic straight line ,their coordinates with relation to xn are allα.When order

A1A2…An－3 ，

a preferred n-dimensional system taking

302010 , −nAAAAAA ，，L

as principal overlapping direction is acquired，because all linear combinations among

302010 , −nAAAAAA ，，L

is a sufficient-order generic point，and A0 ，A1 ，A2 ，…，An–3 are on M，both dimensions of

the sufficient-order generic point and dimensions of M are n-3, so M is a sufficient-order generic

point of the n-dimensional system，and A0 is its opposite trace.

Now ，three intersectant generic planes can be regarded as trace of M，viz. they are all in the

shape of plane. And from theorem 2 ，that three generic planes pairwise intersect determines three

generic straight lines. Because r = 3，then the rank of coefficients matrix of equations of a random

generic straight line is 2，assume the rank of its skew matrix is ( ) 3 , 2 , 1 ，=′′′ jr j when

1 ≤′′′r , it has 1 ≤′′′jr ，so the n-dimensional system M suits can be a convergent oblique pattern；When 2 =′′′r ，at least the rank of skew matrix of one generic straight line's equation is

55

2 =′′′jr ，from theorem 2，M only suits a emanative oblique pattern. ▌

Theorem 4 :under proper oblique axes transform，assume equations determined by m (m＞3)

generic planes' intersection are consistent system of equations ( ) 3 . 21 01

>==+∑=

mmjaxaMn

i

jiji 　，，　，　：　 L

and assume the rank of coefficients matrix of M is r，at least the intersection of three generic planes

satisfies theorem 3，then,

(1) when r = m，only three generic planes can be changed into the shape of plane；

(2) When 3＜r＜m，at most m-r+3 generic planes can be changed into the shape of plane；

(3) When r =3，sometimes, the m generic planes can all be changed into the shape of plane.

Proof：assume intersectant generic planes satisfying theorem 3 is at least the following

0

0

0three

331313

221212

11 1111

=+++

=+++

=+++

axaxaF

axaxaF

axaxaF

nn

nn

nn

L

L

L

：　　　　　　　　　　

：　　　　　　　　　　

：：　　　　　　　

then under proper oblique axes transform

(1) when r =

m，according to linear algebra theory，the dimension of the intersection of a

random other generic plane and the three generic planes is n- 4，apparently，the intersection can

only be a difficient-order generic point，other generic planes can only be regarded as trace of the

difficient-order generic point，viz. their orders are all J = n - 4，but from linear algebra theory，their actual dimensions are all W =n - 1，then according to dimension theorem，their directviewing

dimensions are all Z = W – J =3, viz. they are all difficient-order generic solid. So，only the three

generic planes can be changed into the shape of plane.

(2) when 3＜r＜m，assume the rank of coefficients matrix of the front r generic planes '

intersection ( ) 21 01

rkaxan

i

kiki ，，　，　 L==+∑=

is r，then from question (1)，only three generic planes can be changed into the shape of plane，and

assume they are separately F1，F2，F3，oter r -3 generic planes F4，F5，…，Fr can't be changed

into the shape of plane. Besides，there are still m-r generic planes Fr +1, Fr +2 ,…, Fm ，their normal

direction vectors are all a combination of normal direction vectors of front r generic planes. Assume

the normal direction vector of a certain generic plane such as Fr +1 and the normal direction vector

of F4，F5，…，Fr are linearly independent，viz. the normal direction vector of Fr +1 can only be

regarded as a combination of normal direction vectors of F1，F2，F3 ，and because the rank of

coefficients matrix of system of equations

( )

=+

==+

∑∑

=++

=n

i

riir

n

i

ii

axa

axa

1

1,1

1

0

3 , 2 , 1 0 τττ

is 3 ，the rank of front n -1 columns of submatrixes is 2 ，the rank of skew matrix is r＂= 2 or

r＂=1.So，the intersection of the four generic planes also satisfies the conditions of theorem 3，then

they intersect a same generic point，viz. Fr +1 can also be changed into the shape of plane. By the

56

same token，if the intersection of an other random generic plane Fr +s (1＜s＜m-r) and F1，F2，F3

satisfies theorem 3,it can also be changed into the shape of plane.

When the intersection of all the intersections of m-r generic planes and F1，F2，F3

( )

( )

−==+

==+

∑∑

=++

=n

i

riir

n

i

ii

rmaxa

axa

1

,

1

,,2,1 0

3 , 2 , 1 0

Lδ

τ

δδ

ττ

.

(3) When r =3，if change all of r of question (2) into 3,then it can be proved. ▌

§§§§2 Shape of non-linear figures

In preferred n-dimensional system，directviewing figures of non-linear generic curved surfaces

are not all in the shape of curved surface with zero thickness, they are always in the shape of

directviewing solid. So do directviewing figures of generic curves (non-linear)，they are not all in

the shape of curve.

For example，in example 2 of the third chapter，no matter what oblique axes transformation we

use, we can't change generic sphere

into the shape of directviewing sphere，it is

always in the shape of global solid. In fact，they are traces of difficient-order generic

points，so they can be called difficient-order

generic sphere. Other generic curved surfaces

also have sphere cases. For this kind of generic

curved surfaces，we can call it solid (actually it

isn't solid) generic curved surface(the above

difficient-order generic spheres can all be

called solid generic sphere) ， in order to

distinguish from another kind of generic

curved surface which is hollow generic surface

from direct view,such as

2

323121

2

4

2

3

2

2

2

1 4222224233 Rxxxxxxxxxx =−−−+++ . figure 30

Let's look at figure 30，the cut trace of the hollow generic sphere cut by generic coordinate

surface is a elliptic hollow generic curve，because the principal overlapping direction of the

four-dimensional system is 321 2 eee ++ ，so a random point of the generic curve represents a

straight line which is parallel to the principal overlapping direction.

In fact，the generic curve is a cylindrical surface of three-dimensional space,its directrix is a

circle and its generating line is parallel to vectore 321 2 eee ++ . And because the hollow

generic sphere can be regarded to be composed of a series of cut traces aquired by generic

coordinate surface translating along up and down directions of upright axis，from the figure it is

observed that, every cut trace is a elliptic generic curve，and every generic curve represents a

cylindrical surface of three-dimensional space which is parallel to the generating lines. The hollow

2

2

4

2

3

2

2

2

1 Rxxxx =+++

57

generic sphere can become "hollow"，it is because of the existence of these generating lines.

Under the nonsingular linear transform

=′

+=′

+−−=′

++−=′

2

1

2

12

1

2

1

2

12

1

2

1

2

1

34

413

4212

4211

xx

xxx

xxxx

xxxx

equation 2

323121

2

4

2

3

2

2

2

1 4222224233 Rxxxxxxxxxx =′′−′′−′′−′+′+′+′

can be changed into 2

2

4

2

3

2

2

2

1 Rxxxx =+++ ,

this is just the generic cylindrical surface of example 5 in the third chapter.And under the inverse

transform

′

′+′+′=

=

′−′=

′+′−′−=

4

3214

3

212

3211

2

1

2

1

2

1

2

1

2

12

1

2

1

2

1

x

xxxx

x

xxx

xxxx ，

the latter can be changed into the former.So，some(not all) solid generic curved surface can be

changed into hollow generic curved surface，but all hollow generic curved surface can be changed

into solid ones.

The cases of generic curve are such in essence(for example,the above generic directrix of

figure 30 is a hollow generic ellipse；The cut trace on generic coordinate surface of example 2 and

example 5 in the third chapter is solid generic circle). Since then if generic curved surface and

generic curve referred to are hollow if without special versions，otherwise they are referred to as

solid generic curved surface and solid generic curve.

Whether it is solid or not, both the determinant of generic curved surface and generic curve

are acceptable for dimension theorem. For example,because the directviewing dimension of the

above solid generic sphere is Z=3，its order is J=0，then its actual dimension is

W = Z + J = 3 + 0 = 3，

and the directviewing dimension of the above hollow generic sphere is Z = 2，its order is J = 1，then

its actual dimension is W = 2 + 1 = 3，it is obvious that their actual dimensions are identical. For

the convenience，sometimes for generic curved surfaces whose actual dimension are W = n -1

and generic curves whose actual dimensions are W = n -2 ，whether they are solid or not，we all

separately call them sufficient-order generic curved surfaces and sufficient-order generic curves.

58

Exercises

4．1 Please find out a preferred four-dimensional system and make generic plane

3x1 + 2x2 - 4x3 + 5x4 + 3 = 0

in the shape of plane in the system.

4．2 Find out a six-diemensional system and make generic plane

2x1 -3 x2 + 5 x3 + 5 x4 - 4 x5 - 2 x6 – 5 = 0

in the shape of plane in the system.

4．3 Find out a preferred six-dimensional system and make generic plane

04432 6543 =+−+− xxxx

in the shape of plane in the system.

4．4 The following generic straight lines suit convergent-oblique pattern or emanative-oblique

pattern preferred five-dimensional system?Why? Find out a suited preferred five-dimensional and

make generic straight line

=−++−−

=+−−−+

0446543

0232425

54321

54321

xxxxx

xxxxx

in the shape of line in the system.

4．5 Find out a suited six-dimensional system and make generic straight line

=+−+−−+

=+++−−+

01364643

02432354

654321

654321

xxxxxx

xxxxxx

in the shape of line.

4．6 Find out a preferred five-dimensional system and make

=−+−++

=+−+++

0242322

0334433

54321

54321

xxxxx

xxxxx

in the shape of straight line in the system.

4．7 Find out a preferred six-dimensional system which makes

=+++−++

=+−+−++

=+++−++

04332

0236422

02432

654321

654321

654321

xxxxxx

xxxxxx

xxxxxx

in the shape of point(Can this system be a convergent-oblique pattern?If it can be a

convergent-oblique pattern then it should make six-dimensional system be found a

convergent-oblique pattern).

59

The Fifth Chapter: Making of figures in preferred

n-dimensional system

Figures of preferred n-dimensional system can be made out by three kind of graphic methods

according to the analyticity of their equations and different needs.

In the third and fourth chapter，we have discussed questions of graphic rules (viz. three kind of

pictographies) and the shape of figures，and elementarily know some methods of making figures.

Here，we will discuss more about methods of making figures in preferred n-dimensional system. It

has three aspects of contents for concreteness：(1) cutting-trace method；(2) leading-axis method；

(3) synthetical pictography.

§§§§1 Cutting trace method

In three-dimensional space，a surface of revolution can be made by a curve rotating round a

coordinate axis.But in n-dimensional system，most punctiform figures are nonzero-order generic

points.Under the rotation，once generic point leaves its original position ，directions of most

nonzero vectors belonging to the generic point will change，which may make generic point lose the

possibiltiy of being punctiform. Thus，it will lose equivalence relation with the generic point on

its original position.So，in a general way, attempting to use rotation method to make figures is

impracticable in preferred n-dimensional system.

But，we can image preferred n-dimensional system to be composed of the translation trace of

generic coordinate surface along upright axis. And the projection of generic coordinate surface with

relation to upright axis is zero，so，coordinates of random figures of generic coordinate surface with

relation to upright axis are uniquely determined. That is to say，we can get a series of cut traces of

some figures by translating generic coordinate surface along upright axis and then make out these

figures.

Example 1: Assume there is a generic circular conical surface in four-dimensional space

2

4

2

3

2

2

2

1 xxxx =++ ，

please make out its figure.

Solution：Substitute x4 = x40 into the equation of generic circular conical surface，it gets

2

40

2

3

2

2

2

1 xxxx =++ ，

and make it simultaneous with x4 = x40，then the cut trace

obtained is a generic circle

=

=++

404

2

40

2

3

2

2

2

1

xx

xxxx

(when x40 = 0，the cut trace is a generic circle retracting into a

point)，then the generic circle is equivalent to a sphere of figure 31

three-dimensional space whose centre is on the origin and its radius is |x40| (figure 31). Choose two

points on the sphere

−− 40

4040 ,0,2

,2

xxx

A

60

and B ( 0，0，x40，x40 )

and order AB

then a preferred four-dimensional system is acquired，at the same time ,cut trace of the generic

circular conical surface at x4 = x40 is acquired.

When x4 changes from zero to positive and negative direction continuously，it will get the

trace of generic circular conical surface on a series of generic planes x4 = x4i

=

=++

i

i

xx

xxxx

44

2

4

2

3

2

2

2

1

(i =1 ， 2 ， …)，then the continuously

varying cut traces form the directviewing

figure of the generic circular conical

surface(figure 32).

In the equations of the generic circular

conical surface，order x4 = Ct of which，C=3×105(km/s)，t is time(unit：s=1 sec.)，then

it has

222

3

2

2

2

1 tCxxx =++

this is the equation of light cone[9]，we call it “Einstein light cone”.When x4≤0，every

point of axis x4 represents a black hole

translating to the origin figure 32

along axis Ct，and every generic circle represents that because of the great attractive effect of the

black hole near the black hole,light ray retracts to the centre of sphere in the shape of spherical

wave at the speed of three hundred thousand km per second. When x4 = 0，spherical wave retracts to

a point，it represents light ray has been inhaled into the black hole，there may be catastrophe at the

time，it may change into the white hole or other shiny

celestial body.Then，when x4≥0，every point of axis

x4 represents that the white hole or new celestial

body translates along the positive direction of axis

Ct with time going on，now each generic circle

represents that the spherical wave generated by the

white hole or new celestial body quickly expands at

the speed of enlarging the radius three hundred

thousand km per second.

Figures of other generic curved surface can be

made by this method. In fact ，for a curved surface，making out at least four or five its cut traces and

connecting them by smoothed curve is ok.

Both generic sphere figure 33

22

4

2

3

2

2

2

1 Rxxxx =+++

(refer to it in figure 15 of example 2 of the third chapter)of the third chapter and generic cylindrical

61

surface 22

3

2

2

2

1 Rxxx =++

(refer to it in figure 22 of example 5 of the third chapter) adopted this making method.

Figure 33 is a generic paraboloid of four-dimensional space

4

2

3

2

2

2

1 2xxxx =++

Figures in a preferred four-dimensional system are also made by the above methods.

§§§§2 Leading-axis method

Example 2: Please find out the directviewing figures of generic plane F

F：2x1 – 5x2 + 9x3 + 6x4 – 2x5 – 3x6 – 2x7 – 3 = 0

of seve n-dimensional space.

Solution：when assume the six posterior variables are all zero at the same time ,then it gets a

point of x1 (1.5，0，0，0，0，0，0)，

Considering it is advisable to put the figures being about to be found out of F in first quartile

of plane x1ox2，so，it is better to put the oblique axes of preferred seve n-dimensional in the second

or the fourth quartile. Then we can find out the following solution vectors

6214

5213

4212

3211

624

924

324

224

eeeOA

eeeOA

eeeOA

eeeOA

−+−=

−+−=

++−=

++−=

，

，

，

in the leadingout group

F′：2x1 – 5x2 + 9x3 + 6x4 – 2x5 – 3x6 – 2x7 = 0

(O is the origin)，order OA1A2A3A4

it gets the generic coordinate plane of preferred seve n-dimensional system which changed F′

into the shape of plane (because F′is the leadingout group of F，then figures of F are changed into

the shape of plane at the same time).Point (1.5，0，0，0，0，0，0) of axis x1 on the generic coordinate

plane is the trace of generic point A of F，find out the trace (4，1，0，0，0，0，0) of another point

of F on plane x1ox2 and mark out B ，connect AB ，then the directviewing figure of generic

straight line

=

=−−−++−

0

03326952

7

654321

0x

xxxxxxL ：　

(viz.the cut trace of F cut by generic coordinate plane) determined by intersection of generic plane

F and generic coordinate plane is finished.

Order x7 = 6，substitute it into the equation of F and make it simultaneous with the equation，then the cut trace

=

=−−−++−

6

015326952

7

654321

1x

xxxxxxL：　

of F cut by generic plane x7 = 6 is acquired. In L1，because generic plane

F11：2x1 -5x2 + 9x3+6x4-2x5-3x6-15=0

doesn't include variable x7，then it is parallel to axis x7 and vertical to generic coordinate plane，so

62

it is called projective generic plane of L1. Make the equation of F11 simultaneous with axis x7=0，then the projection

=

=−−−++−′

0

015326952

7

654321

1x

xxxxxxL：　

of L1 on generic coordinate plane is acquired. Choose two projection traces

Jc′(2.5，-2,0,0,0,0,0) and Jd′(5,-1,0,0,0,0,0)

of L1 on L1′，choose proper direction and unit length and find out the directviewing figures of x7，lead Jc′、Jd′to generic point C、D and make

(Jc，Jd are separately opposite traces of C and D )，connect AC，BD，CD，then the directviewing

figure of generic plane is finished(figure 34).

figure 34

It is observed from the example，under many cases we don't be anxious to find out the figure

of upright axis xn，but first find out the cut trace of the figure be made ,it is cut by generic

coordinate plane xn = 0 ;Then find out one or two projection trace of the figure's cut trace on plane

x1ox2 , the cut trace is cut by generic plane xn =xn0（xn0 ≠0）;Then in order to lead the

projection trace to the blank of the figure and and make the figures have pleasing shape and occupy

smaller chart size, choose proper direction and unit length and make out the figure of upright axis

xn. Under this meaning，upright axis xn can also be called leading axis，and the draughting can be

called leading-axis method accordingly.

§§§§3 Synthetical graphic method

Synthetical graphic method is to synthetize three kind of graphic methods. Using more than

two kinds of graphic methods for a figure of n-dimensional space belongs to synthetical graphic

method.

In example 4 of the fourth chapter，the projection trace of generic point B on generic straight

line ，：　 0223

074232

54321

54321

=++++

=+++++

xxxxx

xxxxxL

is far away from B ，if we solely use ordinary graphic method，it must make the chart size too

large，and then bring unnecessary trouble. So，we simultaneously used ordinary graphic method and

76eJJJJ ddcc =′=′

63

indirect method for L，separately mark out the trace of point A and point B ,the coordinates of

opposite trace(from the coordinates of Ja we know generic point A is on the generic coordinate

plane，refer to it in figure 26 of the fourth chapter). Let's look at the following example.

Example 3: assume in six-dimensional space

=+++−+−

=++−+−−

=++−+−−

0242 22

048443244

012 422

654321

654321

654321

xxxxxx

xxxxxx

xxxxxx

M：　

is regarded as difficient-order generic plane，find out its figure according to ordinary graphic

method and direct graphic method.

Solution：from linear algebra theory，we know the actual dimension of M is W=3，because

Z=2，then from dimension theorem we know the order of the difficient-order generic plane to be

found out is J=1，viz. it is regarded as the trace of one-order generic point. And one-order generic

point represents a straight line，so we can assume four one-order generic vertex

of the one-order generic plane are separately lines:

=−

=+

=+++−+−

=++−+−−

=++−+−−

=−

=+

=+++−+−

=++−+−−

=++−+−−

=

=+

=+++−+−

=++−+−−

=++−+−−

=

=+

=+++−+−

=++−+−−

=++−+−−

0 2

06

0242 22

048443244

012 422

0 2

016

0242 22

048443244

012 422

0

02

0242 22

048443244

012 422

0

012

0242 22

048443244

012 422

6

1

654321

654321

654321

4

6

1

654321

654321

654321

3

6

1

654321

654321

654321

2

6

1

654321

654321

654321

1

x

x

xxxxxx

xxxxxx

xxxxxx

A

x

x

xxxxxx

xxxxxx

xxxxxx

A

x

x

xxxxxx

xxxxxx

xxxxxx

A

x

x

xxxxxx

xxxxxx

xxxxxx

A

：　

：　

：　

：　

Regard variable coefficients of the four straight lines' equations as elements of the following

determinants，then，it gets direction vectors of the four straight lines

64

5432

654321

56152

100000

000001

211122

443244

142211

eeee

eeeee

−++=−−

−−−

−−−

=

e

S

Now consder compressing M along vector S (viz. is regarded as single principal overlapping

direction of preferred six-dimensional system)，and make the directviewing shape of M of host

system be in accord with that in the first generic plane of M

F1：x1 -x2 -2x3 +2x4 - 4x5 + x6 +12=0.

Therefor，find out points (-12,0,0,0,0,0)， (4，-14,15，0,0,0)， (-19，5，0，6，0，0) and (-21,11，0,0,-5,0) on F1 and make them have common generic，viz.

make use of transform

+−−

+−

−

−−

−

+−

215

214

213

215

214

213

1195

576

141615

1195

576

141615

eee

eee

eee

eee

eee

eee

○○○□□□

then a couple of preferred six-dimensional systems of

emanative-oblique pattern satisfying the two requests

are acquired. figure 35

Find out the directviewing figure A1A2A3A4 of generic plane F1 according to ordinary graphic

rules in host system，the figure is in accord with the

figure of M in direct view，then in fact it is regarded as

figure of difficient-order generic plane M (figure

35).In the figure，among the four difficient-order

generic vertexes , the traces of A1，A2 are separately

midpoint of M

A11 (-12,0,0,0,0,0) and A21 (-2,10,0,0,0,0)，

the opposite traces of A3 and A4 are separately points

of M

A31 (-16,-2,0,0,0,2)and A41 (-6,8,0,0,0,2)，

Separately mark the four points(trace or opposite

trace)into guest system，and then separately choose a figure 36

point on four straight lines denoted by the four difficient-order generic vertexes which are parallel

to each other

A12（－12，2，15，6，－5，0）；

A22（－2，12，15，6，－5，0）；

A32（－16，0，15，6，－5，2）；

A42（－6，10，15，6，－5，2）.

65

vertexes are also marked into guest system.

Because four three-dimensional subsystems

Ox1x2x3、Ｏx1x2x4、Ox1x2x5、Ox1x2x6

of the six-dimensional system we imagined are all right handed systems，then plane A12A22A32A42

should be on top of plane A11A21A31A41 ，when connect the eight vertexes in order to form

hexahedron to be soved，the edge line near the observer should be drawn in real line，the other three

edge lines behind the hexahedron should be drawn in broken lines(figure 36).

Example 4: Diagrammatize the following nonlinear programming questions by ordinary

graphic method and indirect graphic method：

Objective function： t =5x1+6x2+4x3+3x4-2x5

constraint condition：

≥

≤++++

≤++++

≤+++

≤++++

≤++++

0,,,,

3322234

3963 25

21 4 43

334 52

44536 2

54321

54321

54321

4321

54321

54321

xxxxx

xxxxx

xxxxx

xxxx

xxxxx

xxxxx

In order to introduce some graphic methods of points when the direction of oblique axis is

opposite，in this question, the coefficient of x5 is turned into negative number which is contrary to

routine.

Solution：：：：choose the following three points in figures (it is a generic plane,it is also called

objective generic plane) denoted by objective function

(4,-2,-2,0,0,0)，(6,-3,0,-4,0,0)，(4,-2,0,0,4,0)，

use transform

+−

−

−

215

214

213

244

364

242

eee

eee

eee □□□

then generic coordinate plane of a six-dimensional system is acquired（refer to it in figure 37）.

First regard every equation of constraint condition as equality，that is to say，regard them as

equations of generic plane（be called constraint generic plane）.

The cut traces acquired by intersection of every constraint generic plane and every

two-dimensional coordinate plane is called track lines in every two-dimensional coordinate palne.

For example，the trace line of generic plane (is called the third constraint generic plane,by

analogy,other generic plane can be called the first,the second and the fourth,the fifth constraint

generic palne)denoted by the third equation of constraint condition on plane x1ox2 is trough point A

(7,0,0,0,0,0) and point B (0,5.25,0,0,0,0)，a triangluar area is formed between AB and the original

point O(refer to it in figure 37)，traces(include their intersection points ) of other constraint generic

planes on plane x1ox2 are all on the outside of the area △OAB（to lessen confusion of figures，other trace lines are omitted）.

As will be readily seen，the triangle and its inner points satisfy all the constraint conditions，most points on the outside of the triangluar area don't satisfy any constraint condition，only few

66

figure 37

points can satisfy the constraint condition. So，we call the triangle and its inner area feasible

region，the area outside the triangle unfeasible region.

For preferred n-dimensional system of emanative-oblique pattern，it isn't difficult to find out

trace lines and of every constraint generic plane on every two-dimensional coordinate plane and

intersection points of these trace lines，But it isn't easy to find them out in n-dimensional system of

convergent-oblique pattern.

For example，on plane x1ox4，we easily found out the trace line of the third constraint generic

plane，the trace line is through point A and point C (0,0,0,5.25,0,0).Because traces of other generic

planes are all on the outside of feasible region，then they are omitted.

But for plane x1ox3，there is some difficult. In fact，the figure of axis x3 is also through the

origin，in order to distinguish them, just arrange them parallelly on a straight line which is through

the origin. So，we regard the figure of axis x4 as axis x3 for the moment: first make out the trace

point D(0,0,6.6,0,0,0) of the second constraint generic plane on axis x3，then make DD′to x4

through D ,and make the line segment throughing the origin and denoting endpoints of x3、x4

parallel and intersect axis x4 at D′. Originally，the position of D′is the position of point on axis

x4，but now it reverses the position of the host and the guest，it become a point of axis x3. The other

end of the trace line is point (33,0,0,0,0,0) of x1.

The trace line of the third constraint generic plane intersects axis x1 at point A，the other end is

point (0,0,21,0,0,0) of axis x3. The two trace lines intersect at point 0,0,0,7

39,0,

7

36L (figure

37).

On plane x2ox3，the second and the third constraint generic planes separately intersect axis x2 at

point B and point (0,16.5,0,0,0,0)，and they intersect axis x3 at point D and point (0,0,21,0,0,0) ，and the two trace lines intersect at point G(0,4,5,0,0,0)；On plane x2ox4，the trace line of the third

constraint generic plane separately intersect axis x2 and axis x4 at point B and point C；

On plane x2ox5，the trace line of the second constraint generic plane intersect axis x2 and axis

x5 at point (0，16.5,0,0,0,0), and point (0,0,0,0,8.25,0)，The trace of the third constraint generic

plane is through point B of axis x2 and parallel to axis x5，the two trace lines intersect at point E

(0,5.25,0,0,5.625,0).

67

Trace lines of every constraint generic planes on plane x3ox5 and plane x4ox5 and the

draughting of intersection point are the most difficult.

Originally，coordinates of points on positive direction of axis x5 will lessen values of objective

function，but considering the demand in futural nonlinear programming，it's necessary to introduce

graphic methods under this case.

Here，we have two means of resolution ：One is to make a plane rectangular coordinate

system additionally，take x3 as cross axis，take x5 as vertical axis，order x1＝x2＝x4＝0 in equations

of every constraint generic plane，then find out their trace lines and intersections in the plane

rectangular coordinate system；The other is borrowing-axis method，viz. take x1 or x2 as axis x3 or

axis x5 for the moment.

The anterior method is familiar to people，not only people who learn analytical geometry but

also those who learn descriptive geometry，they are all familiar with this method. So，let's

emphasize on borrowing-axis method：borrow the positive direction of axis x1 as axis x5，according to the solution of trace point D，axis x4 is again borrowed as axis x3.

figure 38

Put scales of axis x5 on top of axis x1，mark out trace points of every constraint generic plane

according to the new scales，connect every corresponding trace point and gain trace lines of every

constraint generic plane on plane x3ox5. Trace lines of the second and the fourth constraint generic

plane on plane x3ox5 separately intersect axis x3 at point D (0,0,6.6,0,0,0) and point

Q(0,0,39,0,0,0)，they intersect x5 at point T(0,0,0,0,8.25,0) and point U(0,0,0,0,6.5,0)，the two trace

lines DT and QU intersect at point

0,

13

81,0,

13

21,0,0R

(figure 38). Because the distances of some points are too small to see ，we magnify contents of the

circle in figure 38，for the convenience of calculating and drafting ，change axis x1 as axis x5，change scales of axis x3 into the same as those of axis x5（refer to it in figure 39a）.RS and RV are

line segments which represent coordinates of R with relation to axis x3 and axis x5. Traces of other

68

generic planes and intersections are all on the outside of

feasible region enclosed by quadrilateral ODRU，so they

are omitted.

Figure 39b introduced resetting method of these trace

points and intersections： because the trace points of

intersections solved in the question are near from each

other，it is difficult to describe the principle clearly ，then

choose a random point N as example to describe it. In the

figure, WN and PN separately represent coordinates of

point N with relation to axis x3 and axis x5. The resetting

process is equivalent to make the borrowed axis x1 revolve

aroundthe origin，during the revolve ，point P arrives at

point P′ along arc ⌒PP ′，point N arrives at N′along

arc ⌒NN ′ . The position of point N′is the last correct

position of N. figure 39

On plane x3ox4，we again use borrowing-axis method（borrow axis x1 as axis x3，the process is

omitted），trace lines of the first constraint generic plane separately intersect axis x3 and axis x4 at

point

0,0,

3

440,0,0,point and 0,0,0,

3

22,0,0 ，Trace lines of the second constraint generic

plane separately intersect the two axes at point D (0,0,6.6,0,0,0) and point (0,0,0,33,0,0)，Traces of

the third constraint generic plane separately intersect the two axes at point (0,0,21,0,0,0) and

point C (0,0,0,5.25,0,0).

The three trace lines DH、HI、CI intersect at point H、I . Of which，coordinates of point H are

；，　，　，　，　，　 009

22

9

5500 54321 ====== txxxxx

coordinates of point I are

.0021

82

21

11300 54321 ====== txxxxx ，　，　，　，　，　

Other trace lines and intersections are on the outside of feasible region enclosed by pentagon

ODHIC，so they are omitted.

On plane x4ox5，trace lines of the third constraint generic plane is straight lines which is though

point C of axis x4 and parallel to axis x5；Trace lines of the fourth and the fifth constraint generic

plane are separately connecting lines among point

(0，0，0，13，0，0), point (0，0，0，6.6，0，0)

of axis x4 and point (0，0，0，0，6.5，0), point (0，0，0，0，16.5，0)

of axis x5.The three trace lines intersect at point

J (0，0，0，5.25，3.375，0) and point K(0，0，0，5，4，0)

(trace lines and intersections of other constraint generic planes on plane x4ox5 are all on the

outside of feasible region，so they are omitted).

Under higher dimensions，methods in plane rectangular coordinate system can be supplement.

Figure 40 represents the trace lines' distribution of every constraint generic plane on plane x2ox3.

It is observed that，only trace lines BG ,DG of the second ,the third constraint generic plane and

their intersections are on the inside of feasible region，other trace lines and intersections are all on

the outside of feasible region.Then it is observed that，graphical method has great advantage，

69

considerable useless calculation can be dispose of by

graphical method.

Trace line S′of objective generic plane S on plane x1ox2 （refer to it in figure 41）is called objective trace line，it is the

cut trace of intersection of objective generic plane and plane

x1ox2.

Along with values of objective function increasing from

zero step by step，S intersects a series of planes parallel to

plane x1ox2，projection of its cut trace on plane x1ox2 is

equivalent to S′translating along the direction of 5e1+6e2

graduallly，during the translating it will intersect every vertex

of feasible region. figure 40

In posterior nine chapters，there are constant distance theorem and optimum point （substitute

a point of feasible region into objective function，t will get the maximum value，the point is called

figure 41

optimum point ）theorem：Under all oblique axis transforms which can make S into the shape of

plane，directviewing distance （directviewing distance is the distance that we can directly see with

our naked eyes，it isn't always the actual distance）of every vertex of feasible region to objective

trace line is invariable；The farthest vertex from directviewing distance to objective trace line is

optimum point to be solved. According to the two theorems，the directviewing distance between

vertex L to S′is the farthest（point G takes second place, point A takes third place），so it is the

optimum point to be solved. Substitute coordinates of L into objective function，it gets t=48，this is

the optimal solution to be solved.

Choose proper direction and unit length and make out the figure of upright axis t，lead point

L to point L′.Parallelogram MNPW is the figure of objective generic plane S（figure 41）.

Now，we use simplex method which hides slack variable and be introduced in the ninth

chapter to test the above results. (The contents is useful for people who are engaged in operational

research or mathematical statistics，people of other majors need not read it)：

Arrange constraint conditions into the following ：

70

( )1

22234330

63 25390

4 43210

4 52 330

536 2440

54321

54321

4321

54321

54321

−−−−−≤

−−−−−≤

−−−−≤

−−−−−≤

−−−−−≤

xxxxx

xxxxx

xxxx

xxxxx

xxxxx

In objective function，the coefficient of x2 is maximum，so，choose x2 as swapin variable,objective

function will has great increment . order x1= x3= x4= x5=0，it has

( )′

−≤

−≤

−≤

−≤

−≤

1

3330

2390

4210

2330

440

2

2

2

2

2

x

x

x

x

x

Now only choosing 4

21

3

33,

2

39,

4

21,

2

33,44Min 2 =

=x

can make formula (1)′be true，so the third equation of formula （1）is regarded as equation of the

greatest constraint force，then it is regarded as equality which hides slack variable，translate x2 to

the left end of the equality，it has

2224333

63 5392

4 3214

4 5 332

536244

54312

54312

4312

54312

54312

−−−−≤

−−−−≤

−−−=

−−−−≤

−−−−≤

xxxxx

xxxxx

xxxx

xxxxx

xxxxx

viz. ( )2

8457690

1227570

43214

829450

2082351550

5411

5431

4312

5431

5431

−+−−≤

−−−−≤

−−−=

−+−+≤

−−−−≤

xxxx

xxxx

xxxx

xxxx

xxxx

now a radical feasible solution is acquired（0，5.25，0，0，0，0）this is equivalent to coordinates

of point B in figure 37.

Substitute 43124

1

4

3

4

21xxxx −−−=

into objective function，it gets . 232

5

2

1

2

635431 xxxxt −−++=

In new objective function，the coefficient of x3 is a bit larger，and take x3 as swapin variable.

Order x1= x2= x4= x5=0 ，

it has ( )′

−≤

−≤

−≤

−≤

−≤

2

5 69 0

57 0

21 0

9 45 0

231550

2

3

3

3

3

x

x

x

x

x

71

Now only choosing ,55

69,57,21,

9

45,

23

155Min 3 =

=x

can make formula (2)′be true，so the second equation of formula（2）is regarded as equation of the

greatest constraint force. Regard formula（2）as an equality which hides slack variable，and translate

9x3 to the left end of the equality，it has

8 4 769 5

122 757

4 321 4

8 2 45 9

208515523

5413

5413

4132

5413

5413

−+−≤

−−−≤

−−=+

−++=

−−−≤

xxxx

xxxx

xxxx

xxxx

xxxx

viz. ( )3

1613341980

255161170

419147218

82459

259341800

541

541

5412

5413

541

−+−≤

−−−≤

+−−=

−++=

+−−≤

xxx

xxx

xxxx

xxxx

xxx

then a new radical feasible solution is acquired （0，4，5，0，0，0），it is equivalent to coordinates

of G in figure 37.

Substitute 54139

8

9

2

9

15 xxxx −++=

into the equation of the new objective function，then another new objective function

, 9

38

9

22

9

744 541 xxxt −−+=

In the new equation，the coefficient of x1 is the largest. So，choosing x1 as swapin variable

will make objective function value increase more quickly.Order x2=x3=x4=x5=0 ，it has

683960

644680

281440

45 0

683600

1

1

1

1

1

−≤

−≤

−≤

+≤

−≤

x

x

x

x

Now only choosing 7

36

28

144

68

396,

64

468,

28

144,,

68

360Min 1 ==

−=x

can make the above formulas be true at the same time. So，the third equation of formual（3）again

is regarded as an equality which hides slack variable. Take x2 as swapout variable and change it

with x1，it has

−+≤

−−≤

−+−=

−+=−

+−≤

541

541

2541

5413

541

161319834

25511716

184197214

82459

25918034

xxx

xxx

xxxx

xxxx

xxx

72

viz. ,

6123608283240

2884142344860

184197214

181089702126

612108180720

254

254

2541

2543

254

+−+≤

+−+≤

−+−=

−−+=

+−−≤

xxx

xxx

xxxx

xxxx

xxx

viz. ( )4

17102390

1623132707

9

7

2

14

19

7

367

1

7

6

14

1

7

39

173520

254

254

2541

2543

254

+−+≤

+−+≤

−+−=

+−+=

+−−≤

xxx

xxx

xxxx

xxxx

xxx

then another radical feasible solution is acquired

0 , 0 ,

7

39 , 0 ,

7

36 ，

This is equivalent to coordinates of point L in figure 37. Substitute formula（4）into the posterior

objective function，it gets . 42

748 542 xxxt −−−=

Coordinates of all variables of the new objective function are all less than zero，according to

optimal solution theorem of simplex method introduced in the ninth chapter，it is the optimal

solution to be solved，viz. when

0 , 0 , 7

39 , 0 ,

7

3654321 ===== xxxxx ，

t will take the maximum value 48，it is in accord with the graphic result of figure 41.

73

The Sixth Chapter: Interleaving and distance between two

linear figures.

In the chapter，we begin to discuss ubiety among linear figures.

The ubiety among linear figures is determined by interleaving，distance and included angle. Of

which the third factor，viz.included angle，because its contents are too much，we will discuss it in

the seventh and the eighth chapter.

§§§§1 Interleaving between two linear figures

Assume equations of two linear figures A and B separately are

A： 01

=+∑=

axa ji

n

iji

(j=1,2,…,r1)

B： 01

=+∑=

bxb ki

n

iki

(k=1,2,…,r2)

and assume the rank of coefficients matrixes of A and B are separately r1 , r2，the rank of

coefficients matrixes of simultaneous group(viz. the equation of A∩B)

( )

( )

==+

==+

∑∑

=

=

2

1

1

1

,,2,1 0

,,2,1 0

rk

rj

M

bxb

axa

ki

n

iki

ji

n

iji

L

L

：　

is r，the rank of augmented matrix is R. Investigate the relationship between r1，r2，r and R,it is any

more than the following four kinds：

(1) r1+r2≥r；

(2) r1 , r2≤r;

(3) r1, r2, r≥1, R≤n+1；

(4) r≤R, viz. R-r=1 or R-r = 0.

So，A and B have the following kind of ubiety：

(1)coincident .

(2)parallel.

(3)having common point but not coincident，again it can be divided into four kinds ：

1) Intersect in a w-dimensional (1≤W≤n-2) linear figure.

2) Both A and B are straight lines and intersect at a point.

3) A and B intersect at a point，but one of A and B is a straight line，the other is a

sufficient-order of deficient-order generic plane，we call A and B singly through. The intersection

point is called through point.

4) A and B intersect at a point，but both A and B are deficient-order(the orders are not always

equal，include zero order) generic plane，under proper oblique axes transform, A can be changed

into the shape of line, B can be changed into the shape of plane，A is through B from direct vies；And under another proper transform,B can be changed into the shape of line and A is changed into

the shape of plane, B is through A from direct view.We call A and B mutually through，intersection

74

point is called through point.

(4) A and B have no common point and aren't parallel，we call A and B mutually interleave.

The above four cases can be generalized into two kinds，of which，coincident can be regarded

as a particular case of intersectant，parallel can be regarded as a particular case of mutually

interleaving .

Theorem 1 When R=r=r1≥r2 or R=r=r2≥r1 ，

A coincides with B.

Proof： Because R=r，then M has solution，it represents that A and B have common point.

and r=r1≥r2，

then M=A∩B=A，

now AA ⊆ and BA ⊆ ,

coordinates of all points on A satisfy equations of B, so，all points on A are on B at the same time，viz. A coincides with B.

It can be proved by the same token，when R=r= r2≥r1，A also oincides with B. Particularly，

when R=r= r1= r2， A=B . ▌

Theorem 2: when R＞r= r1≥r2 or R＞r= r2≥r1，A is parallel to B.

Proof ： because R＞r，then M has no solution，it represents that A and B have no common

point.

But because r= r1≥r2，

then M'=(A'∩B')=A'

(A'，B'，M' are separately leadingout groups of equations A, B,M ), now

'' AA ⊆ ， '' BA ⊆

so from theorem1，A'coincides with B'. In geometry，the coincidence of two figures are regarded as

a particular case of parallel of two figures，viz. if two figures are coincident then they must be

parallel，so，A'∥B'. But A'，B'

are separately equivalent to be figures of A and B translating to the origin，so

A'∥A， B'∥B

then it has A∥B.

It can be proved by the same token，when r=r2≥r1 ，it also has A∥B. ▌

Theorem 3: when R=r＞r1，r2 ，

it has the following four cases：

(1) r≤n-1, A and B intersect at a W-dimensional (1≤W≤n-2) linear figure；

(2) when r=n, r1=r2=n-1，A and B are two straight lines intersecting at a point；

(3) r =n, r1=n-1, r2＜n-1 or r2=n-1, r1＜n-1，A and B singly through；

(4) r =n, 1＜r1, r2＜n-1 ，A and B mutually through.

Proof：because R=r，

then M has solution，and r＞r1 , r2，

it represents that A and B have common point but not coincident under the following four cases.

(1)because r≤n-1，

then the dimension of linear figure M as intersection of A and B, W≥1，and because

r1, r2≥1 and r＞r1，r2，

Then each equation of A and B at least includes an equation of generic plane，Viz. r≥2，then the

75

dimension of M is W≤n-2，so，all common points of A and B filled a W-dimensional (1≤W≤n-2)

linear figure.

(2) Because r1= r2=n-1，then both A and B are straight lines，and because r=n，then M has the

only group of solutions，it represents A and B have the only common point，the solution of M is

coordinates of the common point.

(3) If r1=n-1，r2＜n-1，

then A is a straight line, B is a sufficient-order or deficient-order generic plane under proper oblique

axis transform.Contrarily，if

r2=n-1, r1＜n-1，

then B is a straight line，A is a sufficient-order or deficient-order generic plane under proper

transform. And because r=n，then A and B have the only common point. When r1＞r2，A runs

through B, When r2＞r1，B runs through A. The solution of M is through point coordinate.

(4) Because r1，r2＜n-1，

Then bothA and B are not straight lines,and because r1，r2＞1 ,then both A and B aren't

sufficient-order generic plane(they aren't n-4-dimensional generic solid, yet).But because r=n，then

A and B have the only common point. When change A into the shape of line and change B into the

shape of plane under proper oblique axis transform，A runs through B from direct view；But when

change B into the shape of line and change A into the shape of plane under another proper

transform, B runs through A from direct vies. Then A and B run mutually through，the solution of M

is through point coordinate.

Now let's prove that，the proper oblique axis transform of changing one of A and B into the

shape of line and changing the other into the shape of plane is exist.

Because the actual dimension of A、B can be separately marked as

W1=n- r1， W2=n- r2， Assume we will change A into the shape of line and change B into the shape of plane，viz.

assume Z1=1，Z2=2 (Z1，Z2 is separately the directviewing dimension of A, B)，then from dimension

theorem，the order of A、B is separately

J1=W1-1=n- r1-1， J2=W2-2=n- r2-2，

Because J1＋J2=n-3+n -( r1+ r2)，

then，when r1+ r2=n，J1＋J2=n-3，on A, B，except the common point M，separately choose another

J1=p，J2=s points whose coordinates are the same with those of M with relation to xn (or the same

with relation to any other coordinate axis)

M11，…，M1p and M21，…，M2s，

and make vectors MM11，…，MM1p, MM21，… , MM2s

linearly independent and satisfy conditions of theorem 2 of the third chapter. When r1+ r2＞n，

J1＋J2＜n-3，

we can choose another m= r1+ r2-n points M1，…，Mm except A,B and make them satisfy the above

conditions along with the above points. Order MM11…M1pM21…M2s(when r1+r2=n )

or MM11…M1pM21…M2sM1…Mm(when r1+r2>n )

Viz. A can be changed into the shape of line and B can be changed into the shape of plane.

76

It can be proved by the same token，proper transform of changing B into the shape of line and

changing A into the shape of plane is exist. ▌

Theorem 4: when R＞r＞r1,r2，A and B are mutually interleaving.

Proof：because R＞r，then M has no solution，it represents A and B have no common point.

And r＞r1 ，r2, then A and B aren't parallel. So A and B are mutually interleaving. ▌

Example 1: Try to judge the interleaving of the two planes in four-dimensional space

=++++

=++++

0222

0322

4321

4321

1xxxx

xxxxI ：　

=+−+

=−−+

013

04322

421

321

2xxx

xxxI ：　

Solution： Because R=r=4,r1=r2=2, then we know the two planes are mutually through from

question (4) of theorem 3. The through point to be solved is

−−

19

6

19

26

19

9

19

10,,, First order I1 be changed into the shape of line，I2 keep the shape of plane，so J1=1，J2=0.Choose

a point

−

19

6

19

83

19

47

19

28,,,

on I1,choose a vector 2e1+2 e2-3 e3 through the point and the

through point，and order

3 e3□2 e1+2 e2，

Viz. make I1 run through I2〔refer to it in figure 42，the quadrilateral

of the figure represents I2,the through point is located at

intersection point of the two diagonals of the quadrangle, its

coordinates of the left two vertexes are separately（-2,1,-2,0）

and

−−

36

17,

18

19,

3

5,

4

5, figure 42

the coordinates of the two adjacent vertexes are known，coordinates of the other two vertexes are easily to be

known〕;But，because the rank of coefficients matrix of the

anterior two columns of I2 is 1，the rank of coefficients matrix of

the anterior three columns is 2，then another point which can

change I2 into the shape of line and satisfy theorem 2 of the

third chapter isn't exist except the through point，from theorem 2

of the fourth chapter，we choose axis x2 as upright axis，change

x4 into vertical axis，except the through

point，choose another point on I2 figure 43

−−

19

9

19

20

19

9

19

1,,,

77

choose a vector 3 e1+ e4+2 e3

between the two points and use transform

2 e3□-3 e1- e4，

then a preferred four-dimensional system of keeping I1 in the shape of plane、change I2 into the

shape of line，making I2 run through I1 is acquired〔refer to in figure 43，quadrangle represents I1，the left two vertexes are separately (-3.5，-2.5，0，6.5) and (-3.5，-2.5，1，4.5)〕.

§§§§2 Orthodromic space and normal space

We know，equations of linear figures can be divided into (they are system of linear equations)

the homogeneous and the nonhomogeneous. In linear algebra theory，figures denoted by system of

homogeneous linear equations can be called subspace，the relations between them can be easily

solved. But because figures denoted by system of non-homogeneous linear equations aren't

subspace，relations between them or relations between them and figures denoted by system of

homogeneous linear equations are not easy to solve. So，system of linear equations need to be given

more geometric concepts in order to offer people convenience for further study .

Still assume equations of two linear figures A and B are consistent system of linear equations

A： 01

=+∑=

axa ji

n

iji

(j=1,2,…,r1)

B： 01

=+∑=

bxb ki

n

iki

(k=1,2,…,r2)

and assume the ranks of coefficient matrixes of A and B separately are r1，r2，the rank of

coefficient matrixes of simultaneous group

( )

( )

==+

==+

∑∑

=

=

2

1

1

1

,,2,1 0

,,2,1 0

rk

rj

M

bxb

axa

ki

n

iki

ji

n

iji

L

L

：　

is r. And assume α,βare solution vectors of system of non-homogeneous linear equations A，or α,β∈A,but apparently，α-β=γ∉A.

And in the derivative group of A

A′： 01

=∑=

xa i

n

iji

(j=1 ，…，r1) ，

it has γ∈A′，because end points of α,βare on A，and γ is parallel to the line which runs

through end points of αandβ (refer to it in figure 44)，so，for the convenience of description ，we give the following defination：

Defination 1: A random solution vector of the derivative group

A′： 01

=∑=

xa i

n

iji

(j=1 ，…，r1)

of system of non-homogeneous linear equations A is called a orthodromic direction vector of A( be

called orthodromic vector for short). All orthodromic vectors of A formed solution space of A′，

78

so，we call A′orthodromic direction space of A(be called orthodromic space for short).

Defination 2: Assume

A′： 01

=∑=

xa i

n

iji

(j=1 ,…，r1)

B′： 01

=∑=

xb i

n

iki

(k=1，…，r2)

separately are orthodromic spaces of A and B ，then we call M′=A′∩B′，Viz.

( )

( )

==

==′

∑∑

=

=

2

1

1

1

21 0

21 0

r,,,k

r,,,j

M

xb

xa

i

n

iki

i

n

iji

L

L

：　

is common orthodromic space of A and B (be called

common-orthodromic space for short).

We have known that，so-called sum of subspaces is all

gatherings Q1＋Q2

of α1＋α2 （α1∈Q1，α2∈Q2）of two random subspaces Q1，Q2，of linear space Vn.

If every decompose formula α=α1＋α2 ofα∈Q1＋Q2

is unique，then Q1＋Q2 is direct sum，marked as figure 44 Q1

•

+ Q2 .

Sum of subspace or direct sum is still subspace.

In linear algebra theory，if Vn is n-dimensional euclidean space，and

Vn =Q1

•

+ Q2,

then we call Q1，Q2 orthogonal complement of each other，marked as

Q1=⊥2Q ，Q2=

⊥1Q .

Because all linear combinations among normal vectors of A also form subspace，and the

subspace and orthodromic space of A are orthogonal complement of each other，so we give the

following：

Defination 3: Assume Vn is euclidean space，A′is orthodromic space of linear figure A of

Vn. If ⊥

•

′+′= AAVn ，

then we call ⊥′

1A normal direction space of A(be called normal space). Especially，for normal

space M′of common-orthodromic space of A and B⊥⊥

(when M has solution, ⊥′

1M is also normal

space of M)，we call it common-orthodromic normal space of A and B.

Assume a group of vectors α1，α2，…，αn- r1 of A are linearly independent

and all their linear combinations form orthodromic space of A，of which

79

{ } ( )1 21 21 rn,,,,,, n −== LL ταααα ττττ ,

then the equation of ⊥′

1A is

( )1

1

,,1 0 rnxAn

i

ii −==′ ∑=

⊥Lτατ：　 .

Defination 4: Assume

( )1

1

,,1 0 rnxAn

i

ii −==′ ∑=

⊥Lτατ：　 .

( )2

1

,,1 0 rnxBn

i

ii −==′ ∑=

⊥Lδβδ：　

are separatly normal space of A and B, then we call T=⊥⊥ ′′

11 BA I ，

viz.

( )

( )

−==

−==

∑∑

=

=

2

1

1

1

,,1 0

,,1 0

rnx

rnx

Tn

i

ii

n

i

ii

L

L

δβ

τα

δ

τ

：　

is common normal space of A and B ( be called common-normal space for short).

Because the dimension of linear subspace is n-r(r is the rank of coefficient matrix of equation

of the subspace)，so，from defination 2 it canimmediately get：

Theorem 5: Random two linear figures A and B have n-r-dimensional common-orthodromic

space (r is the rank of coefficient matrix of simultaneous group). ▌

From defination 3 it canimmediately get：

Theorem 6: Random two linear figures A and B have r-dimensional common-orthodromic

normal space (r is the rank of coefficient matrix of simultaneous group).

Theorem 7: Random two linear figuresA and B have r1＋r2-r-dimensional common-normal

space(r1，r2 are separately the rank of coefficient matrix of system of equations A and B,r is the rank

of coefficient matrix of simultaneous group).

Proof：From defination 3，a random normal vector of normal spaceis always a orthodromic

vector of a linear figure，so⊥′

1A and ⊥′

1B separately have n- r1 and n- r2 normal vectors. Of

which，n-r vectors of common-orthodromic space M′of A and B are common-normal vectors of

⊥′1A and

⊥′1B ，so，in all （n-r1）＋（n- r2）normal vectors of

⊥′1A and

⊥′1B ，n-r can be linearly

represented by other normal vectors， so the rank of coefficient matrix of equations of

ommon-normal space T is (n-r1)＋(n- r2)-(n-r)=n-r1-r2+r,

so，the dimension of T is ＷＴ =n-(n- r1- r2+r)= r1+ r2-r. ▌

For example，when A and B are coincident or parallel，if r1＞r2，then the orthodromic space

A′of A is common-orthodromic space of A and B, the normal space ⊥′

1A of A is

80

common-orthodromic normal space of A and B, and the normal space ⊥′

1B of B is common-normal

space of A and B.

Proof：From theorm 1、theorem 2，it has r= r1＞r2，so M′=A′∩B′=A′，

viz. A′ is common-orthodromic space of A and B, and ⊥′

1A is common-orthodromic normal space

of A and B；From theorem 7，the dimension of common-normal space T is

r1+r2-r=r2，

so it has T=⊥⊥ ′′

11 BA I =⊥′

1B ，

viz. ⊥′

1B is common-normal space of A and B. ▌

And for example，when A and B are mutually interleaving， the dimension of their

common-orthodromic space is

0≤W≤n-3，

the dimension of common-orthodromic normal space is 3≤r≤n，

the dimension of common-normal space is 1≤WＴ ≤n-2.

Proof ： Assume r1≥r2，from theorem 4 , it has R＞r ,and r＞r1,r2，so we know by deduction

that r1≤n-1， r2≥2

(if r1 n-1, r2 2，for example r1=n or r2=1，then it has

r= r1 or R=r= r1+ r2，

this is in contradiction with the conditions of theorem 4).When r1= r2=2，it has R=4， r=3，so it has

W=n-3；

When r1= r2=n-1，it may has R=n+1，r=n，then it has W=0，

so it has 0≤W≤n-3.

and，W=n-r，sowe can know by deduction that the dimension of common-orthodromic normal space

⊥′1M is 3≤r≤n. Then from theorem 7，the dimension of common-normal space is r1+ r2-r，when

r1= r2=2, r=3，

then it has r1+ r2-r=1;

and when r1= r2=n-1,r=n，so it has

r1+ r2-r=n-2.

Then we know 1≤WＴ ≤n-2. ▌

When A intersects with B ，from theorem 3,there are four cases：

(1) it has 1≤W≤n-2-dimensional common-orthodromic space，2≤r≤n-1-dimensional

common-orthodromic normal space and 0≤WＴ ≤n-3-diemensional common-normal space.

All the following three cases have zero-dimensional common-orthodromic space and

n-dimensional common-orthodromic normal space，but the dimensions of common-normal space

are not the same，of which

(2) When two straight lines interleave, there is n-2-dimensional common-normal space.

(3) When A and B are singly through，0≤WＴ ≤n-3.

(4) When A and B are mutually through，0≤WＴ ≤n-4.

81

Under the above four cases，the dimension of common-orthodromic space has been given in

theorem 3，because the rank of common-orthodromic normal space is n-r, then we can know by

deduction that its dimension is r.

So difference of the dimension of its common-normal space is not discussed，readers can try to

write out the proof process by yourselves.

§§§§3 Exterior sumof two linear figures To study the distance between two linear figures，we need extend the concepts of sum of

subspace .

3.1 Concepts of exterior sumof two linear figures

From structural theory of solutions of system of linear equations，assume r0 is a particular

solution vector of system of non-homogeneous linear equations A,r is a random solution vector of

A，when we take all solution vectors ηof derivative group A′, then

r=r0+η

will take all solution vectors of A. Because for every r, r0 is the same，then the decompose formula is unique. So when A, A′are

separately regarded as gatherings of all solution vectors of A and A′，｛r0｝is regarded as much

multiplicity gathering including countless same elements r0（r0 is A particular solution vector of

A）,viz. { } } {

boundless

0000

48476

L rrrr ，，　，　= ，

then all gathering of r=r0+η (r∈A, r0∈｛r0｝,η∈A′) can be written into the form which likes

direct sum { } ArA ′+=•

0 (1)

it is called exterior direct sum of A′and gathering｛r0｝.Because 0∉A，then the exterior direct sum

isn't subspace，it doesn't close for addition and amount multiplication of linear space Vn.

From the above analysis we know，relation between linear figure and its orthodromic space is

the relation between exterior direct sum of subspace and the subspace. Then，we extend and apply

concepts and property of subspace to linear figures denoted by system of non-homogeneous linear

equations，then we can learn and study more of their geometric property.

Definition 5 ： Assume equations of two linear figures A and B are system of

non-homogeneous linear equations，A∩B isn't empty，then

A+B=S

is called the exterior sumof linear figures A and B . Of which，S is also a system of

non-homogeneous linear equations.

Assume S′=A′+B′is orthodromic space of S(equation of S is non homogeneous，A′and B′are separately orthodromic spaces of A and B)，when assume α′∈A′，β′∈B′,then δ′=α′+β′∈S′.

assumeδ∈S ，andδ0 is a particular solution vector of S，｛δ0｝is a much multiplicity gathering

including countless identicalδ0 .

Because for every solution vectorδ of S，decompose formula δ=δ0＋δ′

is unique，then from formula (1)，for allδ∈S it has

{ } SS ′+=•

0δ

82

further it has { } ( )BAS ′+′+=•

0δ

viz. { } ( )BABA ′+′+=+•

0δ (2)

This is the formula of relation between exterior sumof A and B and sum of their orthodromic

space . If assume α∈A, β∈B ,then the corresponding decompose formula can be marked as α+β=δ0＋α′+β′.

3.2 Property of outer sum Now，we regard δ0 as a particular solution vector of A( it is called particular solution for

short)，when β′=0 ，from decompose formula

r=r0+η ，

it has α+β=α∈A;

when δ0+α′∈A∩B

it has α+β=β∈B；

But ，when BAI∉′+≠′′ αδβα 0 and 0, ，it has α+β∉A ， andα+β∉B，

But it also has α+β=δ∈S,

viz. α+β∈A+B .

So， A ⊆ A+B， B ⊆ A+B ，

and A∩B ⊆ A+B.

If regardδ0 as a particular solution of B，result is the same. So，as long asδ0 is a particular solution

of one of system of non-homogeneous linear equations A or B ，then it is also a particular solution

of A+B at the same time. Of courseδ0 can be neither a particular solution of A nor a particular

solution of B, but it is still a particular solution of exterior sumA+B .

Exterior sumdoesn't close for addition and amount multiplication. At the same time，exterior

sumis not fit for commutative laws outside of A′+B′ ，viz.

{ } ( ) { }( ) { }( )

{ } ( ) { } ( )ABBA

ABBABA

′+′+=′+′+

+′+′≠′++′≠′+′+••

•••

00

000

but

δδ

δδδ

So，braces of formula (2) can' t be got rid of easily.

Besides，exterior sumstill has the following property(all the following discussion has

assumed that A∩B is not empty)：

Theorem 8: Assume one of linear figures A and B is subspace，the other is not subspace A′and B′ are separately orthodromic space of them，then it has

A+B=A+B and A+B=A′+B′,

viz. A+B=A′+B′.

Proof：assume equation of A is non homogeneous, equation of B is homogeneous, viz . B=B′

and assumeα∈A, β∈B , apparently,β=β′∈B′.

And assumeδ0 is a particular solution vector of A，then from decompose formula

r=r0＋η

and α+β=δ0＋α′+β′

we know α=δ0＋α′(α∈A,α′∈A′)

83

viz. α+β=α+β′=α+β

so it has A+B=A+B

and assumeβ0 ∈A∩B , thenβ0∈A , then from decompose formulaβ0=δ0+α0′，at the same

timeβ0∈B ，viz. β0∈B′, assume α0=-α0′, apparentlyα0∈A， so，δ0 is regarded as linear

combination ofα0∈A′and β0∈B ′,

viz. δ0=α0＋β0，

so it has α+β=δ0＋α′+β′=α0＋α′+β0＋β′

but α0＋α′∈A′，β0＋β′∈B′, then it has

A+B=A′+B′，

so from transitive law，it has A+B=A′+B′ ▌

Theorem 9: Assume both equations of linear figures A and B are non homogeneous，A′and

B′ are separately orthodromic spaces of them，if A′∩B or B′∩A is not empty，thenit has

A+B=A′+B′.

Proof： if A′∩B is not empty then B′∩A is also not empty，so it has

A+B=A′+B=A+B′，

but from theorem 8, it has

A′+B=A′+B=A′+B′，

A+B′=A+B′=A′+B′，

then A+B=A′+B′ ▌

3.3 Equation of outer sum From concept of exterior sumwe know，orthodromic space of exterior sumof linear figures is

sum of orthodromic space of linear figures. And from defination 3，we know linear figures have the

same normal space with their orthodromic space，and normal space has nonnegligible

contribution to determine equations of linear figures.So we give ： Theorem 10: Assume both Q1 and Q2 are subspace of linear space Vn,⊥

1Q and⊥2Q are

separately their orthogonal complement，thenit has

( ) ⊥⊥⊥=+ 2121 QQQQ I (3)

( )⊥⊥⊥ =+ 2121 QQQQ I (4)

are called De Morgan law of intersection and sum of subspace. Proof：similar to De Morgan law of intersection and union of gathering，assume

u∈(Q1＋Q2)⊥，

then u∉Q1＋Q2，

so u∉Q1，and u∉Q2，

it has u∈ ⊥1Q and u∈ ⊥

2Q ，

then u∈ ⊥1Q ∩ ⊥

2Q ，

so it has (Q1∩Q2)⊥⊆ ⊥

1Q ∩ ⊥2Q Contrarily，assume u∈ ⊥

1Q ∩ ⊥2Q ，

then u∈ ⊥1Q and u∈ ⊥

2Q ，

it has u∉Q1, and u∉Q2，

and because we have assumed

84

u∈ ⊥1Q ∩ ⊥

2Q ,

so in random decompose formula

u=α+β

of u it also hasα,β∉Q1, and α,β∉Q2，

then u∉Q1＋Q2，

so u∈(Q1＋Q2)⊥，

then it has ⊥

1Q ∩ ⊥2Q ⊆

( Q1＋Q2)⊥.

From the above we know (Q1∩Q2)⊥

= ⊥

1Q ∩ ⊥2Q It can be proved by the same token

⊥1Q +

⊥2Q =(Q1∩Q2)

⊥. ▌ Theorem 10 represents, that common-normal space of two linear figures is normal space of

exterior sumof the two linear figures. From formula (3), it has

Q1＋Q2= (Q⊥

1∩Q⊥

2)⊥

So in formula(2) ( )⊥⊥⊥ ′′=′+′ BABA I (5)

viz. { } ( )⊥⊥⊥ ′′+=+ BABA I。

0δ (6)

formula (6) can be regarded as general formula of equation of outer sum.From theorem 7，when

assume the ranks of A , B and coefficients matrix of equations of A∩B are separately r1, r2 and

r， ( )⊥⊥⊥ ′′ BA I of formula (6) are r1+ r2-r generic planes taking a group of linearly independent

solution vectors of common-normal space of A and B ，and coordinates of δ0∈{δ0} are

coordinates of a common point of these generic planes. When both equations of A and B are homogeneous , viz . A=A′，B=B′，thenδ0=0，it

represents these generic planes take the origin as a common point.So，formula (5) can be regarded

as a particular case of formula (6)，it is general equation of equation of sum of subspace.

§§§§4 Distance between two linear figures Apparently，when A and B are coincident or intersectant, the distance is zero.So we only discuss

the distance when A and B are parallel or mutually interleaving.

4.1 Distance between two parallel figures Assume A∥B ，and r1≥r2(r1, r2 are separately the ranks of coefficients matrix of equations of

A and B，the following is the same). Because ⊥′

1A is common-orthodromic normal space of A and

B ，then ⊥′

1A ⊥A， ⊥′

1A ⊥B.

order ⊥′

1A separately interleave with A and B ，then

M1= ⊥′

1A ∩A

is a point M1（x11, x21,…, xn1）;

85

and M0 =⊥′

1A ∩B

is a r1- r2-dimensional linear figure. Of which，when r1= r2 ，M0 is also a point M0 (x10 , x20 ,…, xn0). Now，the length of line segment M1 M0 is the distance between A and B ； When r1＞r2 ，make r1- r2 generic planes taking r1- r2 linearly independent orthodromic vectors of M0 as normal

vector and through point M1,at the same time, it intersects M0 at point M2 (x12 , x22 ,…, xn2)， Because M1，M2 are separately on A ，B ，and both M1，M2 are on ⊥′

1A ， then line segment

M1M2⊥A and M1M2⊥B，so，the length of line segment M1M2

( ) ( ) ( )xxxxxx nnMM 12

2

2122

2

1112

2

21 −−− +++= L

is the distance between A and B. Besides，because ⊥′

1B is common-normal space of A, B，then coordinates of point M2 can be

acquired by translating ⊥′

1B , making it through M1 and intersecting B，the result is the same.

4.2 Distance between two mutually interleaving figures Assume A and B are mutually interleaving，and r1≥r2，translate A ( be changed into A0 ) and

make it interleave B，then the exterior sum BAS += 0 of A0 and B is parallel to A , then we kan

find out the distance between A and S according to the solution of distance between two parallel

figures，this is the distance between two linear figures A and B . Example 2: Please find out the distance between straight line

=+−−+

=−+++

=+++−

=−++−

=+−++

052

04272

01322

03232

0223

4321

4321

4321

4321

4321

xxxx

xxxxB

xxxx

xxxx

xxxx

A

：　

：　

of four-dimensional space.

Solution：：：：Because r = 4, R =5 , then A and B are mutually interleaving.

And because A has a orthodromic vector

18e1+ e2-11 e3- e4

B has two orthodromic vectors 3e1-4e3+11e4 , 3e1-4e2- e4

So，equation of common-normal spaceT is

=−−

=+−

=−−+

04

011

01118

421

431

4321

xxx

xxx

xxxx

T：　

and one solution vector of T is

3 e1+2 e2+5 e3+ e4 ,

choose a point (9，－14，0，0)

of B , through the point and taking vector ｛3，2，5，1｝

as normal vector,and then make a generic plane

S：3 x1+2 x2+5 x3+ x4+1=0 ，

It is the equation of exterior sum SBA =+0

of A0 (A0 is the new generic plane formed of intersection of A s translation and B ) and B . Because

86

S∥A and 3＞1 (3 is the rank of coefficient matrix of A，1 is the rank of coefficient matrix of

S )，then 01118 4321 =−−+′⊥xxxxA ：　

is also the common-orthodromic normal space of A and S. And the intersectionsof ⊥′

1A and A, S

are separately

=−−+

=++++

=−−+

=+++−

=−++−

=+−++

01118

01523

01118

01322

03232

0223

4321

4321

0

4321

4321

4321

4321

1

xxxx

xxxxM

xxxx

xxxx

xxxx

xxxx

M

：　

：　

then M1 is a point , 447

1397,

447

278,

447

391

149

381

−− ，：　M

M0 is a plane. Taking two orthodromic vectors 2e2＋e3-9e4 of M0 and e1-7e2+11 e4

as normal vector, through point M1 , then make generic planes

2 x2+ x3-9 x4+27=0

and x1-7 x2+11 x4-28=0 ,

they intersect M0 at

=−+−

=+−+

=−−+

=++++

028117

02792

01118

01523

421

432

4321

4321

2

xxx

xxx

xxxx

xxxx

M ：　

at the same time the intersection point is

−−

5811

17863,

5811

5104,

5811

4487,

1937

7922M

Then | M1 M2|≈0.32

is the distance between straight line A and plane B . Coordinates of the above point M2 can be acquired by translating T through M1 and making

it intersect S (The result is the same with that of the above method).

Exercises 6.1 Please find out interleaving and distance between the two linear figures of four-dimensional

space

=−+++

=−+−+

=++−+

=++++

05223

030432

072255

053234

4321

4321

2

4321

4321

1

xxxx

xxxxL

xxxx

xxxxL

：　

：　

.

87

The Seventh Chapter: Included angle questions and their

linear solutions between two linear figures

Included angle questions have the most abundant and interesting contents of the position

relations among linear figures.

§§§§1 Diversity of included angle questions bertween two linear figures

of higher space

This question has two meanings，one is the nonuniqueness of included angle amounts between

two linear figures，the other is the nonuniqueness of solutions of included angle questions .

1.1 Nonuniqueness of included angle amounts

1.1.1 Two kind of projection methods

In the first chapter we have referred to：∀α∈Vn ， if ασα，then σ is a identical

transform.

When Vn is regard as a n-dimensional euclidean space (the following is the same)，then σ

can be regarded as projection(be referred to orthographic projection) of Vn to itself，

viz. σ(Vn)=Vn.

Contents of projection not only include image setσ(Vn) (is also called projection of (Vn ) and

image source set(is also called passive set, space been projected or figures been projected for

short)，but also include two contents of set (be called path set、projection direction set、projection

direction space) of projection direction(also be called projection path) and space of accepting or

containing source gatherings(be called contain set、acceptance set or acceptance space).Because of

the nonsingularity of identical transformation(or be called nondegenerate)，so，its projection

direction is a zero vector.

When σ is a (orthographic)projection of Vn to one of its subspace Vm (m＜n)，Viz. when σ(Vn)=Vm ，σis a singular linear transform. For example，when assume

ei∈Vn (i=1 ，2 ，… ，n)， e j∈Vm (j=1 ，2 ，… ，m；m＜n)，

the corresponding transform can be marked as

( )( )

( )( )

( )

=

=

=

=

=

+

0

0

1

22

11

n

m

mm e

e

e

e

e

e

e

e

σ

σ

σ

σ

σ

LL

LL

From singularity of σ(or be called degeneracy)，we can know by deduction that projection

directions (can be called degenerate direction)are nonzero vectors. All linear combinations among

projection direction also form subspace of Vn ，viz. projection direction space(also be called path

set).Because under σ,random projection direction is perpendicular(or only satisfy that inner

product is zero) to image (also be called projection of random element)of random element of Vn ，so，projection direction space is normal space

⊥mV of Vm . Then，the whole process of the

88

projection can be described as：

Image set of passive set ．．．．．．．．．．．．．．．．．．．．V．n． from path set ．．．．．．． ⊥mV to contain set ．．．．．．．．．．V．m． is ．．σ．(．V．n．)． Passive set．．．．．．．．．．、．path ．．．．

set．．．、．contain set．．．．．．．．．． and image set．．．．．．．． are four principal elements of forming projection，every ．．．．．element of ．．．．．．．．．the four should not be absent．．．．．．．．．．．．．．．．．．．．．．．．.

For the conveniece，we can apprehend projection concepts like this： σcan be decomposed into n-m unidirectional stagebystage projections，viz. assume σ′(Vn)=Vn-1，

then σ2′(Vn)=σ′〔σ′(Vn)〕=σ′(Vn –1 )= Vn -2，

then σ(Vn)= σn-m′ (Vn)= Vm.

All the above discussions can be regarded as projecion of Vn to one of its subspaces.The

following will discuss the second kind of projection：projection between two different subspaces of

Vn .

Assume nsnp VVVV ⊂⊂ 21 ， ， and Vp1≠Vs2 ， dimensions of Vp1 and Vs2 are separately p

and s，then projectionσ0 from Vp1 to Vs2 is involved in the first kind of projection σ，viz.

because σ(Vn)= Vs2 andσ0 (Vp1) ⊆ Vs2，

then σ0 (Vp1) ⊆σ(Vn) .

When we carry the first kind of projectionσ，we also carry the second projectionσ0 at the

same time，when the first kind is finished，the second kind is finished at the same time.

Assume Vp1 is sufficient-order or difficient-order (include zero-order)generic plane through the

origin and in the ordinary position，Vs2 is sufficient-order or difficient-order generic coordinate

plane，and assume the equation of Vs2 is

( )，　：　 1-2

0

0

2

1

2 ns

x

x

x

V

n

s

s

s ≤≤

=

=

+

+

LL

and Vp1 is all the linear combinations of linearly independent vectors

r1，r2，…，rp (2≤p≤n-1) ,

of which r i =｛ri1 ，ri2 ，…，rin ｝i=1 ，2 ，… ，p.

then the corresponding expressions of the orthographic projection from Vp1 to Vs2 is

( )( )

( )

，：　

2211

22221212

12121111

0

+++=

+++=

+++=

spsppp

ss

ss

rrrr

rrrr

rrrr

eee

eee

eee

L

LLLL

L

L

σ

σ

σ

σ

When we orderly carry unindirectional stagebystage projection taking directions parallel to

xn , xn－1 ，…，xs+1 as projection direction，coordinates of ri orderly become zero from back to front，column vectors of coefficient matrix

89

=

pnpp

n

n

rrr

rrr

rrr

r

L

LLLL

L

L

21

22221

11211

of ri fade away from back to front. By n-s stagebystage projections，the back n-s coordinates of ri

all become zero，at last，matrix r become coefficient matrix of σ0(r)

=′

pspp

s

s

rrr

rrr

rrr

r

L

LLLL

L

L

21

22221

11211

When s＞p，r′is row-nonsingular matrix( the rank is p)；When s=p ,r′is a p-order square

matrix

(the rank is still p).So，when s≥p，projection or ri in Vs2 is still linearly independent. But when

s＜p ，r′is column-nonsingular matrix (the rank is s)，image σ0(ri) of vector groupri also

become linearly dependent vector group, so, the dimension of imageσ0(Vp1) of Vp1 also drops to s .

1.1.2 Defination of included angle between two linear figures

In the following，let's first discuss a kind of very important projection—rightangle projection，it is the most important foundation of all included angle questions (include defination and solution

method,etc)between two linear figures of higher space.

Theorem 1：Assume is a plane of n-dimensional space， 1π is a coordinate plane， ⊥= 12 ππ is a n-4-dimensional generic coordinate coordinate plane which is orthogonal

complement with 1π of each other，H1 and H2 are two straight lines ofπ，1π , H21 and H12 , H22

are separately projections of H1，H2 onπ1 and 2π . If

H11⊥H21 and H12⊥H22

(or only satisfy that the inner product is zero)，then

H1⊥H2 .

Proof：Assume h1=｛α11，…，αn1｝, h2=｛α12，…，αn2｝

are separately orthodromic vectors of H1，H2，and assume equation ofπ1 is

=

=

=

0

0

0

4

3

1

nx

x

x

L

：　π

then equation of 2π is

=

=

0

0

2

1

2x

x：　π

so projections of h1，h2 on1π ， 2π are separately

h11 ={α11，α21，0…，0}， h21={α12，α22，0…，0}

h12={0，0，α31，…，αn1}，h22={0,0, α32，…，αn2}

If H11⊥H21，H12⊥H22

Then h11⊥h21 ，h12⊥h22

So α11α12＋α21α22=0

90

and α31α32＋…＋αn1αn2=0

then α11α12＋…＋αn1αn2=0

so h1⊥h2

viz. H1⊥H2. ▌

Theorem 1 it can also be called rightangle projection theorem，it is brought forward and

proved by Mr Jian Zhaoquan of Beijng Institute of Technology[13]，then it is called Jianshi solution

theorem in this book.

Projection process of theorem 1 is equivalent to interchange path gathering and contain

gathering，viz. first take 2π as path gathering，take 1π as contain gathering，and finish the

projection from πto 1π ；And then take 1π as path gathering ，take 2π as contain gathering，and

finish the projection fromπ to 2π contrarily.

For Jianshi solution theorem,we can extend it like this：

Theorem 2：assume Vm(m＜n) is subspace of euclidean space Vn，Vmo is a sufficient-order or

difficient-order generic coordinate plane which has the same dimension with Vm.

H1，H2，…，Hm

is m straight lines of Vm， H11，H21，…，Hm1

and H12，H22，…，Hm2

are separately projections of H1，H2，…，Hm

on Vmo and ⊥

moV . If H11，H21，…，Hm1

and H12，H22，…，Hm2

are perpendicular to each other(or only satisfy that the inner product is zero)，then

H1，H2，…，Hm

are perpendicular to each other.

Proof：：：：divide H1，H2，…，Hm in pairs into ( )

2

1−mm groups，then prove it according to

method of theorem1. ▌

But，the inverse of theorem1, theorem 2 don't exist.

Proof：because theorem 2 is the generalization of theorem 1，so it is ok to

prove that the inverse of theorem1 don't exist .

Assume H3，H4 are anther two straight lines of π which are perpendicular to each other，

h3=｛β11 ，…，βn1｝, h4=｛β12 ，…，βn2｝

are separately their orthodromic vectors. Because H3，H4 and H1，H2 have coplanarπ，then

both h3 and h4 can be linearly represented by h1，h2，viz.

h3=λ11 h1+λ12 h2, h4=λ21 h1+λ22 h2

(λ11，λ21，λ12，λ22 all nonzero numbers).So， β11=λ11α11+λ12α12 , …，βn1=λ11αn1+λ12αn2 β12=λ21α11+λ22α12, …，βn2=λ21αn1+λ22αn2.

because h3⊥h4，

and α11α12＋…＋αn1αn2=0，

then β11β12+…＋βn1βn2=λ11λ21(α211＋…＋α2

n1)+λ12λ22（α212＋…＋α2

n2）=0，

viz. λ11λ21(α211＋…＋α2

n1)= -λ12λ22（α212＋…＋α2

n2）.

And because projectons of h3 ，h4 on 1π are separately

h31=｛β11，β21，0，…,0｝， h41=｛β12，β22，0，…,0｝，

91

and α11α12＋α21α22=0，

if h31⊥h41，

then β11β12+β21β22=λ11λ21 (α211＋α2

21)+λ12λ22 (α212＋α2

22) =0,

viz. λ11λ21 (α211＋α2

21)= -λ12λ22 (α212＋α2

22)，

so it must has ,2

2

2

12

2

22

2

12

2

1

2

11

2

21

2

11

nn αα

αα

αα

αα

++

+=

++

+

LL

(now π and 1π are coincident or parallel)，but when

,2

2

2

12

2

22

2

12

2

1

2

11

2

21

2

11

nn αα

αα

αα

αα

++

+≠

++

+

LL

it must has λ11λ21 (α211＋α2

21) ≠ -λ12λ22（α212＋α2

22），

viz. β11β12+β21β22≠0 ，

so now h31 is not parallel to h41 .

By the same token，h3 ，projections of h4 on 2π are not always perpendicular . ▌

It is not difficult to know according to the above reasons that not random two straight lines

perpendicular to each other are useful for included angle questions.

From theorem 2 ,it can immediately get

Theorem 3：Assume Vm(m＜n) is a subspace of euclidean space Vn，Vmo is a sufficient-order

or difficient-order generic coordinate plane which has the same dimension with Vm.

H1，H2，…，Hm

are m straight lines of Vm， H11，H21，…，Hm1

and H12，H22，…，Hm2

are separately projections of H1，H2，…，Hm

on Vmo and ⊥0mV . If H1，H2，…，Hm

are perpendicular to each other. And their projections H11，H21，…，Hm1

on Vmo are also perpendicular to each other，then their projections

H12，H22，…，Hm2

on ⊥0mV must be perpendicular to each other(or only satisfy that the inner product is zero).

Contrarily，if H1，H2，…，Hm

are parallel to each other. And their projections

H12，H22，…，Hm2

on ⊥0mV are also parallel to each other(or only satisfy that the inner product is zero)，then their

projections H11，H21，…，Hm1

on Vmo must be parallel to each other. ▌

Theorem 2: and theorem 3 can be called generalized rightangle projection theorem by a

joint name .

We know：zero vector is parallel to random vectors，it is linearly dependent with random

vectors at the same time.Verticality can determine the changelessness of projection dimensions ，

92

and correlativity can determine the change of projection dimensions. In the proof of the above three

theorems，we stipulate that projection of two vectors can only satisfy the inner product is zero，in

order to emphasize its verticality and ignore its correlativity.

Because the inverses of theorem1, theorem 2 aren't exist，then vector group h1，h2，…，hm

parallel to each other of Vm which satisfy theorem 3 is uniquely determined.

According to principle of the above theorem 3，we give the following:

Defination 1:Assume the dimension of two subspaces Vp1 and Vs2 is W=p ，if projections σ(ri ) of a group of vectors ri (i=1 ，2 ，… ，p) parallel to each other of Vp1 on Vs2 are also parallel

to each other(aslo including some or all projections that only satisfy the inner product is zero to

each other)，then these vectors are called vectors that can determine included angles between Vp1

and Vs2，be called angle vectors for short.

In fact，to a considerable degree,the process of determining included angles φi between two

linear figures is the process of determining the group of angle vectors ri or projectionsσ(ri )

(i=1 ，2 ，… ，p) of the group of angle vectors .

Defination 2: assume the angle vectors between Vp1 and Vs2 are

r1，r2，…，rp

and ri∈Vp1 (i=1 ，2 ，…，p)，then the ratio

( )

pir

ri

i

i,,2,1 cos

L　== ϕ

σ

of module |σ(ri ) | of their projections on Vs2 to module | ri | of their own is cosine of included

angle φi between Vp1 and Vs2 .When Vp1 and Vs2 are separately regarded as orthodromic spaces of

linear figures A and B (equations of A and B can be non homogeneous)，because

A≌Vp1， B≌Vs2

(“≌”represents congruence[12]

)，then included angle between Vp1 and Vs2 is the included angle

between two linear figures A and B.

When all |σ(ri ) | = | ri | (i=1 ，…，p )

are true at the same time，the included angle between two linear figures A and B is zero，so A and B

are coincident or parallel，only at this time，the zero included angle between two linear figures is

considered to be meaningful；But when |σ(ri ) | = | ri | (1≤i≤p)

aren't true at the same time，the zero included angle between two linear figures is considered to be

meaningless，we only choose those nonzero included angles φi (1≤i≤p) to define the included

angle between two linear figures. The purpose of making the rule is to ensure the result consistency

of included angles under different solution methods.

The projection process of the above defination1, defination 2 can be described as：

Passive set Vp1⇒ path set ⇒⊥2sV contain set Vs 2⇒ image setσ(Vp1).

1.1.3 Common vector and uncommon vector————————sameness of two linear figures' dimension

In n-dimensional euclidean space，system of orthogonal vectors formed of n vectors are called

a group of orthogonal bases；Orthogonal bases formed of unit vectors are called orthonormal basis，be called orthogonal-unit vector.

Because the subspace Vm(m≤n) of euclidean space Vn is also a euclidean space under the

defined inner product，so，we can also choose m unit orthogonal vectors ε1，ε2，…，εm of

Vm to form orthogonal-unit vector of Vm，and this group of orthogonal-unit vectors can be extend

93

to the orthogonal-unit vectorsε1，ε2，…，εm of Vn.

Defination 3：：：：Assume both Vp1 and Vs 2 are subspaces of euclidean space Vn，εi1∈Vp1 andεj2∈Vs2 (i=1 ，…，p; j=1 ，…，s) are separately orthogonal-unit vectors of Vp1 and Vs2. If εi1 ,εj2∈Vp1∩Vs2(1≤i≤p； 1≤j≤s) ，

viz. εi1 ，εj2 are also orthogonal-unit vectors of Vp1∩Vs2， thenεi1 ，εj2 are also called common

orthogonal-unit vectors of Vp1 and Vs2， be called common vector for short.

Otherwise，if εi1 ,εj2 ∉Vp1∩Vs2 (1≤i≤p；1≤j≤s) ，

then they are called uncommon vectors of Vp1 and Vs2.

We know，when two linear figures are parallel or coincident (r=p or r=s，r is the rank of

coefficient matrix of equations' simultaneousness of two figures，refer to it in theorem 1 and

theorem 2 of the sixth chapter),their included angle is zero(it is known)，if order their dimensions

be same，it is ok to substitute common-orthodromic space for figures of high dimensions. So we

only discuss how to change the dimensions of two linear figures from different to same when they

are neither coincident nor parallel(viz. r＞p, s).

when two linear figures are neither coincident nor parallel，those angle vectors only determine

included angles between two linear figures can be changed into common vectors (from defination 3,

we can infer that the number of common vectors included in the two subspaces are the same)of

orthodromic space of the two linear figures by unit change ，and those orthogonal-unit vectors

that determine nonzero angles between two linear figures are uncommon vectors of orthodromic

space of the two figures.

Assume the dimensions of two linear figures A and B are separately p and s，and p＞s，p-s=k，and assume they have r-dimensional common-orthodromic space M，and A′, B′, M′are

separately orthodromic spaces of A , B , M . Then ，A′and B′separately have r common vectors，p-r and s-r uncommon vectors. From defination 3 and defination of orthodromic space，we know

M′is all linear combinations of the r(r＞p, s) common vectors. Assume A0′and B0′ are

separately all linear combinations of p-r and s-r uncommon vectors，then all nonzero angles

between A′and B′are the same with those between A0′and B0′.

But how to separately change A′ and B′ into A0′ and B0′? It is ok to separately delete r

common vectors from A′and B′. The method is separately taking A′and B′as passive set，taking M′as path set，taking M

⊥′as contain set，then finishing the projection from A′, B′through M′to M

⊥′，and then image sets A0′and B0′are separately acquired；Or，make A and

B separately intersect M⊥，it also can get

A∩M⊥

=A0， B∩M⊥

=B0，

apparently，the included angle between A0′and B0′is also the included angle between A0 and B0 .

This step can be called deleting common vector for short(If the dimension of common-orthodromic

space is zero,then this step can be omitted.).

From defination of the included angle，the dimensions of two linear figures should be equal，but，although by deleting common vectors ,dimensions of A0′and B0′separately been changed

into p-r=p0, s-r=s0，it still p0＞s0，and p0-s0=k，then the included angle between A0′and B0′can't be determined yet.

Because p0-s0=k，it shows that there are k unwanted uncommon vectors in A0′， they

should be deleted from A0′. How these k uncommon vectors appeared unwantedly? Originally，in

normal space⊥′

0B of B0′，there are k vectors linearly dependent with orthodromic vectors of A0′，

94

make ⊥′

0B intersect A0′， it gets A0′∩ ⊥′0B =M0′. Because the dimension of B0′is s0 ， then

the rank of ⊥′

0B is s0 ，then the rank of M0′is s0+n-p0=n-k ，viz. the dimension of M0′is k，and M0′is on A0′，k orthogonal-unit vectors of M0′are just the unwanted uncommon vectors of

A0′ .

Taking A0′as passive set，M0′as path set， ⊥′0M as contain set，finish image set A1′of

projection from A0′through M0′to ⊥′

0M ， or，make A0 intersect ⊥0M ，then it gets

A0∩ ⊥0M =A1.

Now，because the rank of A1 changes from n-p0 of A0 to n-p0+k=n-s0 ，then，then dimension

of A1 changes from p0 of A0 to n-(n-s0)=s0 which is the same with that of B0 . Apparently，the

included angle between A1′ and B0′ is equal to the included angle between A1 and B0，it is

equal to the included angle between A and B at the same time.

From the above description，we have the following

Theorem 4：：：：Random two linear figures of different dimensions can be changed into linear

figures of same dimensions in the case of keeping included angle changeless. ▌

Inference: After two linear figures A and be changed into A1 and B0 of same dimensions, the

amount of uncommon vectors they include has the following two cases：s0≤1；s0＞1. ▌

From defination of included angle between two linear figures，when s0＞1，the amount of

nonzero angles between A and B is also more than 1.This is so-called nonuniqueness of the amount

of included anges.

1.2 Nonuniqueness of the solution of included angle questions

Because of the nonuniqueness of the amount of included angles，it determines that methods of

solving included angle questions are not unique.

There are two kinds of methods of solving included angle quesions among linear figures，one

is called linear solution，the other is called nonlinear solution.

Definition 4: Adopting methods of solving system of linear equations，separately find out one

of their orthodromic vectors(or normal vector) on two linear figures(or on their normal space)，and

method that use included angles (or cosine of the included angles)between the two vectors to define

included angles between the two linear figures are called linear solutions of included angle

questions.

Definition 5: Method that need analyzing the relation between nonlinear equation and system

of linear equations of two linear figures to determine included angles between the two linear figures

are called nonlinear solution of included angle questions.

In graph field,Mr Jian Zhaoquan of Beijing Institute of Technology takes the lead in using unit

circle to determine included angles [13]

between two planes of higher space，he opened the door of

using nonlinear solution. So，nonlinear solution is also called Jianshi solution solution or JianShi

method. Linear solution is comparatively easy，but it can't solve all included angle questions；Jianshi

solution solution can solve all included angle questions，but the method is very complex.So，the two

methods can learn from others's strong points to offset one's weakness and supplement each other，

95

make included angle questions among linear figures of high space that trouble people for a long

time be readily solved.

For the range of application of linear solution and Jianshi solution solution，we have the

following：

Theorem 5：Two linear figures A and B have the same amount of uncommon vectors, when

s0≤1, they are fit for linear solutions，when s0＞1, they aren't fit for linear solutions.

Proof：when s0≤1，the amount of nonzero angles between two linear figures is zero or one 1，

of which,when s0=0, the included angle between two linear figures must be zero，so the two linear

figures are parallel or coincident. From theorem 1 and theorem 2 of the sixth chapter，this kind of

questions are already fit for linear solutions.

When s0=1，the amount of nonzero angles between two linear figures is also 1，from theorem

4，both uncommon vectors of two figures are uniquely determined，their included angles (must be

nonzero) are included angles between the two linear figures.In fact, the proof process(belonging

to linear operation process)of theorem 4 has preliminarily solved methods of determining the two

uncommon vectors.

If s0＞1，then the amount of nonzero angles of two linear figures is also more than 1.From

structure questions of solutions of system of linear equations we know，the amount of solution

vectors involved in a random system of elementary solutions of equations of A1 and B0 is more than

1. The method of theorem 4 can only determine the amount of solutions in their elementary solution

analysis，but it can't ultimately give respective and uniquely determined solution. That is to say，directions of each uncommon vector can't be uniquely determined by methods of solving system of

linear equations. So，included angle between the two figures A and B aren't fit for linear solution.

▌

Included angle questions aren't fit for linear solution can all be solved by Jianshi solution

solution.

The contents of Jianshi solution solution are too many，we will speciallly introduce them in the

eighth chapter，here we only introduce linear solution.

§§§§2 Linear solution of included angle questions between two linear

figures

Linear solution can be divided into positive angle method and complementary angle method.

2.1 Positive angle method

1°The included angle φ.between two generic planes

F1：a11x1+…+an1xn+a1=0

F2：a12x1+…+an2xn+a2=0

Because direction of normal vector of generic plane in n-dimensional space is uniquely

determined，then included angle questions between two generic planes is generalization of dihedral

angel questions in three-dimensional space. So，we use included angle between normal vectors of

F1，F2 to define included angle between two generic planes.Then

2

2

2

12

2

1

2

11

211211cos

nn

nn

aaaa

aaaa

++++

++=

LL

L

ϕ (1)

this is the cosine formula of included angle φbetween two generic planes F1，F2.

96

(1)If two generic planes are perpendicular，then their normal vectors are also perpendicular.

Then the necessary and sufficient conditions of F1⊥F2 is

a11a12+…+an1an2=0;

(2)If two generic planes are parallel ，then their normal vectors are parallel.So the necessary

and sufficient conditions of F1∥F2 is

2

1

12

11

n

n

a

a

a

a==L .

2°The included angle φ. Between two straight lines

( )

( )1,,2,1 0

1,,2,1 0

1

2

1

1

−==+

−==+

∑∑

=

=

njbxbH

njaxaH

n

i

jiij

n

i

jiij

L

L

：　

：　

From linear algebra theory，one system of elementary solution of derivative group H1′only

includes one solution，assume the solution is xi=αi(i=1 ，…，n)，of which

( )

1,1,11,11,1

11,11,111

11

−−+−−−

+−

+−=

nnninin

nii

i

i

aaaa

aaaa

LL

LLLLLL

LL

λα

(i=1 ，…，n. λ are nonzero numbers，the following is the same)，so，the orthodromic vector(also

be called direction vector) of H1 is

1,1,1,1

1111

1

1

−−−

=

nnnin

ni

ni

aaa

aaa

LL

LLLLL

LL

LL eee

H λ

={α1 , … ，αn} (2)

The orthodromic vector H2 = {β1 , … ，βn}

of H2 can be got by the same method，Then

22

1

22

1

11 ||cos

nn

nn

ββαα

βαβαϕ

++++

++=

LL

L

(3)

this is cosine formula of included angle between two straight lines H1, H2.

(1)If two straight lines are perpendicular， then their orthodromic vectors are also

perpendicular. So,the necessary and sufficient conditions of H1⊥H2 is α1β1+…+αnβn = 0 ;

(2)If two straight lines are parallel，then their orthodromic vectors are also parallel.So，the

necessary and sufficient conditions of H1∥H2 is

n

n

β

α

β

α==L

1

1 .

3°The included angle between two intersectant or mutually interleaving planes

97

( )

( )2,,2,1 0

2,,2,1 0

1

2

1

1

−==+

−==+

∑∑

=

=

njbxbI

njaxaI

n

i

jiij

n

i

jiij

L

L

：　

：　

which have one-dimensional common-orthodromic space is φ.

Apparently，the rank of coefficient matrix of common-orthodromic space

( )

( )

−==+

−==+

∑∑

=

=

221 0

221 0

1

1

n,,,jbxb

n,,,jaxa

Hn

i

jiij

n

i

jiij

L

L

：　

is n-1. Assume the equation of common-orthodromic space H⊥

=F is

F：c1x1+…+cnxn=0

Apparently，F is a generic plane.

Because F∩I1=H1 and F∩I2=H2 are separately straight lines

( )

( )

=

−==+

=

−==+

∑∑∑∑

=

=

=

=

0

1,,2,1 0

0

1,,2,1 0

1

12

1

11

n

i

ii

n

i

jiij

n

i

ii

n

i

jiij

xc

njbxb

H

xc

njaxa

H

L

L

：　

：　

Then，this question is changed into question 2°.The included angle between H1,H2 is

included angle φ.Between two intersectant or mutually interleaving planes I1，I2 .

4°Included angle questions between two n-r+1-dimensional linear figures which have

n-r-dimensional common-orthodromic space.

Assume ( )　；　：　 111,,2,1 01

−<<−==+∑=

nrrjaxaAn

i

jiij L

and ( )　；　：　 111,,2,1 01

−<<−==+∑=

nrrjbxbBn

i

jiij L

have n-r-dimensional common-orthodromic space

( )11 1,,2,1

0

0

1

1 −<<−=

=

=

∑∑

=

= nrrj

xb

xa

Mn

i

iij

n

i

iij

；：　 L

98

and find out the equation of common-orthodromic normal space M⊥

is

( )∑=

⊥ −==n

i

iik rnkxcM1

,,2,1 0 L：　 ,

because intersections of M⊥

and A，B are separately straight lines

( )

( )

( )

( )

−==

−==+

−==

−==+

∑∑∑∑

=

=

=

=

rnkxc

rjbxb

H

rnkxc

rjaxa

H

n

i

iik

n

i

jiij

n

i

iik

n

i

jiij

,,1 0

1,,1 0

,,1 0

1,,1 0

1

12

1

11

L

L

L

L

：　

：　

So，this question can be changed into question 2°.The included angle between H1,H2 is

included angle between A，B .

Example 1: Please find out the included angle between generic plane

F： 2x1+x2+2x3+x4=0

of four-dimensional system and generic coordinate plane

x4=0 .

Solution：：：：From formula (1)，the cosine of included angle between normal vectors ｛2,1,2,1｝

and ｛0,0,0,1｝ of two generic planes is

， 3162.010

1

21212

1101020102cos

22222≈=

+++

×+×+×+×+×=ϕ

so φ≈71°33′54″.

Example 2: It is known that two planes of four-dimensional system

=

=−++

=

=+++

0

06453 and

0

04375

4

321

2

4

321

1

x

xxxI

x

xxxI

：　

：　

intersect on straight line

=

=−++

=+++

0

06453

04375

4

321

321

x

xxx

xxx

H：　 ，

find out their included angle φ.

Solution：From formula (2),it gets H={13,-11,4,0},

then the normal space of H is a generic plane

F：13x1-11x2+4x3=0.

Make F intersect I1 , I2 , then it separately gets straight lines

99

=

=+−

=−++

=

=+−

=+++

0

0 41113

0645 3

0

0 41113

0437 5

4

321

321

2

4

321

321

1

x

xxx

xxx

H

x

xxx

xxx

H

：　

：　

and from formula(2),it separately gets H1={61,19,-146,0}； H2={32,20,-49,0},

then，the cosine of included angle between two planes I1，I2 is

( ) ( )

， 9624.097147350

9486

4920321461961

004914620193261cos

222222≈=

++++

×+−×−+×+×=ϕ

so φ≈15°45′42″.

In three-dimensional space，equations of I1、I2 don't include x4=0，so the normal vector of the

two planes is uniquely determined. In three-dimensional space,we take included angle between

normal vectors of the two planes as included angle between the two planes，the result is identical

with the above result.

Example 3: Find out the included angle between planes

=++++

=++++

=++++

=++++

033423

0332

0122

013

4321

43212

4321

43211

xxxx

xxxxI

xxxx

xxxxI

：　

：　

of four-dimensional system.

Solution：Because the rank of coefficient matrix of simultaneous group is 3，the rank of

augmented matrix is 4，then I1，I2 are mutually interleaving. And，their common-orthodromic

space is straight line

=+++

=+++

=+++

0 32

02 2

03

4321

4321

4321

xxxx

xxxx

xxxx

H：　

(“3x1+2x2+4x3+3x4=0” is already be regarded as unwanted equation，it is deleted from H ).Then

find out the normal space H⊥

=F of H is generic plane F：x1-x4=0 ，

and order F separately intersect I1，I2 at straight lines

=−

=++++

=++++

=−

=++++

=++++

0

033423

0332

0

0122

013

41

4321

4321

2

41

4321

4321

1

xx

xxxx

xxxx

H

xx

xxxx

xxxx

H

：　

：　

find out H1={1,-5,1,1,}； H2={1,5,-4,1},

100

Then ，the included angle between I1，I2 can be denoted by vectors H1，H2，viz.

， 7781.04328

27cos ≈

−=ϕ

so φ≈38°54′47″.

2.2 Complementary angle method

5°The included angle between straight line

( )1,,1 01

−==+∑=

njaxaHn

i

jiij L：　

and generic plane F：b1x1+…+bnxn+b=0

is φ.

Included angle questions between straight lines and generic planes is generalization of those of

three-dimensional space. So, the complementary angle of included angle between normal vectors of

straight line and generic plane is included angle between straight line and generic plane.

Assume from formula (2),we find out orthodromic vector of H is

H={α1,…,αn}

then ( )4 sin22

1

22

1

11

nn

nn

bb

bb

++++

++=

LL

L

αα

ααϕ

this is the sine formula of included angle between straight line H and generic plane F.

(1)if straight line is parallel to generic plane，then the normal vectors of the straight line and

the generic plane are perpendicular. Then the necessary and sufficient conditions of H∥F is α1b1+…+αnbn=0 ；

(2)If straight line is perpendicular to generic plane，then the normal vectors of the straight line

and the generic plane are parallel. Then the necessary and sufficient conditions of H⊥F is

n

n

bb

αα==L

1

1 .

6°Included angle questions between generic palne and a random linear figure.

Assume the common-orthodromic space of a generic plane

F：a1x1+…+anxn+a = 0

and a linear figure(can be regarded as a difficient-order generic plane)

( )11,,1 01

−<<==+∑=

nrrjbxbLn

i

jiij ；　：　 L

is

( )

−<<−==

=

∑∑

=

=

11 1,,2,1 0

0

1

1

nrrjxb

xa

Mn

i

iij

n

i

ii

；

：　

L

and the rank of coefficient matrix of M is r+1 , and assume the equation of common-orthodromic

normal space M⊥

to be found out is

101

( )1,,1 01

−−==+∑=

⊥rnkcxcM

n

i

kiik L：　 ,

From theorem 3 of the sixth chapter ，we know M⊥∩L=H is a straight line

( )

( )

−−==

==+

∑∑

=

=

1,,1 0

,,1 0

1

1

rnkxc

rjbxb

Hn

i

iik

n

i

jiij

L

L

：　

Then，included angle questions between generic plane F and difficient-order generic plane L

can be changed into included angle questions between generic plane F and straight line H，viz. the

included angle between F and H is equal to the included angle between F and L. According to

method of the above question 5°,the included angle φ. Between F and L can be aquired.

7°The included angle φ.between straight line ang a random linear figure.

Assume the equations of straight line H and difficient-order generic plane L are separately

( )1,,1 01

−==+∑=

njaxaHn

i

jiij L：　

( )11,,1 01

−<<==+∑=

nrrkbxbLn

i

kiik ；　：　 L ,

and assume the equation of normal space H⊥

=F of H to be found out is

F：c1x1+…+cnxn= 0

Apparently，F is a generic plane. Then according to method of question 6, we can first find out

the included angle ϕπ

−2

between F and L ，its complementary angle φ is the included angle

between H and L.

8°Included angle questions between plane

( )2,,1 01

−==+∑=

njaxaIn

i

jiij L：　

which has one-dimensional common-orthodromic space and difficient-order generic plane

( )11,,1 01

−<<==+∑=

nrrkbxbLn

i

kiik ；　：　 L .

Apparently，common-orthodromic space is a straight line

( )

( )

−<<==

−==

∑∑

=

=

11,,1 0

2,,1 0

1

1

nrrkxb

njxa

Hn

i

iik

n

i

iij

；　

：　

L

L

Assume we find out H⊥

=F is generic plane

F：c1x1+…+cnxn= 0 ,

102

because F∩I=H0 is also a straight line

( )

=

−==+

∑∑

=

=

0

2,,1 0

1

1n

i

iik

n

i

jiij

xc

njaxa

H

L

：　

then this question is changed into question 7°.The included angle between H0 and L is the

included angle between plane I and difficient-order generic plane.

9°Included angle questions between n-r+1-dimensional linear figure

( )1,,1 01

−==+∑=

rjaxaAn

i

jiij L：　

which has n-r-dimensional common-orthodromic space and a random difficient-order generic

plane ( )21,,1 01

−<<==+∑=

nsskbxbLn

i

kiik ；　：　 L .

Apparently，the rank of coefficient matrix of common-orthodromic space

( )

( )

−<<==

−==

∑∑

=

=

21,,1 0

1,,1 0

1

1

nsskxb

rjxa

Mn

i

iik

n

i

iij

；　

：　

L

L

of A , L is r，assume the equation of M⊥

to be found out is

( )rncxcMn

i

ii −==+∑=

⊥ ,,1 01

Lτττ：　 .

then because M⊥∩A=H is a straight line

( )

( )

−==

−==+

∑∑

=

=

,1, 0

1,,1 0

1

1

rnxc

rjaxa

Hn

i

iik

n

i

jiij

L

L

τ：　

then the question is changed into question 7°.The included angle between H and L is included

angle between A and L.

In fact ,if we cancel the restrict to r,s，then from theorem 4，question 9°can generalize

all contents of linear solution. For example， in 9°, ofder r=2，s=1 ，viz. it is question 1°；Order

r=n, s=n-1 , viz. it is question 2°；order r=n，s=1 ，viz. it is question 5°etc.

Example 4: Find out the included angle φ.Between generic plane

F：3x1-5x2+2x3+x4-2x5-6 = 0

of five-dimensional space and plane

=++−−+

=−−+++

=++−−+

013 25

0233

035432

54321

54321

54321

xxxxx

xxxxx

xxxxx

I：　

103

Solution：：：：Because common-orthodromic space is a straight line

=+−−+

=−+++

=+−−+

=−++−

0325

033

05432

02253

54321

54321

54321

54321

xxxxx

xxxxx

xxxxx

xxxxx

H：　

from formula (2) it gets H={73,8,-407,329,-153}

So，the equation of H⊥

=F0 is

F0：73x1+8x2-407x3+329x4-153x5 = 0

make F0 intersect I at straight linez

=

=

=

=

++−−+

−−+++

++−−+

−+−+

0

0

0

0

1325

233

35432

153329407873

54321

54321

54321

54321

0

xxxxx

xxxxx

xxxxx

xxxxx

H ：　

And from formula (2) ,it gets

H0={15201,-14920,2325,-8201,-17347}

from formula(4) sin 802447.043827257236

151346sin ≈

×=ϕ

then we get the included angle between F and I is φ≈53°21′52″.

Example 5: find out the included angle between straight line

=++−−

=+

=−−

=+

0354352012

0

03453

035

54321

53

421

21

xxxxx

xx

xxx

xx

H：　

and plane

=++−−+

=−−+++

=++−−+

01325

0233

035432

54321

54321

54321

xxxxx

xxxxx

xxxxx

I：　

of example 4.

Solution：：：：the direction of H is vector

=

−−

−−=

354352012

10100

034053

00035

2380

1

54321 eeeee

H

={3,-5,2,1,-2}，

then H⊥

is the derivative group of equations of generic plane of example 4

F′：3x1-5x2+2x3+x4-2x5 = 0

So，included angle between H and I is the complementary angle of included angle between F and

104

I ，viz. φ≈90°－53°21′52″≈36°38′8″.

Exercises 7.1 Please give voice to the four elements that compose projection in turn. 7.2 Please give voice to contents and the relations of two kinds of projections. 7.3 Please give voice to what's rightangle projection? 7.4 Prove：If a plane of 4-dimensional space ang a generic plane have no common point,then

they must be parallel(clue：use theorem 2 of the sixth chapter). 7.5 Prove：If a plane of 4-dimensional space and a straight line have no common point, then

they are not always parallel(clue：make out another plane through the straight line interleaving with

the known plane，…，use theorem 2 and theorem 4 of the sixth chapter). 7.6 Please find out the included angle φ.Between two generic planes F1：7x1+4x2-5x3+2x4+2x5-x6+3 = 0

F2：4x1-3x2+6x3-5x4-2x5+3x6-5 = 0 7.7 Please find out the included angle φ.Between two straight lines

=++−+

=+−++

=+−−+

=+−++

=−+−+

=++−−

012322

02535

03543

014235

023433

03322

4321

4321

4321

2

4321

4321

4321

1

xxxx

xxxx

xxxx

H

xxxx

xxxx

xxxx

H

：　

：　

7.8 Please find out the included angle φ.Between two planes

=++−−

=++−−

=−+++

=+−−+

06246

054353

0233

035432

4321

43212

4321

43211

xxxx

xxxxI

xxxx

xxxxI

：　

：　

7.9 Please find out the included angle φ.Between the plane

=−+++

=+−−+

0233

035432

4321

43211 xxxx

xxxxI ：　

of exercise 7.8 and the straight line of exercise 7.7

=+−++

=−+−+

=++−−

014235

023433

03322

4321

4321

4321

1

xxxx

xxxx

xxxx

H ：　

105

The Eighth Chapter: Jianshi solution solution of included

angle questions between twolinear figures

Jianshi solution solution is the nonlinear solution referred to in the seventh chapter.

According to graphic principle, Mr Jian Zhaoquan of Beijing Institute of Technology use unit

circle method to solve included angle questions [13]

between two planes of n-dimensional space.

This is a breakthrough of great significance in studying included angle questions between two

linear figures of higher space. Now，we introduce this method(also be called Jianshi solution

solution) by principle of analytical geometry，and extend this method to included angle questions

between two random linear figures.

§§§§1 Orthogonal transform and principal axis questions

1.1 Orthogonal matrix and orthogonal transform We know，assume Q is n-order square matrix of real number field，if

Q′=Q－1 ，

viz. inverse matrix of Q is equal to its transposed matrix，then Q is called a orthogonal matrix. If the

matrix of linear transform σis a orthogonal matrix，then σ is called a orthogonal transform. Orthogonal transform can be divided into orthogonal-unit vector transform and coordinate

transform of vectors(sometimes it includes coordinate transform of points). Of which，when assume

e1,…，en are a group of orthogonal-unit vectors of Vn，and α=｛x1,…，xn｝ are nonzero vectors of Vn，then

′

′

=

=

nnn e

e

e

e

Q

e

e

MMM

111

σ (1)

(Q is a orthogonal matrix) is called orthogonal transform of orthogonal-unit vector e1 ，… ，en to

e1′，…，en′， when keep α motionless and only rotate coordinate system

=

nn x

x

Q

y

y

MM

11

：　σ (2)

is called orthogonal transform of coordinates of vectorα，it is also called orthogonal linear

transform of x1，…，xn to y1，…，yn.

The second kind of orthogonal transform is used the most in this chapter，it is mainly used for

some linear figures to be changed from general position to particular position.But this time we

always use its inverse transform，so，we write inverse transform of formula (2) underneath:

′=

−

nn y

y

Q

x

x

MM

111：　σ (3)

The core of orthogonal transform is orthogonal matrix，but for constructing orthogonal

106

matrix，we usually choose a group of linearly independent vectors and orthogonalize them，and then

extend them to orthogonal basis of Vn ， last unitize them. This process is called

orthogonal-unitization of vectors.

1.2 1.2 1.2 1.2 Concepts and property of exterior product among vectors————

orthogonal-unitization of vectors

Assume α1，…，αm (m≤n) are linearly independent，then there are two approaches of

separately find out a group of orthogonal vectors β1，…，βm .

The first approach is called Schmidt orthogonal process(Schmidt)，the second is called exterior

product method(accordingly,the first can be called inner product method).

Let's introduce the first approach.

Order β1=α1，

and assume β2=α2+kβ1，

in order to determine k, makeβ1 intersectβ2，and it need to make inner product

(β1，β2)=(β1,(α2+kβ1)=(β1，α2)＋k（β1，β1）=0.

becauseβ1 ≠0, (β1，β1)≠0，then only choose

( )( )11

21

,

,

ββ

αβ−=k

it can get ( )( )11

2122

,

,

ββ

αβαβ −= ，

similarly, it can get ( )( )

( )( ) 2

22

32

1

11

31

33,

,

,

,β

ββ

αββ

ββ

αβαβ −−= ， … … … …，

( )( )

( )( ) 1

11

1

1

11

1

,

,

,

,−

−−

−−−−= m

mm

mmm

mm βββ

αββ

ββ

αβαβ L ， β1，…，βm are not only intersect with each other，but also can be mutually linearly

represented withα1，…，αm.

Now let's introduce exterior product method.

Defination: assume αi=｛αi1 ,αi2 , … ,αin ｝, i=1 ，… ，n .

Then the exterior product of n-1 linearly independent vectorsα1，…，αn－1 can be represented as

,12,11,1

22221

11211

21

121 n

nnnn

n

n

n

n λα

ααα

αααααα

ααα ==×××

−−−

−

L

LLLL

L

L

L

L

eee

(4)

in the formula，λis a nonzero number，λαn is called the exterior product

of α1，…，αn－1 ，and α1，…，αn－1 are all called multiplication factors of αn.

Theorem 1：Exterior productαn separately intersect its multiplication factorsα1,…,αn－1 .

107

Proof：：：：because the coordinates of αn={αn1,…, αnn}

are algebraic cofactors of the first row's elements of determinant (it is also called exterior product

determinant )of formula(4)，viz.

( )

nnjnjnnn

njj

njj

j

jn

,11,11,12,11,1

21,21,22221

11,11,11211

1

11

−+−−−−−

+−

+−

+−=

ααααα

αααααααααα

λα

LL

LLLLLLL

LL

LL

(j=1，…，n )，and the inner product of a random vectorαi(i=1，…，n-1)αn ofα1，…，αn－1 can

be represented as

( ) 0

1,

,12,11,1

................................................................21

11211

...............................................................21

2211 ==+++=

−−− nnnn

inii

n

inii

nninninini

ααα

ααα

ααα

ααα

λαααααααα

L

LLLL

L

LLLL

L

L

L

(i=1 ，… ，n-1)viz. αn and αi are orthogonal. ▌

It should be pointed out that，direct addition and multiplication operation are unsuitable for exterior product determinants，viz. when both determinants |A| and |B| of n×n matrixes A and B are

exterior product determinants,it has

|A|+|B|≠|A+B| and |A|·|B|≠|A·B|

So，when more than two exterior product determinants add or multiply，it should first carry

internal arithmetic of exterior product determinants ，then add or multiply.

As will be readily seen，formula (4) is generalization of defination of exterior product between

two vectors in three-dimensional space. In the past，when determine normal vector (refer to it in example 4 of §2 in the third chapter)of

generic plane F：3x1+2x2+x3+3x4+5x5+x6+6=0

and orthodromic vector (formula (2) of §2 in the seventh chapter)of straight line，we have used

this kind of exterior product of vectors.

Now let's discuss the process of orthogonalizing linearly independent vectors α1，…，αm (2≤m≤n-1)

by exterior product method.

Because αi=｛αi1 ,αi2 , … ,αin ｝( i=1 ，… ，m) ,

then when 0

21

22221

11211

≠

mmmm

m

m

ααα

αααααα

L

LLLL

L

L

we can assume α1×…×αm×em+1×em+2×…×en－1=

108

( ) 11

1

1111

1

1

1,1,1

11,11,1111

1111

1

1

1

++

−−

−+

−+

−+−

=−=

==

mm

mnmmm

nm

nm

mn

mnnmmmmmm

nnmm

nnmm

βλ

ααα

ααα

ααααα

ααααα

L

LLLL

L

L

LLLLLL

LLOLLLL

LLLLLL

LL

LLLLLLL

LL

LL

eee

eeeee

(both λm+1 and the followingλ1，…，λm, λm+2,…，λn are nonzero numbers). Assume αi∈Vm( i=1 ，… ，m)，Vm ⊂ Vn,

then βm+1∈ ⊥mV .

When separately regard α1，…，αm

as normal vectors of m generic planes，the intersection of orthodromic spaces of the m generic

planes is ⊥

mV . So，βm+1 is a common-orthodromic vector of the m generic planes. Assume βk=｛β1k ,…，βnk ｝k=m+1，m+2，…，n

in turn.Then we can separately find out all the orthodromic vectors which is perpendicular to each

other λm+2βm+2=α1×…×αm×βm+1×em+2×…×en-1 ， λm+3βm+3=α1×…×αm×βm+1×βm+2×em+3×…×en-1 ， … … … … … … ， λnβn =α1×…×αm×βm+1×…×βn－1 ，

of ⊥

mV Separately substitute βn ,β1，β2，…，βm－1 for e1，…，em in turn，then λ1β1=βn×e2×e3×…×em×βm+1×…×βn－1， λ2β2=βn×β1×e3×…×em×βm+1×…×βn－1， … … … … … … ， λmβm =βn×β1×β2×…×βm－1×βm+1×…×βn－1

can be separately found out，When 0

21

22221

11211

=

mmmm

m

m

ααα

αααααα

L

LLLL

L

L

,

order λm+1βm+1=e1×e2×…×en－m－1×α1×…×αm， … … … … … … ， λmβm =βm+1×…×βn－1×βn×β1×β2×…×βm－，

then allβ1，β2，…，βm can be found out in turn.

It can be verified that，β1，β2，…，βn are not only orthogonal with each other，but also β1，β2，…，βm can be linearly represented with：α1，…，αm (butβ1，β2，…，βm here

109

are not required to be consistent with the group of orthogonal vectors aquired by inner product

method).

Last，use formula nii

i

i 1 ，，，　 L==β

βε

and separately unitize β1，β2，…，βn，

then a group of orthogonal-unit vectorsε1，ε2，…，εn are acquired.

As will be readily seen，exterior product method also offer another method to solve system of

homogeneous linear equations. In one of system of elementary solutions of system of

homogeneous linear equations acquired by this method，when the amount of solution vectors is

more than one，every solution vector is orthogonal with each other.

Example 1: Please change the two planes

=++++

=++++

0222

0322

4321

4321

1xxxx

xxxxI ：　

=+−+

=−−+

013

04322

421

3212 xxx

xxxI ：　

of four-dimensional space from general position to particular position.

Solution：：：： translate I1，I2 to the origin，it gets

=+++

=+++′

022

022

4321

43211 xxxx

xxxxI ：　

=−+

=−+′

03

0322

421

3212 xxx

xxxI ：　

first find out two solution vectors

{ } { }3,0,1,13,0,1,1

121

112

0100

1221

12123 =−−−=−==

eeeeeee

β ,

{ }4,11,6,6

3011

1221

12124 −=

−

=

eeee

β

of I1′, order λ1β1=β4×e2×β3= -β4×β3×e2=

{ }1,2,0,311

0010

3011

41166=

−

−−−=

eeee

(λ1=11) and order -11β2=β4×β1×β3=

110

{ }. 3,6,14,511

3011

1203

41166−−−−=

−

−=

eeee

assume

{ }

{ }

{ }

{ }4,11,6,6209

1

3,0,1,111

1

3,6,14,5266

1

1,2,0,314

1

4

44

3

33

2

22

1

11

−==′

−==′

−−−==′

==′

β

β

β

β

β

β

β

β

e

e

e

e

then it gets orthogonal matrix

−

−

−−−

=

209

4

209

11

209

6

209

611

30

11

1

11

1266

3

266

6

266

14

266

514

1

14

20

14

3

Q

use fourmula (3),it gets

+−−=

−−=

++−=

+++=

−

43214

4213

4322

43211

1

209

4

11

3

266

3

14

1209

11

266

6

14

2209

6

11

1

266

14209

6

11

1

266

5

14

3

yyyyx

yyyx

yyyx

yyyyx

：　σ

chooseβ1，β2 as new normal vector of I1′，then it gets

=−−−

=++′

036145

023

4321

4311 xxxx

xxxI ：　

and two normal vectors of I2′are orthogonal with two normal vectors of I1′，then I2′is also

changed into

=+−+

=−+′

041166

03

4321

4212 xxxx

xxxI ：　

111

substitute the above transforms into new equations of I1′，I2′，it separately gets

=

=′′

=

=′′

0

0

0

0

4

3

2

2

1

1

y

yI

y

yI

：　

：　

this is the equations when they are in particular position.

Under most cases，we can only change one of two linear figures from general position to

particular position，and the other is still on general position after orthogonal transformation.

Because orthogonal transformation keep the inner product changeless，then the included angle

betweeen two figures is changeless.

1.3 Principal axis questions [14]

Principal axis questions can be generalized in a ward，that is change quadratic form to

principal axis .

So-called change quadratic form to principal axis is to find out a orthogonal transformation(3)

for the quadratic form

( ) ( )

=

n

nn

x

x

Axxxxf MLL

1

11 ,, (5)

given in real number field. Of which，A=(aij )n×n a real symmetric matrix，and substitute

formula(3) and its transpose into formula(5),make

( )

′

n

n

y

y

QQAyy ML

1

1 (6)

has the simplest form.

22

22

2

11 nn yyy λλλ +++ L (7)

In the first chapter we have introduced concepts of characteristic root and characteristic vector

of matrix A in linear algebra theory，but the contents it introduced are not integral ，so give

complementary introduction as the following:

Characteristic roots of real symmetric matrix are all real numbers.

If ξ，η are characteristic vectors correspondent with two different characteristic roots of

real symmetric matrix A，thenξ，ηare orthogonal.

If matrix B=QAQ′，

of which Q is orthogonal matrix，A and B are real symmetric matrix，then we call A and B

contragradient similar. If A and B are contragradient similar，characteristic polynomials of A and B

are the same，then A and B have the same characteristic roots. For a random real symmetric matrix

A，there must be a orthogonal matrix Q making QAQ′be a diagonal matrix，viz.

=′

n

QQA

λ

λO

1

(8)

112

So in formula(3)，formula(6)，as long as find out proper orthogonal matrix Q and change A into

diagonal matrix，then quadratic form(5) is changed into the form of quadratic sum of formula(7).

But，finding out orthogonal matrix must first find out all characteristic roots of A，then we can

find out all characteristic vectors belonging to different characteristic roots ，last unitize them，it is

very troublesome to do this.In fact，all characteristic roots (include multiple roots)of A are just

elements of principal diagonal of diagonal matrix in formula（8）.So，as long as all characteristic

roots(include multiple roots) of A are found out ，and be used as coefficients of corresponding

variables of formula(7)，this solved questions of changing quadratic form into quadratic sum and

relieved the trouble of finding out orthogonal matrix.

Using elementary transform，change characteristic matrix AI −λ of real symmetric matrix A

into diagonal matrix，elements of principal diagonal are called invariant divisors of AI −λ . Divide

invariant divisors of AI −λ into different products of one-order factor powers of λ，all one-order

factor powers (the same are counted by times they appeared ) are called elementary divisor of

AI −λ .

By elementary divisors of AI −λ , we can find out all characteristic roots(include multiple

roots) of A.

Example 2: change quadratic form

1484525 323121

2

3

2

2

2

1 =+++++ xxxxxxxxx

into the form of quadratic sum (it is also called normalized form).

Solution：：：：the characteristic matrix of the matrix

=

524

222

425

A

of the quadratic form is

−−−

−−−

−−−

=−

524

222

425

λ

λ

λ

λ AI

make elementary transform

( ) ( )( ) ( )( )

( ) ( )( )( )

( ) ( )( )( ) ( )( )

−−

−⇒

−−

−−−⇒

−−

−−−

−−−

⇒

−−−−

−−−

−−−

⇒

−−−

−−−

−−−

⇒

−−−

−−−

−−−

10100

010

001

10100

1120

001

10100

1120

524

91120

1120

524

425

222

524

524

222

425

λλ

λ

λλ

λλ

λλ

λλ

λ

λλλ

λλ

λ

λ

λ

λ

λ

λ

λ

onλΙ-A so，the three characteristic roots of A are separately 1，1，10，from formula(6) and

formula (8)， it gets .110 2

3

2

2

2

1 =++ yyy

113

§§§§2 Projective generic elliptic cylinder and projection of generic circle

2.1 Projection of sufficient-order generic circle

In example 5 of the third chapter，we have introduced generic cylindrical surface

22

3

2

2

2

1 Rxxx =++

the generic cylindrical surface is also called projective generic cylindrical surface. When a generic

sphere ( ) 22

404

2

3

2

2

2

1 Rxxxxx =−+++

intersect generic plane x4=x40 ，the cut trace is a generic circle.What about the projection of the

generic circle on generic coordinate plane x4=0 ? Substitute x4=x40 into equation of the generic

sphere，then a generic cylindrical-surface equation through the generic circle is acquired,its

generating line is perpendicular to the generic coordinate plane，and make it simultaneous with

generic coordinate plane，then

=

=++

04

22

3

2

2

2

1

x

Rxxx

is the proetion equation of the generic circle on the generic coordinate plane.

When generic plane which through generic centre of sphere and intersect generic sphere is on

general position，the cut trace is a generic circle. The following is an example of finding out

projection of the generic circle on generic coordinate plane.

Example 3: Assume center of a generic phere of four-dimensional space is on the origin，its

radius is 1. A generic plane

F：2x1＋x2＋2x3+x4=0

intersect the generic sphere at generic circle

=+++

=+++

022

1

4321

2

4

2

3

2

2

2

1

xxxx

xxxxK：　

Please find out the projection of the generic circle with relation to generic coordinate plane x4=0.

Solution：：：： Apparently，the generic circle is on an oblique position with relation to generic

coordinate plane，so，its projection on the generic coordinate plane is a generic ellipse. First find out

equation of projective generic elliptic cylinder (in the following it is called projective generic

cylinder surface by a joint name or be called generic cylinder surface for short)which includes the

generic circle and the generating line is perpendicular to generic coordinate plane.And make the

equation of the projective generic cylinder surface with the equation of generic coordinate plane，viz. projective equation of the generic circle on generic coordinate plane is acquired.

Equation of projective generic cylinder surface can be determined in this way：arrange F and

transpose it，then it gets F0：x4= -2x1-x2-2x3 ， apparently， F0 and F have the same solutions，viz. F=F0. Sustitute F0 into the equation of the

generic sphere，it gets

1484525 323121

2

3

2

2

2

1 =+++++ xxxxxxxxx

Apparently，figures denoted by the quadratic form equation include the generic circle K，or

generic circle K is on the figures denoted by the the quadratic form . We also observe that，the

114

quadratic form doesn't include variable x4，so，the quadratic form is equation of projective generic

cylinder surface which includes generic circle K and its generating line is perpendicular to generic

coordinate plane x4=0 .

figure 45

Make equation of the projective generic cylinder surface simultaneous with equation x4=0 of

generic coordinate plane，it gets

=

=+++++′

0

1484525

4

323121

2

3

2

2

2

1

x

xxxxxxxxxK ：　

this is the projective equation of generic circle K to be found out on generic coordinate plane(refer

to it in figure 45，the four-dimensional system in figure is generated by transform 2e3 □ e1+2e2，the ratio of directviewing unit length of every axis is |σ(e4)|∶|σ(e1)|∶|σ(e2)|=1∶1∶1).

Usually，for random generic circle

=+++

=+++

0

1

2211

22

2

2

1

nn

n

xaxaxa

xxK

L

L

：　

its generic plane equation can be arranged as

( ) ( )0 1

1111 ≠++−= −− nnn

n

n axaxaa

x L

substitute it into generic sphere equation of K, then it gets equation of projective generic cylinder

surface

( ) 12

111

121211112121

2

12

2

12

22

2

22

12

2

1

=+++++

+

+++

++

+

−−−−−−

−−

nnnnnn

n

n

n

n

nn

xxaaxxaaxxaaa

xa

ax

a

ax

a

a

LL

L

or ∑ ∑∑−

=

−

=

−

= ≠

=+

+1

1

1

1

1

1

2

2

2

, 12

1n

i

n

i

n

j ji

jiji

n

i

n

i xxaaa

xa

a

115

and make it simultaneous with xn=0 ，then it gets projective equation of the generic circle K on

generic coordinate plane

=

=+ +′ ∑ ∑ ∑−

=

−

=

−

= ≠

0

, 12

11

1

1

1

1

1

2

2

2

n

n

i

n

i

n

j ji

jiji

n

i

n

i

x

xxaaa

xa

a

K ：　

(an≠0).

2.2 Projection of difficient-ordergeneric circle

Difficient-order generic circle is the cut trace got by difficient-order generic plane cutting

generic sphere.

Example 4: Assume the cut trace of generic sphere

12

6

2

5

2

4

2

3

2

2

2

1 =+++++ xxxxxx

of six-dimensional space cut by a difficient-order generic plane

=

=

−−−++

−+−−+

0

0

2223

22

654321

654321

xxxxxx

xxxxxx

is a two-order generic circle

=

=

−−−++

−+−−+

=+++++

0

0

2223

22

1

654321

654321

2

6

2

5

2

4

2

3

2

2

2

1

xxxxxx

xxxxxx

xxxxxx

L：　

Please find out projection of L on difficient-order generic coordinate plane

=

=

0

0

6

5

x

xI：　 .

Solution：：：：From the question，equation of difficient-order projective generic cylinder surface

found out should not include variable x5,x6，but it should include the given difficient-order generic

circle L. Regard the front four terms of every equation of difficient-order generic plane which L is

onas constant，then，the posterior two variables can be found out according to Cram's rule. Because

the determinant of coefficient of the posterior two variables is

, 411

22−=

−−

−=D

( )( )

( )( )

4321

4321

4321

4321

4321

4321

535-7x

22231

2

3535

12223

2 and

6

5

xxx

xxxx

xxxxD

xxxx

xxxx

xxxxD

x

x

+−−=

−++−−

−−+−=

+−−−=

−−++−

−−−+−=

then it gets

−++=

−++=

43216

43215

4

5

4

3

4

5

4

74

3

4

5

4

3

4

5

xxxxx

xxxxx

116

apparently，the new equation has the same solution with equation of difficient-order generic plane.

Substitute it into equation of generic sphere and make it simultaneous with equation of I，then

projective equation of L on difficient-order generic coordinate plane is

=

=

=−−+−

−+++++

′

0

0

830343050

465025252545

6

5

43423241

3121

2

4

2

3

2

2

2

1

x

x

xxxxxxxx

xxxxxxxx

L ：　

Usually，for random difficient-order generic circle

( )

−≤≤==

=+++

∑=

n

i

iij

n

nrrjxa

xxx

1

22

2

2

1

22,,1 0

1

；　L

L

(difficient-order generic plane∑=

=n

i

iij xa1

0 doesn't include unwanted equation) ， when the

determinant of coefficient 0

,2,1

22,22,1

11,21,1

≠=

+−+−

+−+−

+−+−

nrrrnrrn

nrnrn

nrnrn

aaa

aaa

aaa

D

L

LLLL

L

L

of the posterior r variables of difficient-order generic plane

( )∑=

−≤≤==n

i

iij nrrjxaL1

1 22,,1 0 ；　：　 L

we find out

nrrkrn

rn

i

iirrkenrrn

njjkrn

rn

i

iijjkrnjrn

nkrn

rn

i

iikrnrn

k

aaxaaa

aaxaaa

aaxaaa

D

LL

LLLLLLL

LL

LLLLLLL

LL

,1

1

,1,1

,1

1

,1,1

11,1

1

11,11,1

++−

−

=−+−+−

++−

−

=−+−+−

++−

−

=−+−+−

∑

∑

∑

−

−

−

=

(j，k=1 ，… ，r；2≤r≤n-2)，then it gets

　，，　，

D

Dx

D

Dx

D

Dx r

nrnrn === +−+− L2

21

1

or

+++=

+++=

+++=

−−

−−+−

−−+−

rnrrnrrn

rnrnrn

rnrnrn

xdxdxdx

xdxdxdx

xdxdxdx

L

,2211

2,2221122

1,2211111

10

L

LLLLL

L

L

：　 (9)

117

in the formula ( )∑=

+−−=

r

j

ijjk

kj

ik aAD

d1

11

(i=1 ，… ，n-r；k=1 ，… ， r)，of which，Ajk is algebraic cofactor of elements of the j-th row、the k-th column of determinant D (1≤j ，k≤r)，or

D

Dd ik

ik = ,

of which

nrrkrnirrkrnrrn

nkrnikrnrn

ik

aaaaa

aaaaa

D

LL

LLLMLLL

LL

,1,1,1

11,111,11,1

++−−+−+−

++−−+−+−

−

−

=

(i=1 ，… ，n-r；k=1 ，… ， r)，apparently ，L10 and L1 have the same solutions.

Substitute formula(9) into equation of generic sphere

122

2

2

1 =+++ nxxx L

it gets ， 11 1 1 1

2 =+∑ ∑∑∑−

=

−

=

−

= =

rn

i

rn

i

rn

j

r

k

jijkiki xxddx

Pick up i=j terms of the formula and separately incorporate them with every terms of ∑−=

rn

i

ix1

2then it

gets 111 1 1 1

2

1

2 =+

+∑ ∑∑∑∑−

=

−

=

−

= = ≠=

rn

i

rn

i

rn

j

r

k ji

jijkiki

r

k

ik xxddxd (10)

From formula(10)，equation of projective generic cylinder surface doesn't include variable

xn－r+1 , xn－r+2 , … , xn ,

So，the generic generating line of the generic cylinder surface is perpendicular to generic plane

xn－r+1=0 , xn－r+2=0 , … , xn=0 ,

So，projection of the difficient-order generic circle on difficient-order generic coordinate plane

=

=

=

+−

+−

0

0

0

2

1

2

n

rn

rn

x

x

x

LL

：　

is system of equations acquired by simultaneousness of formula(10) and equation of L2

=

=

=

=+

+

+−

+−

−

=

−

=

−

= = ≠=

∑ ∑∑∑∑

0

0

0

11

2

1

1 1 1 1

2

1

2

n

rn

rn

rn

i

rn

i

rn

j

r

k ji

jijkiki

r

k

ik

x

x

x

xxddxd

LL

(11)

Before the operation of the above projection，we should first examine dimensions of the two

linear figures L1, L2 are equal or not. If dimensions of the two linear figures are not equal，then

uncertainty of included angles in operation of solving included angles will appear，such solution

results are nonsensical. Now，we can adopt method of theorem 4 of the seventh chapter，first make

the dimensions of the two figues identical .

118

When determinant of coefficient of posterior r terms of L1 is D=0，we can choose normal

space ⊥1L of L1 and make it intersect generic sphere，find out projection of the cut difficient-order

generic circle on L2 according to the above method，and then solve included angle questions

between L1 and L2 by complementary angle method.But，the dimension of L2 is not equal to the

dimension of ⊥1L ，then we should find out projection of difficient-order generic circle of

⊥1L

on ⊥2L ，then the question become using included angle between normal spaces

⊥1L ， ⊥

2L of L1，L2 to define included angle between L1，Ｌ2.

§§§§3 Jianshi solution solution of included angle questions between two

linear figures

3.1 Principle and steps of Jianshi solution solution

From Jianshi solution theorem and generalized rightangle projection theorem(refer to it in the

seventh chapter)，m orthogonal-unit vectors of Vm can be projected into right angles in another Vmo，but，the m orthogonal-unit vectors satisfying the above conditions are not always easy to uniquely

etermined. So，when Jianshi solution study included angle questions between two planes，he first

make a circle whose radius is 1 on one of the plane，it is called unit circle，its projection on the

other plane is usually a ellipse(sometimes it is a circle，sometimes it is a line segment，sometimes it

is even a point). The unit circle always has such two radii perpendicular to each other，their

projections on another plane are not only perpendicular to each other ，but also uniquely become

semi-major axis and semi-minor axis (both are principal axes)of the ellipse. Then，of two random

radii perpendicular to each other of the unit circle，only one of projections of the two radii is the

longest,the other is the shortest.Jianshi solution use the length of the two semi-major axis and

semi-minor axis as cosine of the included angle between two planes.This principle is also called

Jianshi solution principle.

Jianshi solution principle can be made the following generalization：：：：the projection of

difficient-order unit circle denoted by quadratic equation with s unknown of a s-dimensional

linear figure of n-dimensional space on another s-dimensional linear figure is usually a

difficient-order generic ellipse. The difficient-order generic ellipse altogether has s radii

perpendicular to each other，their projections on another linear figure are not only perpendicular to

each other but also uniquely become s semiaxes (all are principal axes)of the difficient-order

generic ellipse.

According to the above principle，Jianshi solution method can be divided into the following

steps：

First，examine two linear figures through the origin or not and their dimensions are the same

or not.If they are not through the origin and their dimensions are not the same，then translate them

to the origin ang make their dimensions the same.

Second，assume equations of the two linear figures has already separately been system of

homogeneous linear equations

∑∑

=

=

=

=

n

i

iij

n

i

iij

xbL

xaL

1

2

1

1

0

0

：　

：　

( j =1 ，… ，r )

119

(the ranks of coefficient matrixes of Ｌ1，Ｌ2 are r)，find out a orthogonal transformation σby

method of the first section of this chapter，and separately change L1，Ｌ2 into

0

0

2

1

1

=′

=′′

+−

=

∑jrn

n

i

iij

yL

yaL

：　

：　

( j =1 ，… ，r )

Third，arrang equation of L1′into the form of formula (9)，substitute it into equation

122

2

2

1 =+++ nyyy L

of unit generic sphere，equation(10) of projective generic cylinder surface which included

difficient-order generic circle

( )

==′

=++

∑=

n

i

iij

n

rjya

yy

1

2

2

2

1

,,1 0

1

L

L

and its generic generating line is perpendicular to difficient-order generic coordinate plane L2′is

acquired，and make it simultaneous with equation of L2′,then equation(11) of difficient-order

generic circle is acquired.

Fourth，find out a real symmetric matrix A from formula(11) ，and find out all characteristic

roots λ1 ，λ2，…，λn-r (includemultiple roots) of A.

Fifth，change (10) into the form of quadratic sum

122

22

2

11 =+++ −− rnrn zzz λλλ L

from formula(6) and formula(8).

Sixth，it is known by deduction from ellipse questions，halves of every principal axis length of

generic ellipse separately are

,1

,,1

,1

21 rn−λλλL

this is cosines of all angles φ1, φ2，…，φn-r between L1，Ｌ2，viz.

,1

cos

,1

cos

,1

cos

2

2

1

1

rn

rn

−

− =

=

=

λϕ

λϕ

λϕ

LLL

Whenλ1 ，λ2，…，λn-r are not equal to 1 at the same time，chooseλi≠1 (1≤i≤n-r) viz.

angles of φi≠0 to define included angles between L1，L2.

3.2 Included angle questions between two planes

According to the priciple and steps of Jianshi solution method，assume two planes I1，I2 are

separately changed into

120

( )

( )nkyI

njyaI

k

n

i

iij

,,4,3 0

2,,1 0

2

1

1

L

L

==′

−==′ ∑=

：　

：　

after translation and rotation.make generic sphere (unit generic sphere)whose centre on the origin

and the radius is 1 intersect I1′， it gets

( )

−==

=+++

∑=

n

i

iij

n

njya

yyy

L

1

2

2

2

2

2

1

2,,1 0

1

L

L

：　

then the question become question of finding out projection of unit circle. assume projection of unit

circle Lon I2′is ellipse

( )

==

=++′

nky

NyyPyMyL

k L,4,3 0

12

221

2

1：　

it can be changed into normalized form (viz. the form of quadratic sum) after a orthogonal

transformation

( )

==

=+′′

nkz

BzAzL

k ,,4,3 0

12

2

2

1

L

：　

From ellipse questions，we know lengths of its semi-major axis and semi-minor axis are separately

BA

1 and

1,

then BA

1cos ,

1cos 21 == ϕϕ

this is cosine of included angle to be found out between I1，I2.

For the convenience of the latter operation，we give the following：

Theorem 2:(also be called coefficient theorem)：

there is the following relation

( )

( )

−+++=

−+−+=

22

22

2

1

2

1

MNPNMB

MNPNMA

(12)

between quadratic form ( ) 1, 2

221

2

121 =++= NxxPxMxxxf

and its normalized form ( ) 1, 2

2

2

121 =+= ByAyyyf .

Proof：：：： Because the real symmetric matrix of f(x1 , x2 ) is

NP

PM

2

2

assume its two characteristic roots are separately A and B ，viz. it has

121

( )

−+±+=2

2

2

1, MNPNMBA ▌

In order to save length of writing，in the following examples both the two planes are assumed

through the origin ，and one plane has become horizontal coordinate plane x1ox2.

Example 5: Please find out included angles φ1，φ2 between two planes of four-dimensional

space

=+++

=+++

022

022

4321

43211 xxxx

xxxxI ：　

=

=

0

0

4

3

2x

xI ：　

Solution：：：：because the determinants of coefficient of the posterior two variables of I1 are

zero，then the relation between x3，x4 and x1，x2 can't be uniquely determined.But，the included

angle between normal space ⊥

1I of plane I1 and I2 is the cosine of the included angle between I1

and I2. So，the cosine of included angel between ⊥

1I and I2 is the sine of included angle between I1

and I2.

Choose two orthodromic vectors ｛2，2，-3，0｝and ｛1，1，0，-3｝of I1 as normal vector

of ⊥

1I ，then it gets

=−+

=−+⊥

03

0322

421

3211 xxx

xxxI ：　

because the determinant of coefficient of two posterior terms is

930

03=

−

−=D

( )( )

( )( ) 21

21

21

2

21

21

21

1

330

223

663

022 but

xxxx

xxD

xxxx

xxD

+=+−

+−−=

+=−+−

+−=

then substitute . 3

1

3

1 ,

3

2

3

2214213 xxxxxx +=+=

into the equation 12

4

2

3

2

2

2

1 =+++ xxxx

of unit generic sphere，and make it simultaneous with I2，then the projection is a ellipse

=

=

=++

0

0

9141014

4

3

2

221

2

1

x

x

xxxx

From coefficient theorem〔viz. formula(12)〕it gets

( )

19514

951414141014142

1 22

=+=

=−= −+−+=

B

A

so，the equation of projection is changed into

122

=

=

=+

0

0

19

19

4

3

2

2

2

1

y

y

yy

so , 6882.019

3sin , 1sin 21 ≈== ϕϕ

then φ1=90°， φ2≈43°29′

viz. the maximum included angle between plane I1 and I2 is 90°，the minimum included angle is

about 43°29′.

Example 6: Please find out the included angles φ1，φ2 between two planes

=

=

=+++

=+++

0

0

022

022

4

3

4321

4321

1

x

xI

xxxx

xxxxI

：　

：　

of four-dimensional space.

Solution：：：： the determinant of coefficient of two posterior terms of I1 is

, 321

12==D

and ( )( )

( )( ) , 3

21

22 , 3

22

122

21

2121

21

211 x

xx

xxDx

xx

xxD −=

+−

+−=−=

+−

+−=

then substitute x3= - x1 , x4= - x2

into equation of generic sphere and make it simultaneous with I2

=

=

=+

0

0

122

4

3

2

2

2

1

x

x

xx

of which,the generic cylinder surface has been normalized form，so

, 2

1coscos 21 == ϕϕ φ1=φ2=45°

Example 7: In example 1，I1，I2 are separately changed itno

=

=′

=

=′

0

0

0

0

4

32

2

11

y

yI

y

yI

：　

：　

please find out their included angle φ1，φ2.

solution：：：：Because ⊥′

1I =I2′，then substitute the equation of I2′into the equation of generic

sphere and make them simultaneous according to method of example 5，then it gets that the

projection of unit circle on its personal plane is also a unit circle

123

=

=

=+

0

0

1

4

3

2

2

2

1

y

y

yy

so sinφ1= sinφ2=1；φ1=φ2=90°

In the above examples，we all writen out the projective equation of unit circle on another plane

(or the plane of its own). In fact，as long as directly using formula(12) to change equation of

projective generic cylinder surface into normalized form，then questions will be solved.So in the

following discussions，under most cases,we can only consider transform of equation of projective

generic cylinder surface，and needn't write out the simultaneous equations of them intersecting a

plane.

Example 8: Please find out the included angle between two planes of five-dimensional space

=+−+

=−++

=−++

022

02_3

023

5431

54321

5421

1

xxxx

xxxxx

xxxx

I ：　

=

=

=

0

0

0

5

4

3

2

x

x

x

I ：　

solution：：：：determinant of coefficients of three posterior terms of I1 is

, 1

122

111

230

−=

−

−

−

=D

( )( )

( )( )

( )( ) , 16x8

222

2311

30

, 115

122

1231

20

, 3

122

1123

23

but

21

1

21

21

3

21

1

21

21

2

2

1

21

21

1

+−=

−−

−−

+−

=

+−=

−

−−−

−+−

=

=

−−

−−−

−+−

=

x

x

xx

xx

D

xx

x

xx

xx

D

x

x

xx

xx

D

then substitute x3= -3x2 ，x4=5x1-11x2，x5=8x1-16x2，

into equation of generic sphere，it gets

138736690 2

221

2

1 =+− xxxx

from formula(12)

124

( )

( ) 1719.4742221654772

1

8281.22221654772

1

≈+=

≈−=

B

A

then it approximately gets

, 0459.01719.474

1cos , 5946.0

8281.2

1cos so

11719.4748281.2

21

2

2

2

1

≈≈≈≈

=+

ϕϕ

xx

then φ1≈53°31′， φ2≈87°22′.

From example 2,example 3 of the seventh chapter and examples of this chapter,It is observed

that ，included angles betweeen two planes can be divided into the following three kinds：

(1) quasi-dihedral angle kind：two planes either mutually interleaving or intersect at a straight

line，their maximum angle is 0°＜φ1≤90°，the minimum angle isφ2=0°.This kind of

questions are equal to dihedral angle questions (include dihedral angle questions)in

three-dimensional space. Solving this kind of questions always adopt linear solution，sometimes

adopt Jianshi solution solution. At the same time，we only use their maximum included angleφ1 to

define included angles between them.

(2) equal-oblique kind：the maximum angle is equal to the minimum angle，viz. φ1=φ2 ，and 0°≤φ1，φ2≤90°. Of which，whenφ1=φ2=0°，two planes are parallel or coincident(fit

for linear solution)；whenφ1=φ2=90°，random vectors of one plane is perpendicular to all vectors

of the other plane，then we call the two planes are hyper perpendicular.When they have common

point(the common point is unique), we call them perpendicularly mutual-through.

(3) unequal-oblique kind：the maximum angle is not equal to the minimum angle，viz.

φ1＞φ2 orφ1＜φ2，and 0°＜φ1，φ2≤90°.

For example，example 2,example 3 of the seventh chapter are examples of quasi-dihedral

angle kind；example 6,example 7 of this chapter are examples of equal-oblique kind，of which，two

planes of example 7 are hyper perpendicular；example 5 and example 8 are examples of

unequal-oblique kind.

Two concentric unit circles on hyper perpendicluar plane have a very interesting phenomenon.

Take I1′，I2′of example as an example，this pair of concentric unit circles are separately

=

=

=+

0

0

1

2

1

2

4

2

3

1

y

y

yy

L：　

=

=

=+

0

0

1

4

3

2

2

2

1

2

y

y

yy

L ：　

Of which，a random point of L1 is vertex of a right-circular cone taking Ｌ2 as the base line，and a random point of L2 is vertex of a right-circular cone taking L1 as the base line. That is to say，the distance of a random point of L1 to L2 is the same，at the same time, the distance of a random

point of L2 to L1 is the same. This phenomenon also exists in higher dimensional space.Now prove it as the following：

125

proof：：：：Assume M ( x1 , x2 ,0,…,0)， N ( 0 , … , 0 , xn-1 , xn )

are separately random points of two unit circles

=

=

=+

0

0

1

3

2

2

2

1

1

nx

x

xx

LL

：　

=

=

=+

−

−

0

0

1

2

1

22

1

2

n

nn

x

x

xx

LL

：　

because vectors

{ }

2 so

1 but

then

,,0,,0,,

22

1

2

2

2

1

22

1

2

2

2

121

=

=+=+

+++=

−−=−=

−

−

−

MN

xxxx

xxxxMN

xxxxOMONMN

nn

nn

nnL

By the same token，for another two points M0，Ｎ0 of L1，L2，it also has

200 == MNNM ▌

In three-dimensional space，two different circles of the same circular cylinder(or right-circular

cone) are perpendicular to each other. This is not only because of two planes the two circles on

paralleling to each other ，but also on the two circles,the distance between every two corresponding

points( it is referred to point intersecting the same generating line of the circular cylinder or the

circular cone) is the same. So，when people judge whether two circles are parallel or not，sometimes they need not consider whether the plane they on is parallel or not，only consider

whether the distances between corresponding points are equal or not. In other words，when

distances between corresponding points of two circles are equal， the two circles must be parallel.

But in n-dimensional space，distances between two random points of two concentric unit circles on

two perpendicular mutual through planes are identical. So，a random section of arc of one circle can

be regarded to be parallel to a random section of arc of anothercircle，we call this phenomenon

hyper parallel phenomenon (or be called pseudo parallel phenomenon)between two circles.

When diameters of two concentric circles aren't the same,hyper parallel phenomenon also

exists. At the same time ，it also exists on difficient-order (the orders are nonzero)generic circle of

same or different diameters of two random hyper perpendicular linear figures，be limited to the

length of writting，here we will not introduce it detailedly.

3.3 Included angle questions between other linear figures

According to principle and steps of Jianshi solution method，we assume equations of two

linear figures has been changed into system of homogeneous linear equations，and at least one

126

linear figure has been changed into a difficient-order(or sufficient-order )generic coordinate plane. Example 9: Please find out included angle φ.Between generic plane

F：2x1+x2+2x3+x4=0

of example 3 and generic coordinate plane x4 = 0.

Solution：：：：From example 3 and example 2，projective generic cylinder surface

1484525 323121

2

3

2

2

2

1 =+++++ xxxxxxxxx

has been changed into the form of quadratic sum

110 2

3

2

2

2

1 =++ yyy

so , 10

1cos ,1coscos 321 === ϕϕϕ

then φ1=φ2=0°，φ3≈71°34′. Choose nonzero angle φ3=φ to define included angle between Fand x4=0 ，the result is the

same with that of example 1 of the seventh chapter.

Example 10: Find out included angle between two two-order generic planes

=−−−++=−+−−+

02223

022

654321

6543211 xxxxxx

xxxxxxL：　

==

00

6

52 x

xL ：　

of six-dimensional space.

Solution：：：： From example 4，equation of projective generic cylinder surface is

830343050465025252545 434232413121

2

4

2

3

2

2

2

1 =−−+−+++++ xxxxxxxxxxxxxxxx

characteristic matrix of the quadratic form is

−

−−−

−−−

−−−

25151725

15251523

17152525

25232545

λ

λ

λ

λ

its invariant divisor are 1，1，λ-8, (λ-8)(λ2-104λ+1004)，

so its elementary divisor is

. 171052 , 171052 ,8 ,8 −−+−−− λλλλ

so the quadratic form is changed into

( ) ( ) 817105217105288 2

4

2

3

2

2

2

1 =++−++ yyyy

viz. 14

17526

4

17526 2

4

2

3

2

2

2

1 =+

+−

++ yyyy

127

so , 2929.017526

2cos , 8619.0

17526

2cos , 1coscos 4321 ≈

+=≈

−=== ϕϕϕϕ φ1=φ2=0°，φ3≈30°28′, φ4≈72°58′.

Choose nonzero angleφ3，φ4 as included angle between L1，Ｌ2.

Like this kind of questions，because the order of the quadratic form is very higher，then

amount of calculation is much. In fact，included angle between normal space ⊥1L of L1 and

⊥2L

of L2 is also equal to included angle between L1 and L2 ，so，we can also use included angle between

⊥1L and

⊥2L to define included angle between L1 and Ｌ2.

For example，first determine four linearly independent solution vectors in L1 ｛3，－5，0，0，0，－1｝， ｛0，5，－3，0，0，4｝， ｛0，0，3，5，0，－4｝， ｛0，0，0，4，－3，－5｝，

then the normal space of L1 is

=−−

=−+

=+−

=−−

⊥

0534

0453

0435

053

654

643

632

621

1

xxx

xxx

xxx

xxx

L ：　

and the normal space of L2 is plane x5ox6，viz.

=

=

=

=

⊥

0

0

0

0

4

3

2

1

2

x

x

x

x

L ：　

because the determinant of coefficient of four anterior terms of ⊥1L is D=180≠0

then we find out

+=

−−=

−−=

−−=

654

653

652

651

4

5

4

3

4

3

4

54

5

4

34

7

4

5

xxx

xxx

xxx

xxx

　

substitute it into the equation of unit generic sphere，it gets the projective generic cylinder surface

equation 4403121 65

2

6

2

5 =++ xxxx

from coefficient theorem，it gets

17526 , 17526 +=−= BA

viz. , 14

17526

4

17526 2

6

2

5 =+

+−

yy

so , 17526

2cos ,

17526

2cos 21

+=

−= ϕϕ

this is completely in accord with the above results.

128

Example 11: Find out included angle between difficient-order generic plane

=−−−++

=−+−−+

02223

022

654321

654321

xxxxxx

xxxxxxL：　

of example 10 and plane

=

=

=

=

0

0

0

0

6

5

4

3

x

x

x

x

I：　

solution：：：：The included angle betweeen plane I and normal space L⊥

of difficient-order generic

plane L is complementary angle of included angle between I and L，so，the cosine of included angle

betweeen I and L⊥

is sine of included angle between I and L.But in example 10，determinant of

coefficient of four posterior terms of L⊥

is

, 45

5340

4053

4003

1000

−=

−−

−−

−

−

=D

Then it gets

−=

+−=

=

−=

216

215

24

213

533

295

54

xxx

xxx

xx

xxx

substitute it into equation of unit generic sphere,then projective generic cylinder surface equation is

, 13

298

9

130951 21

2

2

2

1 =−+ xxxx

from coefficient theorem it gets

126.1489038.492222.98

18

44827

9

884

9

4591309

9

88804

9

1309459

2

1

≈+≈

≈+=

−

+++

=A

B≈98.2222-49.9038≈48.3184 ，

then， , 14390318448

1sin , 08220

126148

1sin 21 .

..

.≈≈≈≈ ϕϕ φ1≈4°43′, φ2≈8°16′.

Example 12: Find out included angle between one-order generic plane

=+++++

=+−++−

=+++++

022

0

032322

654321

654321

654321

1

xxxxxx

xxxxxx

xxxxxx

G：　

of six-dimensional space and

=

=

=

0

0

0

6

5

4

2

x

x

x

G ：　

Solution：：：：Because determinant of coefficient of three posterior terms of G1 is D = 9，then we

129

find out , 3

1 ,

9

1 ,

9

731621514 xxxxxxxx +=+==

substitute it into the equation of unit generic sphere,then equation of projective generic plane is

815418162162140 3121

2

3

2

2

2

1 =++++ xxxxxxx ，

So，the real symmetric matrix is

=

162027

01629

279140

G ，

the characteristic root of G found out is

, 931151 , 931151 , 162 321 −=+== λλλ

so equation of projective generic cylinder surface is changed into the form of quadratic sum

, 8199.0931151

9cos , 668.0

931151

9cos ,

2

1cos Then,

. 181

931151

81

9311512

321

2

3

2

2

2

1

≈−

=≈+

==

=−

++

+

ϕϕϕ

yyy

So， φ1=45°， φ2≈48°05′， φ3≈34°55′.

Kinds of included angles between two linear figures of n-dimensional spacen，similiar to

included angle questions between two planes，they can be divided into three kinds：

(1) quasi-dihedral angle kind，like example 9. This kind of questions usually adopt linear

solutions ；

(2) equal-oblique kind：in example 12，if coefficient of x1 of the first equation becomes 1，it

will hasφ1=φ2=φ3=45°；

(3) unequal-oblique kind，like example 10、example 11、example 12.

If discussions of the seventh chapter are added，it also includes

(4) line angle kind(it is only referred to included angles between two straight lines)；

(5) line-plane angle kind(it is referred to included angles between straight lines and

sufficient-order or difficient-order planes).

3.4 Other questions of Jianshi solution principle————————exterior product method and

included angle

In the above discussions，we have accepted such a fact: in unit difficient-order generic circles

denoted by s-order quadratic equation of a s-dimensional figure in n-dimensional euclidean space

Vn, there must be such s radii lines perpendicular to each other，their projections on another

s-dimensional linear figure are not only perpendicular to each other but also uniquely become s

semiaxes of a difficient-order generic ellipse. But ，what are these s radii lines on earth?How about

their directions(whether they are uniquely determined or not)?Now let's answer this question.

Assume equations of two linear figures L1，Ｌ2 have been changed into homogeneous，and

their dimensions also been changed into the same，of which, L2 has been changed into a

difficient-order generic coordinate plane，viz. assume

( )

( )22,,1 0

22,,1 0

2

1

1

−≤≤−==

−≤≤−==

+

=∑

nssnjxL

nssnjxaL

js

n

i

iij

；　：　

；　：　

L

L

and assume the projective generic cylinder surface equation of unit diffcient-order generic circle

130

( )

−≤≤==

−+++

∑=

n

i

iij

n

nssjxa

xxx

1

22

2

2

1

22,,1 0

1

；　L

L

with relation to L2 found out is ( ) , 11

21 =

nx

x

Gxx ML

Of which，G is s×s a real symmetric matrix. Apparently，at the time of projective generic

cylinder surface equation being changed into the form of quadratic sum，L1 and L2 also go through

orthogonal transformations.

Assume characteristic roots of matrix G (apparently, G is a real symmetric matrix)of the

quadratic form found out are separately λ1,…,λs ，and characteristic vectors belonging to λ1,…,λs of G are separately α1,…,αs (be a system of orthogonal vectors)，all their linear combinations

form a s-dimensional subspace( be called a characteristic subspace of G ). Unitize α1，…，αs and

make become orthogonal-unit vectors e1′,…,es′of the characteristic subspace ，viz. order

, , , 1

1

1

s

s

sα

α

α

α=′=′ ee L

of which e1′=｛β11 ，…，βs1 ｝，

e2′=｛β12 ，…，βs2 ｝， … … …，

es′=｛β1s ，…，βss ｝，.

and assume es+1′= es+1，es+2′= es+2，…，en′= en， then they are expanded into orthogonal-unit

vectors e1′，…，en′of Vn. So it gets orthogonal matrix

,

1

11

111

=

O

L

L

L

sss

s

Qββ

ββ

Assume the rank of coefficients matrix of L1 is A，then it has

=

=

′

0

011

1 MMM

nn x

x

A

y

y

QAL：　

Order AQ′=A0 (Q′= Q－1

is the transpose of Q )，and assume row vectors of A0 are separately

r1′=｛r11 ,…, rn1｝，

r2′=｛r12,…, rn2 ｝， … … … …，

rn-s′=｛r1，n-3，…，rn,n-s ｝，

then the equation of L1 become

( )∑=

−==′

=

′

n

i

iij

n

snjyrL

y

y

AL

1

1

1

01

,,1 0 or

0

0

L

MM

：　

：

131

Accordingly，the equation of L2′become Ｌ2′：ys+j = 0 ( j = 1 ，… ，n-s )

At this time, because equations of difficient-order (sometimes it can be sufficient-order)projective

generic cylinder surface have been changed into the form of quadratic sum，so，according to Jianshi

solution principle (include extended Jianshi solution principle)，directions of s radii lines

perpendicular to each other of difficient-order generic circle cut by L1′can be denoted by s

solution vectors of L1′，and projections (are directions of s semi principal axes of difficient-order

generic ellipse )of the s solution vectors on L2′are separately coincident with axes y1,y2,…, ys.

So，when solve the s solution vectors using exterior product method, in multiplication factors,one

of new orthogonal-unit vectors e1′，e2′，…，es′should be deleted in turn，viz. when assume the

s solution vectors(from defination1 of the seventh chapter，they are also angle vectors of L1′,L2′)

are P1′，ＰＰＰＰ2′，…，ＰＰＰＰs′in turn，it has

k1′P1′= r1′×r2′×…×rn-s′×e2′×e3′×…×es′，

k2′P2′= r1′×r2′×…×rn-s′×e1′×e3′×…×es′， … … … … … … … …，

ks′Ps′= r1′×r2′×…×rn-s′×e1′×e2′×…×es-1′

(k1′,k2′,…,ks′are numbers). Then the cosine of included angle between L1 and L2 can be

denoted as , |}0,,0,1{|

cos 1

1

1

11

11

1P

P

eP

eP

′

•′=

′•′

′•′=

−876

L

n

ϕ

, |}0,,0,1,0{|

cos 2

2

2

22

22

2P

P

eP

eP

′

•′=

′•′

′•′=

−876

L

n

ϕ

, |}0,,0,1,0,,0{|

cos

1

s

sns

s

ss

ss

sP

P

eP

eP

′

•′=

′•′

′•′=

−−876

L

876

L

LLLLLLLL

ϕ

Because P1= P1′Q，P2= P2′Q，…，PS = Ps′Q

are directions of the s radii lines of the difficient-order generic circle before transformation，so，

P1= P1′Q，P2= P2′Q，…，PS = Ps′Q

can also be found out by exterior product method. Because

ki′Pi′= r1′×…×rn-s′×e1′×…×ei-1′×ei+1′×…×es′，

but ( )s,1,2,i , L==′=′i

i

iii QQα

αerr ，

then when assume rj= {a1j , a2j ,…, anj }

( aij are elements of coefficient matrix of L1；i =1 ，2 ，… ，n; j = 1 ，2 ，… ，n-s), it also has

k1P1= r1×r2×…×rn-s×α2×α3×…×αs，

k2P2= r1×r2×…×rn-s×α1×α3×…×αs， … … … … … … … … …，

ksPs= r1×r2×…×rn-s×α1×α2×…×αs-1

( k1, k2,…, ks are numbers)，so it also has

132

. cos

, cos

, cos

22

22

2

11

11

1

ss

ss

sα

αϕ

α

αϕ

α

αϕ

•

•=

•

•=

•

•=

P

P

P

P

P

P

LLL

Because P1′，ＰＰＰＰ2′，…，Ps′are perpendicular to each other，and orthogonal transformation

changes vectors perpendicular to eache other into vectors that perpendicular to each other，then，P1′，ＰＰＰＰ2′，…，Ps′are also perpendicular to each other.

According to the above analysis we know，s angles vectors P1，ＰＰＰＰ2，…，Ps of L1 that

perpendicular to each other and their projections on L2 also perpendicular to each other (or

only satisfy that the inner product is zero)are not only exist ，but also their directions are also

uniquely determined；Angular degrees of included angles between L1 and L2 are also uniquely

determined.

This kind of solution method can be called (in Jianshi solution solution)exterior product

method，accordingly，the method introduced in the above section can be called projection method.

Example 13: Find out included angle between plane

=−+

=−+

03

0322

421

3211 xxx

xxxI ：　

of four-dimensional space and coordinate plane

=

=

0

0

4

3

2x

xI ：　 .

Solution：：：：Because here I1 is just ⊥

1I of example 5，then the projective generic cylinder surface

equation of I1 with relation to I2 is 9141014 2

22

2

1 =++ xxxx

(refer to it in example 5)，its real symmetric matrix is

=

145

514

9

1 G

( )19

19

9

14

9

59

5

9

14

but −

−=

−−

−−=− λλ

λ

λλ GI

(I is a two-order unit matrix)，then two characteristic roots of G are separately

, 1 , 9

1921 == λλ

Regard every element of determinant |λI-G| as coefficients of the following system of equations，it has

=

−+−

=−

−

09

14

9

5

09

5

9

14

21

21

xx

xx

λ

λ

First substitute λ1 into the system of equations，it gets

133

=

−+−

=−

−

09

14

9

19

9

5

09

5

9

14

9

19

21

21

xx

xx

we find out a characteristic vector of λ1 is α1=｛1，1，0，0｝，

secondly substituteλ2 into the above system of equations,it also gets

=

−+−

=−

−

09

141

9

5

09

5

9

141

21

21

xx

xx

we find out a characteristic vector of λ2 is

α2=｛-1,1,0,0｝，

because r1={2，2，-3，0}, r2={1，1，0，-3}，

then it gets { }2,4,3,33

0011

3011

03223

4321

2211 =

−

−

−=××=

eeee

rrP α ，

{ }0,0,1,19

0011

3011

03229

4321

1212 −=−

−=××=

eeee

rrP α ，

Because ( ) , ,,2,1 cos siii

ii

i L=•

•=

α

αϕ

P

P

then it gets , 19

3

4169911

33cos 1 =

++++

+=ϕ

, 122

11cos 2 =

−−=ϕ

then φ1≈46°31′， φ2＝0°.

(φ1 , φ2 are just complementary angles of included angles between the two planes of example 5)

Choose nonzero angleφ1≈46°31′to define included angle between I1 and I2 .

Now let's see the result after orthogonalization：Unitize α1,α2, it gets

, 0,0,2

1,

2

10,0,

2

1,

2

121

−=′

=′ e , e

order e3′= e3 , e4′= e4 ， Then it gets orthogonal matrix

−=

1

12

1

2

12

1

2

1

Q

134

because

−

−

−=′

1

12

1

2

12

1

2

1

321

322QA

then it gets { } { }23,0,0,20,23,0,4 21 −=−= r , r ，

So， { }2,22,0,36

0010

23002

023046

4321

1211 =−

−

′′′′

=′××=

eeee

errP

{ }0,0,1,018

0001

23002

0230418

4321

2211 −=−

−

′′′′

=′××=−

eeee

errP

[under new coordinates, of P1, e1′=｛1,0,0,0｝，of P2 , e2′=｛0,1,0,0｝]. it gets

, 1cos , 19

3cos 21 == ϕϕ

this is in coincident with the result of the above.

The purpose we introducing exterior product method is to give an account of the existence of

the s solution vectors ＰＰＰＰ1，ＰＰＰＰ2，…，ＰＰＰＰs. But in actual application，we still adopt simple and practical

projection method.

Exercises 8.1 Is exterior product operation of vectors in n-dimensional space carried between two

vectors?Why? 8.2 Try to use exterior product method to orthogonalize system of vectors α1={3,2,0,5,3}, α2={5,3,2,1,4}, α3={2,1,2,7,5}, α4={8,-1,2,4,3} . 8.3 Try to use exterior product method to find out a system of elementary solutions of

one-order generic plane . 0322

0235

54321

54321

=+−+−

=++−+

xxxxx

xxxxx 8.4 Use Jianshi solution solution to find out the included angle between generic plane 3x1+5x2-4x3-2x4=0

of four-dimensional space and generic coordinate plane

x4 = 0 ，

and check it by linear solution. 8.5 Find out the included angle between plane

=++−

=−−+

0453

0423

4321

4321

xxxx

xxxx

of four-dimensional space and coordinate plane

135

=

=

0

0

6

5

x

x. 8.6 Find out the included angle between two planes

==

=++−=−−+

00

02532

04254

4

32

4321

43211 x

xI

xxxx

xxxxI ：　：　

of four-dimensional space. 8.7 Find out the included angle between one-order generic plane

=

=

=

=+−++−

=−+−−+

=−+−−+

0

0

0

022322

042332

0652326

6

5

4

2

654321

654321

654321

1

x

x

x

K

xxxxxx

xxxxxx

xxxxxx

K

：　

：　

.

8.8 Change the plane I2 of example 5 into

=

=⊥

0

0

2

1

2x

xI ：　

then it become to find out the included angle between I1 and⊥2I . And check whether the result

is complementary angle of the included angle of example 5 or not. 8.9 Change the coefficient of the first equation x1 of G1 in example 12 into 1，check

whetherthe included angle between new G1 and G2 is equal-oblique kind or not. 8.10 Find the included angle between two planes

.02433

0225

02532

04254

4321

43212

4321

43211

=+−−

=+++

=++−

=−−+

xxxx

xxxxI

xxxx

xxxxI

：　

：　

136

The Ninth Chapter: The application of High Dimensional

Euclidean Geometry

High Dimensional Euclidean Geometry has very extensive application prospect. But be

limited to the length of writing，we only introduce some applications of High Dimensional

Euclidean Geometry in operations research.

§§§§1 Application of High Dimensional Euclidean Geometry in linear

programming

Methods of solving linear programming mianly had two kinds in the past：one is graphical

method，the other is simplex method. Of which,when the amount of variables of graphical method

is not less than 4，people once tried to use methods of high dimensional descriptive geometry [8]

. In

these years，many people are engaged in the study of non-linear programming，of which,(of

constrained extreme value questions)interiorpoint method is regarded as a kind of approximately

perfect method，and be attempted to use to solve linear programming questions [15]

. In the following we will introduce new methods，they are methods of using oblique axes

transform 、oblique axes draughting to solve linear programming，this kind of new method is also

called graphical method of preferred n-dimensional system. The main characteristics of the new methods are rigorousness and integrity of theory，concision

and directviewing of figures，it is easy to learn and master，convenient and practical，accurate and

reliable，it has very extensive application prospect. To verify its correctness and reliability，sometimes the acquired results should be compared to

some above methods .

1.1 Example one

A certain factory has three kinds of products of A，B，C，according to technological

requirements，each product should be manufactured on three equipments of the first、the second、the third in turn，processing time of each product on every equipment ,running time of every

equipment in every day and profit of each product can be achieved can be referred to in the

following table. Three kinds of products are all salable. How to arrange the output of this three

kinds of products，then it can make the factory achieve the maximum profit?

product

time

equipments

A

B

C

running time of every equipment in

every day（h）

the first 1 1/2 1 ≤23

the second 1/2 1 1 ≤23

the third 1 2/3 2/3 ≤22

profit (yuan/piece) 25 20 15

Solution：：：：According to conditions given in the question，assume we should arrange

the amounts of A，B，C to be x, y, z，the objective function is t，then it has the

following linear programming model

137

( ) ( )1 0,,

223

2

3

2

232

1

232

1

condition constraint

152025max function objective

≥

≤++

≤++

≤++

++=

zyx

zyx

zyx

zyx

zyxt

：

：

Now，change coefficients of each equation into integers

( )( )( )

( )2

66223

4622

4622

R

Q

P

zyx

zyx

zyx

≤++

≤++

≤++

first use simplex method to compute it. Order

x = x1, y = x2, z = x3,

and introduce slack variables x4，x5，x6，then the model is changed into normalized form：

objective function：Max t =25x1+20x2+15x3+0x4+0x5+0x6 constraint condition：

=+++

=+++

=+++

66 223

46 22

4622

condition constraint

6321

5321

4321

xxxx

xxxx

xxxx

：　

so it has

−−−=

−−−=

−−−=

3216

3215

3214

22366

22 46

2 246

xxxx

xxxx

xxxx

When order x1 = x2 = x3 = 0,

it gets t = 0，now it gets a radix feasible solution

X (0)

= (0,0,0,46,46,66)

because the coefficient of objective function x1 is larger，then t will increase much after taking x1 as

swapin variable. Order

x2 = x3 = 0

it has

≥−=

≥−=

≥−=

0366

0 46

0246

16

15

14

xx

xx

xx

(3)

Now only choosing 223

66

1

46

2

46min 1 =

= ,,x

can make formula(3) be true. Then take x6 as swapout variable and transpose it with x1,

6321

3215

3214

22663

2246

2462 getsit

xxxx

xxxx

xxxx

−−−=

−−=+

−−=+

138

viz.

−−−=

+−−=

+−+=

6321

6325

6314

3

1

3

2

3

222

3

1

3

4

3

424

3

2

3

2

3

12

xxxx

xxxx

xxxx

substitute it into objective function,it gets

, 3

25

3

5

3

10550 632 xxxt −−+=

When order x2 = x3 = x6 = 0 ,it can get another radix feasible solution

X(1)

= (22,0,0,2,24,0).

In the new objective function，only the coefficient of x2 is more than zero，so again take x2

as swapin variable. In order to determine swapout variable，we can order x3 = x6 = 0 , then it gets

( )4

03

222

03

424

03

12

21

25

14

≥−=

≥−=

≥+=

xx

xx

xx

now only choosing 18

3

2

22,

3

4

24, Min2 =

−=x

can make formula(3) be true，so，take x5 as swapout variable and transpose it with x2 ,

6312

5632

6324

3

1

3

222

3

23

1

3

424

3

43

2

3

22

3

1 getsit

xxxx

xxxx

xxxx

−−=+

−+−=

+−=−

viz.

+−=

−+−=

−+−=

561

5632

5634

2

1

2

110

4

3

4

118

4

1

4

38

xxx

xxxx

xxxx

substitute it into objective function ,it gets

. 2

15

2

55610 653 xxxt −−−=

There is "discriminant theorem of optimum solution "in linear programming，its contents are：substitute operation results of swapin variables into objective function，a newe objective function is

acquired，if variables of the new objective function are all less than zero，the radix feasible solution

found out is optimum solution[15]

. According to the theorem，variables of the new objective function

are all less than zero.

139

So when order x3 = x5 = x6= 0 ，the radix feasible solution

X(2)

= (10,18,0,8,0,0)

found out is optimum solution. Viz. A should be arranged to be producted 10 pieces every day，B

should be producted 18 pieces，C should be producted 0 piece，the maximum profit achieved is 610

yuan.

Although the above computation process is very cockamamie，the results are correct. But，trouble occurred when we use interior point method

[15] to compute(the detailed computation

process is omitted)，the optimum solution found out is 10 pieces of A，ten pieces of B，eight pieces

of C，"maximum profit" be achieved is 570 yuan，it is 40 yuan less than that of simplex method!

What's the cause? It turns out that，when we compute by equations，the system of equations is

consistent and the coefficients matrix is nonsingular，from linear algebra theory，it has uniquely

determined solution(but it is not always optimum solution of linear programming).

In fact, considering generalized case，this kind of questions can only aquired analogous results.

For example，we use “interior point method” to solve the following models objective function： ( ) ∑−=

=1

1

Max n

i

ii xAXf

constraint condition： ( ) ∑−=

=≤=1

1

,,2,1 n

i

jijij mjaxaXg L

(x1 , x2 ,…, xn-1≥0 )

Order F(X)=-f (X)， qj (X)=-gj (X)+aj≥0 ，

then we can change the modes into the following minimization questions of unconstraint property

( ) , , Min rXP

Of which ( ) ( ) ( )[ ]∑=

−+=

m

j

j XqrXFr,XP1

1

viz. ( ) ∑ ∑ ∑−

= =

−−

=

−−−=

1

1 1

11

1

,n

i

m

j

n

i

jijiii axarxArXP

solve partial derivative for ( ) , r,XP it gets

∑∑=

−

=

−

+−=∂

∂ m

jn

i

jiji

ji

i

iaxa

raA

x

P

12

1

1

order mj

axa

rR

n

i

jiji

j ,,1 2

1

1

L=

−

=

∑−=

it gets the system of linear equations

1,,1 1

−==∑=

njARa j

m

i

iji L

If we don't cosider the restrictionof x1 , x2 , …, xn-1≥0 (viz. don't rank them as constraint

condition or gj(X) doesn't include them), then we can at least find out a set of

R1=b1，R2=b2，…，Rm=bm

140

when m = n-1 and coefficients matrix is nonsingular. extract square root for

mjR

r

j

,,1 L=

and order mjcb

j

j

,,1 1

L==

then it gets a new system of linear equations

mjrcaxa jj

n

i

iji ,,1 1

1

L==−∑−=

viz. ( ) mjrcaXg jjj ,,1 0 L==−−

So it has ( )[ ] ( ) 0 lim0

=−=−−→

jjjjjr

aXgrcaXg

viz. when take limit for obstacle factor r ,it gets system of linear equations

g j (X) = aj j =1, 2, …, m . Goning in a roundabout way only to change sign of inequality of constraint condition into equal

sign，then change inequality into equality!

1.2 Theory 、、、、 steps and assumption of graphical method in preferred

n-dimensionalsystem

1.2.1 Theory The theory of graphical method in preferred n-dimensional system (be called preferred

n-dimensional system method for short) includes graphic rules of preferred n-dimensional system

and two fundamental theorems. Of which，we have detailedly introduced the graphic rules of

preferred n-dimensional system from the first to the fifth chapter，here we only introduce two

fundamental theorems.

Assume the model of linear programming is objective function

S：Max t = f (X)

constraint condition： g j (X) ≤aj，X≥0

in the formula，X = x1, x2,…, xn-1， j =1, 2 , …, m；aj are nonnegative integers. Assume objective generic plane S has been changed in the shape of plane by proper oblique axes

transformation，and Ci (i =1 , 2 ,…, I ) are other vertexes enclosing convex polyhedron of feasible

region except the origin，the trace of generic point that Ci is on is Ci′.

The cut trace of S cut by generic coordinate plane is marked as S t =0 .

When order t = t0 and substitute it into objective function，the equation represents a

generic plane f (X) = t0

parallel to t axis.When introduce leading-axis method in the fifth chapter，we call the generic

plane projective generic plane. Its top intersects objective generic plane S，the cut trace is generic

straight line

( )

=

=

0

0

tt

tXfS ：　π

its underside intersects generic coordinate plane，the cut trace is generic straight line

141

( )

=

=′

0

0

t

tXfS ：　π

cut trace πS ′ is just the projection of cut trace

πS on generic coordinate plane. Assume the

projectionπS ′ just intersects a certain vertex Ci of convex polygon，then the projection also

intersects trace point Ci′of generic point that Ci is on.

Make a parallel line of upright axis t through Ci ,it intersects objective generic plane S at point

Li，then Li is located atπS ，the length of line segment |CiLi| is just the corresponding value of t of

objective function when constraint condition satisfies the coordinates of Ci ，and make a parallel

line of upright axis t through Ci′, it also intersects S at generic straight lineπS ，and the

intersection point Li′is projective

trace of generic point that Li is on.So it is easy to know that

. and , iiiiiiii LCLCLLCC ′′′′ ∥∥ The trace line (it is also the trace line of St=0 )of S on plane x1ox2 is called objective trace line，marked as S{1,2}，its equation is

{ }

( )

=

=

=

=

−

−

0

0

0

00,,,,

1

3

121

2,1

n

n

x

x

t

xxxf

S

L

L

：　

And make perpendicular line Ci′Ki′(Ki′is foot point) to objective trace line S{1,2} through

point Ci′，then Ci′Ki′is parallel to perpendicular line CiKi (Ki is foot point) of St=0 made

through Ci and along the direction parallel to x1ox2，Ki′are projective traces of generic point that

Ki are on，so，

. and , iiiiiiii KCKCKKCC ′′′′ ∥∥

(so， iiii KLKL ′′∥ ，

viz. iiiiii KLCKLC ′′′≌△△ ).

Because figures of Ci Ki are completely coincident with those of Ci′Ki′，under

generalized cases，we can only make diagrammatic representations (Ci′and Ki are usually not

marked out) for point Ci and point Li′of figures be made out，so regard the connecting line

between Ci and Ki′as the connecting line between Ci′ and Ki′, and call |Ci′Ki′|=ki′directviewing distance of Ci to objective trace line S{1,2}.

From theorem 1 of the fourth chapter，we can always change objective generic plane in the

shape of plane by proper oblique axes transform. The question is ，the proper oblique axes

transform is not unoque. Then，under different oblique axes transforms，the trace of the same vertex

142

Ci is different because of the difference of the generic point，are these distances of these different

trace points to objective trace lines(it will not change with the change of oblique axis transform)the

same? If they are not completely identical，it is nonsensical to determine it optimum point or not

by directviewing distance of a certain vertex to objective trace line. So，we give the following:

Constant distance theorem：：：：On the premise of keeping objective generic plane S in the shape

of plane，no matter what different oblique axes transform we take，the directviewing distance of

each vertex of convex polyhedron to objective trace line S{1,2} is changeless. Proof：：：：Assume the objective generic plane is S：A1x1+A2x2+…+An-1xn-1- t = 0 ，

C(x10, x20, …, xn-1,0,0) is a random vertex of convex polyhedron，when C is zero with relation to

oblique axes x30, x40,…, xn-1,0，C is the point of plane x1ox2 ，so no matter what different oblique

axes transform we take，its position on plane x1ox2 is always changeless，so，its distance to objective

trace line is also changeless.

When x30, x40,…, xn-1,0 are not zero at the same time，then under different oblique axes

transforms，the directviewing position of C with relation to plane x1ox2 also has some change，the

changing trace should be a straight line. if we want to prove that on the premise of keeping S in the

shape of plane，no matter what different oblique axes transforms we take ,the directviewing

distance of point C to S{1,2} is always changeless，we can only prove this straight line is parallel to

S{1,2} . Therefor we can assume

S11=α11e 1 +α12 e 2 +α3 e 3 ，S21=α21 e 1+α22 e 2 +α4 e 4 ，… ，

Sn－3,1=αn－3,1 e 1 +αn－ 3,2 e 2 +αn－1 e n－1 ；

S12=β11 e 1 +β12 e 2 +β3 e 3 ，S22=β21 e 1 +β22 e 2 +β4 e 4 ，… ，

Sn－3,2=βn－3,1 e 1 +βn－3,2 e 2 +βn－1 e n－1

are two different sets of solution vectors of S，they are regarded as principal overlapping directions

of the two different oblique axes transforms. S is in the shape of plane under transform

∑

∑

−−

−−

−−

−−

−−

−−

−−−

−−−

22,311,321-n1

222121244

212111233

2

22,311,311-n1

222121144

212111133

1

eee

eee

eee

eee

eee

eee

nnn

nnn

βββ

βββ

βββ

ααα

ααα

ααα

□□□ □□□

：

：

LLLLLLLLLL

LLLLLLLLLL

and vectorOC (O is the origin) has the following change：

,

,

2

1

0,12,3

3

3012201

1

0,11,3

3

301110

2

1

0,12,3

3

3012201

1

0,11,3

3

301110

2

1

ee

ee

−−+

−−

−−+

−−

−

−−

−

−−

−

−−

−

−−

n

nn

n

nn

n

nn

n

nn

xxx

xxxOC

xxx

xxxOC

β

β

β

β

β

β

β

β

α

α

α

α

α

α

α

α

LL

LL□□

so point C has common generic point with point

143

, 0,,0,, 1

0,12,3

3

301220

1

0,11,3

3

3011101

−−−−−

−−

−

−−LLL

n

nn

n

nn xxx

xxxC

α

α

α

α

α

α

α

α　：

of plane x1ox2 and

, ,0,0, ,1

0,12,3

3

301220

1

0,11,3

3

3011102

−−−−−

−−

−

−−LLL

n

nn

n

nn xxx

xxxC

β

β

β

β

β

β

β

β　：

successively，and the vector between the two points is

( ) ( )

( ) ( )2

11

2,3112,30,1

33

12331230

1

11

1,3111,30,1

33

1133113021

e

e

−++

−+

+

−++

−=

−−

−−−−−

−−

−−−−−

nn

nnnnn

nn

nnnnn

xx

xxCC

βα

βαβα

βα

βαβα

βα

βαβα

βα

βαβα

L

L

Because the trace line of objective generic plane on x1ox2 is

{ }

=

=

=

=+

−

0

0

0

0

1

3

2211

2,1

t

x

x

xAxA

S

n

L：　

its orthodromic vector (also be called direction vector) is

A2e1-A1e2

If ( )211221 ee AACC −= λ

(λ is nonzero number)，then the changing trace of C with relation to x1ox2 is parallel to objective

trace line，now we will prove it.

Because S11，S21，…，Sn-3,1 and S12，S22，…，Sn－3,2 are all solution vectors of S，and ｛A1，…，An－1，-1｝ is normal vector of S，it is perpendicular to all solution vectors of S，then it has

A1α11+A2α12+A3α3=A1β11+A2β12+A3β3= 0

A1α21+A2α22+A4α4=A1β21+A2β22+A4β4= 0 … … … … … … …

A1αn－3,1+A2αn－3,2+An－1αn－1=A1βn－3,1+A2βn－3,2+An－1βn－1= 0

, ,

, ,

, , so,

1

4422221

1

4422221

1

3312211

1

3312211

LLLLLLLL

A

AA

A

AA

A

AA

A

AA

βββ

ααα

βββ

ααα

+−=

+−=

+−=

+−=

. , 1

112,32

1,3

1

112,32

1,3A

AA

A

AA nnn

n

nnn

n

−−−

−

−−−

−

+−=

+−=

βββ

ααα

substitute vector 21CC between the above points C1, C2 into them and multiplied by

11

2,3112,30,1

33

12331230

1

)()(

−−

−−−−− −++

−−

nn

nnnnnxx

A

βα

βαβα

βα

βαβαL

144

then the direction of the straight line through points C1, C2 is A2e1-A1e2 ,but in

( )211221 ee AACC −= λ

.

)()(

1

11

2,3112,30,1

33

12331230

A

xx

nn

nnnnn

−−

−−−−− −++

−

−=βα

βαβα

βα

βαβα

λ

L

This proves that the straight line is parallel to objective trace line S{1,2} (it also proves that the

changing line of Ci on plane x1ox2 is indeed a straight line).Then the theorem is proved. ▌

Because the length of line segment |Ci′Li′| is just the value of t of objective function，and △CiLiKi≌△Ci′Li′Ki′，

So . ,,2,1, jimjiLC

LC

KC

KC

jj

ii

jj

ii ≠=′′

′′=

′′

′′；L

That is to say，the directviewing distance Ci′Ki′of vertex Ci to objective trace line S{1,2} is

longer，then the value of the corresponding objective function will be larger.And from constant

moment theorem，it has

Optimum point theorem：：：：Of every vertex of convex polyhedron，the point whose

directviewing distance to objective trace line S{1，2} is the longest is optimum point to be found out.

▌

1.2.2 Method、、、、steps

a. choose proper preferred n-dimensional system

Choose a set of solution vectors

S11=α11e 1 +α12 e 2 +α3 e 3 ，

S21=α21 e 1+α22 e 2 +α4 e 4 ， … … … … … … …，

Sn－3,1=αn－3,1 e 1 +αn－ 3,2 e 2 +αn－1 e n－1 ；

of objective function S：f (X) = t

as principal overlapping direction，make use of transform

−−

−−

−−

−−−− 22311311

22212144

21211133

eee

eee

eee

,n,nnn ααα

ααα

ααα □□□ LLLLL

then we can get a preferred n-dimensional system that can make S in the shape of plane.

b. determine coordinate of every vertex of convex polyhedron and their directviewing position

in preferred n-dimensional system.

For the convenience of discussion，we regard equations P，Q，R of constraint condition of

example 1 as the same with general equations for the moment，we call them constraint generic

plane. Because P，Q，R don't include variable t，and these three equations are consistent(have no

unwanted equations)，so，they intersect at a point on generic coordinate plane，coordinate of the

intersection point is the unique solution

x1=10, x2=10, x3= 8 ,

of system of linear equations with three unknown acquired when equation of constraint condition

145

takes equality .

From the above discussion we have known，this set of solutions is not optimum solution.

The optimum solution is x1=10, x2=18, x3= 0 .

Lead a straight line parallel to axis t through the corresponding optimum point of optimum

solution，coordinate of the intersection point of the straight line with S with relation to axis t is the

optimum solution to be found out. Finding out the intersection point of the straight line with S，the

best method is to change S in the shape of plane.So ，find out a solution vector -2e1+e2+2e3

among S and take it as principal overlapping direction，make use of transform

2e 3 □ 2e1－e 2

then we get the generic coordinate plane of preferred four-dimensional system that makes S in the

shape of plane(figure 46).

figure 46

First find out trace lines of each constraint generic plane on plane x1ox2，then separately

determine coordinates of intersection points of these trace lines according to principle of three kind

of graphic methods and principle of descriptive geometry，we determine them vertexes of convex

polyhedron or not according to whether coordinates of these points satisfy each constraint condition

or not.

Refer to it in figure 46，three trace lines P1P2 , Q1Q2 , R1R2 of P , Q , R on plane x1ox2

separately intersect at three points

( ) ( )； 0,0,18,10 , 0,0,6,26 , 0,0,3

46,

3

46111 CBA −

Three trace lines P2A2 , Q2A2 , R2R3 on plane x2ox3 separately intersect at two points

146

B2(0，20，13，0) and A2 (0，0，23，0)

( Q2A2∥R2R3 ，it has no intersection point in euclidean space)；Three trace lines A2P1 , A2Q1 , R3R1

on plane x3ox1 separately intersect at

A2，B3 (20，0，3，0) and C2 (10，0，18，0) .

Connect A1A2 and B1B2 (or C1C2) it gets the intersection point D .Because the projection of

A2A1 on plane x1ox2 is OA1，so，make a parallel line of axis x3 through point D to OA1, it

intersects OA1，coordinate of the intersection point is just coordinate of D with relation to axis x1,

axis x2 (they are omitted in the figure).Coordinate of D(10，10，8，0) is the unique solution when P,

Q, R take equality. The position and coordinate of the above points are all acquired by methods of

elementary descriptive geometry.

Under general cases，the figure of objective function S needn't be made，optimum point can be

directly found out by method of step c，then it is ok to substitute coordinate of optimum point into

equation of S . But in our discussion，for the convenience of illustrating questions，this step is

indispensable. In the following ,we will make out the directviewing figure of S .

In equation of S，order t =200，it gets a projective generic plane

S′：25x+20 y +15 z = 200 ，

its cut trace on generic coordinate plane is a generic straight line，the generic straight line is

coincident with trace line Sxy′. Now assume trace line of generic straight line that OC1 is on on

plane xoy is OJc ，it is directviewing-coincident with OC1，then trace line OJc intersects Sxy′at

point Jn′，we regard Jn′as projective trace of a generic point N of S .

Choose proper direction and unit length, then make out the figure of upright axis t，lead Jn′to

generic point N，connect ON，and make C1M∥Ot ,it intersects the directviewing extension line of

ON at M . Then separately make EG and FH through two generic points E and F of traces Je，Jf on Sxy，make OMJJJJ hfge ∥∥

( Jg , Jh are separately opposite traces of G，H )，connect GH ( M is on GH))，

generic parallelogram EFGH is the figure of generic plane S.

As will be readoly seen，all points that satisfy constraint condition are located at convex

hexahedron OR1C1Q2A2B3D enclosed by quadrilaterals DC1R1B3 ，DC1Q2A2 ，OR1C1Q2 ，OR1B3A2

and △DA2B3 ， △OQ2A2 ，because points A1，B1，B2，C2 are separately located at the outside out

of triangle or quadrilateral of plane x1ox2 , plane x2ox3 and plane x3ox1 ，their coordinates don't

satisfy all constraint conditions，then they are not on the convex polyhedron.

c.determine optimum point and optimum solution

1) According to method of drawing parallel lines，put down one right angle side of triangular

rule clinging to objective trace line, put another ruler clinging to the other right angle side of

triangular rule，fix the ruler，slide the triangular rule along direction of crossing feasible region

clinging to the ruler，when the right angle side parallel to objective trace line slide on the border of

being away from feasible region，the last vertex of convex polyhedron pressed by the right angle

side is optimum point to be found out.

2) Order t = t0 ( t0 = 0 , 1 , 2 , …) and substitute them into objective function in turn，then

make them simultaneous with x3 = 0, x4 = 0, … , xn-1= 0, t = 0, it gets

147

{ }

( )

=

=

=

=

′

−

−

0

0

0

0,,,,

1

3

0121

2,1

t

x

x

txxxf

S

n

n

L

L

：　

Then S{1,2}′is equivalent to translate objective trace line along direction of crossing feasible

region. When S{1,2}′is about to or just be away from feasible region，the distance between S{1,2}′and the last vertex of convex polyhedron is the nearest，arrange every value of t0 , make its

increase (or decrease)amplitude go down，until the distance between S{1,2}′and the vertex is small

enough to satisfy people，then the value of t0 is optimum solution .

Take figure 46 as an example，apparently optimum point is located at near the other six

points (except the origin)of convex hexahedron OR1C1Q2A2B3D. From constant distance theorem

and optimum point theorem，among the six vertexes ,the distance between C1 and the objective

trace line is the farest ，so it is the optimum point to be found. Substitute coordinate of C1 into

objective function S , it gets t = 610，this is the maximum value of objective function under the case

of meeting constraint conditions.

Or，make a parallel line of axis t through C1 to S, it intersects S at point M， because we

measure that 41 610eMC = ，then we know t = 610 .

1.3 An assumption

By above new graphic methods and steps，we associate the defect of simplex method. At the

same time，we find that ideas of preferred n-dimensional system method happen to have the same

view with that of simplex method in some questions，this idea of happening to have the same view

may help us get over the defect of simplex method. In linear programming，we may as well call every equation served as constraint condition

constraint equation，and deem that every equation has some certain constraint force. Magnitude of

the constraint force is determined by vertexes' amount and vertexes' quality of convex polyhedron

of constraint generic plane denoted by the constraint force. In other words，a constraint generic

plane has more vertexes ,its quality will be better，the constraint force of its equation is greater，contrarily it is less. Of which, the quality of vertex of convex polyhedron is ：substitute coordinates

of these vertexes into objective function, values found out is satisfying.

It is observed from figure 46，generic plane R has C1，D，R1 and B3 four vertexes，according

to the distance between them and objective trace line S{1,2}′，sorting order of quality is 1，2，3，4；Q has C1，D，Q2，A2 four points，sorting order is 1，2，5，6；P only has D，B3，A2 three

points，sorting order is 2，4，6. So，constraint force of equation (R) is the greatest，(Q) takes second

place，constraint force of (P) is the least.

At least computation procedure of simplex method happens to have the same view with

constraint force of constraint equations .

But，at least the introduction of slack variable increases much unnecessary work in solving this

kind of questions. The original idea of introducing slack variable is to change unequal relations into

equal relations.But coefficient of slack variable in objective function is zero ，then slack variable is

out of function for the last operation results. So，hiding slack variable will not influence the

correctness of operation results. Then we regard some constraint conditions as equalities hiding

148

slack variable.Still use the above example：

Arrange the constraint conditions as the following：

( )( )( )

( )′

−−−≤

−−−≤

−−−≤

1

223660

22460

22460

321

321

321

R

Q

P

xxx

xxx

xxx

still take x1 as swapin variable，and order x2 = x3 = 0 it gets

( )′

−≤

−≤

−≤

2

3660

460

2460

1

1

1

x

x

x

Because only choosing

223

66,

1

46,

2

461 =

=x

can make formula(2)′be true，then，(R) is regarded as equation of the greatest constraint force, it

is of the corresponding formula (1)′of the third inequality of formula (2)′. Regard (R) as

an equality hiding slack variable，and change the coefficient of x1 into 1 then move it to the left

end of equal mark，it has

( )′

−−=

−−≤+

−−≤+

3

3

2

3

222

22460

24620

321

321

321

xxx

xxx

xxx

Further arrange formula (3)′into

( )″

−−=

−−≤

−+≤

3

3

2

3

222

3

4

3

424 0

3

2

32 0

321

32

32

xxx

xx

xx

Substitute x1 into objective function ,then it gets

. 3

5

3

10550 32 xx

t −+=

When order x2 = x3 = 0 it gets the initial radix feasible solution X(1)

= (22,0,0,0)，This is just

coordinate of vertex R1 of convex polyhedron in figure 46.

According to differentiation theorem of optimum solution ，coefficient of variable x2 of new

objective function is still more than zero，so the corresponding radix feasible solution X(1)

isn't the

optimum solution.

And take x2 as swapin variable and order x3 = 0，then it gets

149

( )′

≥−=

−≤

+≤

4

03

222

3

4240

320

21

2

2

xx

x

x

Because only choosing 18

3

2

22,

3

4

24, Min2 =

−=x

can make formula (4)′be true，so，(Q) is regarded as equation of the greatest constraint force

except (R),it is of the corresponding formula (1) of the second unequality of formual (4)′. And

regard the second equation of formula (3)＂ as an equality hiding slack variable，change

coefficient of x2 into 1，transpose other equations and further arrange them into

( )5

10

183

2

320

1

32

32

=

−=

−+≤

x

xx

xx

substitute formula (5) into objective function,it gets

t = 610-5x3，

from differentiation theorem of optimum solution，variable coefficients of new objective function

are all less than zero. So when order x3 = 0, it gets radix feasible solution X(2)

= (10,18,0,0)

viz. the optimum solution. Apparently，it is just coordinate of C1 of figure 46，now objective

function will take its maximum value

t = 610 .

If simplex method can be improved according to the above ideas，this age-old method will

become more perfect. So，we extend this method to a generalized case：

assume objective function is

∑−+=

=1

1

Max n

ki

ii xAt

constraint condition is m,,,jaxan

i

jiji L21 1

1

=≤∑−=

xi≥0 i =1 , 2 , …, n-1

In order to change constraint condition into standard model，we first introduce slack variable

and make it become

11,22111

11,222212121

11,12121111

0

0

0

−−−+

−−−

−−

−−−−=+

−−−−=+

−−−−=+

nnmmmmmn

nnn

nnn

xaxaxaax

xaxaxaax

xaxaxaax

L

LLLLLL

L

L

it becomes

150

11,2211

11,22221212

11,12121111

0

0

0

−−

−−

−−

−−−−=

−−−−=

−−−−=

nnmmmm

nn

nn

xaxaxaa

xaxaxaa

xaxaxaa

L

LLLLLL

L

L

after hiding slack variable Apparently，this is still a standard model(onlly at the beginning, basis

vector is a zero vector to all appearances, with the introduction of swapin variable, it becomes a

nonzero vector). So for linear programming，all definations、theorems are still efficient here. But，because initial radix feasible solution changes from representing the origin to representing a vertex

of convex polyhedron located at axis x1, x2 , …, xn-1，proof methods of these theorems have some

small changes：

Differentiation theorems of optimum solution and infinite optimum solutions

1.Assume after several iterations

X(0)

= ( c1, c2 , …, ck , 0 , … , 0 ) ( 1≤k≤m )

become a corresponding feasible basis vector，at the same time the objective function becomes

( )∑−+=

≤≤>+′=1

1

1 0 n

ki

kkii mkbbxAt ；　　

If for all i = k+1 , k+2 , …, n-1 it has Ai′≤0 ，then X(0)

is optimum feasible basis vector or be

called optimum solution，Ai′is called check number.

2. If X(0)

= ( c1, c2 ,…, ck , 0 , … , 0 ) ( 1 ≤ k ≤ m )

is a optimum solution，for all i = k+1 , k+2 , …, n-1 it has Ai′≤0 ，at the same time there is

check number Al′= 0 ( k+1≤l≤n-1 ) of a certain non-radix variable, then the programming has

infinite optimum solutions.

Proof：：：：Use reduction to absurdity. If

X(0)

= ( c1, c2 , …, ck , 0 ,…, 0 ) (1≤k ≤m )

is not optimum solution，there must be a non-radix variable xi ( for example assume it is xk+1 ) that

can be swaped in radix variable，then it gets a new radix feasible solution

X(1)

= ( c1′, c2′, …, ck′, ck+1 , 0 ,…, 0 ) (1≤k ≤ m )

then when substitute radix variable xk+1 = ck+1 newly swaped in into new objective function,it gets

∑ ∑−

+=

−

+=+++ +′+′′=+′′=

1

2

1

2

111

n

ki

n

ki

kkkiikii bcAxabxat

viz. bk+1 = Ak+1′ck+1+bk

Because Ak+1′≤0 , ck+1≥0，then Ak+1′ck+1≤0 .

When 11 ++′

kk cA = 0 and ck+1＞0 ，both X(1)

and X(0)

are optimum solutions(now the

programming has infinite optimum solutions，refer to it in the second section of the theorem). When

11 ++′

kk cA ＜0，it has bk+1＜bk ，so X(1)

is not optimum solution，but X(0)

is optimum solution. Then

the first section of the theorem is proved. ▌

Proofs of the second section of the theorem and differentiation theorem of unbounded

solution (is omitten) are similar to this，forgive not giving uncecessary details.

§§§§ 2 Application of High Dimensional Euclidean Geometry in

non-linear programming

In the above we have introduced method of using preferred n-dimensional system method to

151

solve linear programming，now let's introduce method of using preferred n-dimensional system

method to solve non-linear programming.

Although traditional methods of solving non-linear programming take extreme value theory as

basic theory，in concrete operating process，on one hand they painfully pursue optimal value(viz.

minimum value) of t，on the other hand they don't definitely regard t of objective function t = f (X)

as function of X ，instead they always omit the variable t. A variable is cut down to all

appearances，but both actual amount of calculation and computing results are not satisfying.

In fact，even t is regarded as function of X，and extremum of t exists，the extremum is not sure

to be optimal value to be found out. In example one，we find that mistake occured in interior point

method in linear programming. In fact，extremum questions have no direct、inevitable(only

indirect、accidnetal )relations with optimal value of objective function.

Based on the above reasons，preferred n-dimensional system method doesn't begin with

extremum problems，but deal with some problems according to projective theory of the seventh

chapter in this book.

2.1 Example 2

In order to keep the continuity of ideas，we will discuss it on the basis of example 1. At the

same time, for the conveniece of description，we omit English signs' subscripts of every vertex of

convex hexahedron in example 1，of which ,quadrilateral

DCRB，DCQA and △DAB

are separately formed of simultaneousness of

equations of constraint generic plane and generic

coordinate plane ：

=

=++

=

=++

=

=++

0

66223

0

4622

0

4622

321

321

321

t

xxxR

t

xxxQ

t

xxxP

：　

：　

：　

The two quadrilaterals and a triangle and three

coordinate planes x1ox2, x2ox3, x3ox1 form an enclosed

hexahedron(figure 47).

Just like curve of second order

y = ax2

figure 47

(in plane analytic geometry, it is an equation of parabola taking axis y as principal axis)when a→0，

it can be regarded as a straight line，the cut trace of objective generic plane

25x1+20x2+15x3-t = 0

of example one on generic coordinate palne can be regarded as a generic parabola whose radius of

curvature is infinite. These two terms can be mutually transformed under some certain cases，viz.

this quadratic function can be transformed into linear function：

first rewrite this quadratic function into

02

2

1 =′−′ xxaf：

152

and regard it as trace line of S′on plane x1ox2. And the cut trace ( be marked as S0′ for the

moment)of S′on generic coordinate plane Ox1x2x3 is a parabolic cylinder of three-dimensional

space,its directrix is f，and its generating line is parallel to vector

321 2641

6

41

13eee ′+′+′ , its equation is

=

=

′−′−

′

−′′

0

041

3

412

13 32

2

31

0

t

xx

xxa

S ：　

then the equation of S′is

041

3

412

13 32

2

31 =

′

−′+

′−′−′

xx

xxatS ：

after rotation

=′

=′

−−=′

+−=′

tt

xx

xxx

xxx

33

212

11

41

4

41

5

241

5

41

4

S′becomes

( ) 034541

1

2

1354

41321

2

321 =++−

−+−− xxxx

xxa

tS：

Apparently，when a→0 and t = 0，here S will be coincident with Sπ(figure 48)of example

one graduallly.

Choose a = 0.05，it gets

figure 48

153

( ) 034541

1

2

1354

820

1321

2

321 =++−

−+−− xxxx

xxtS：

Now we make explanations of simulative economic issues for the above discussions：

A certain factory produces three kinds of products A，B，C. Assume the optimal production

plan is to planthe productions of A，B，C to be x1, x2, x3，and need operating the following three

kinds of costs and price system (unit: thousand yuan,the following is the same)at the same time：

the first： 2x1+ x2+2x3 ≤46 (P)

the second： x1+2x2+2x3 ≤46 (Q)

the third： 3x1+2x2+2x3 ≤66 (R)

x1, x2, x3≥0

The amount of profit produced ( assume it is t) is in obedience to equation of the above

objective generic curved surface S. Then it has the following model：

objective function：

( )321

2

321 345

41

1

2

1354

820

1Max xxx

xxxt +++

−+−=

constraint condition：

≥

≤++

≤++

≤++

0,,

66223

4622

4622

321

321

321

321

xxx

xxx

xxx

xxx

How to plan productions(viz. how x1, x2, x3 take values) of the three kinds of products, then it

can acquired maximum profit ?

In the example，objective function is a quadratic function，so it is a very typical non-linear

programming (belong to quadratic programming [15]

)in substance.

2.2 Methods and steps

It is generally agreed that，the core of solving non-linear programming is to determine search

direction and search range. Although it is not always necessary，but in order to illustrate questions

step by step，we also begin with these two questions.

2.2.1 Determination of search direction Assume objective function is

S：t = f (X)(X = xj ，j = 1 , 2 , … , n－1 )，

and when assume α is principal axis direction of generic curved surface t = f (X)，it has

1 and , 0 ≠⋅≠⋅ αα OtOt (viz.：α is neither perpendicular nor parallel to upright axis t ,of

which O is the origin).

Order t= t0, substitute it into the equation of S and make it simultaneous with t = t0，then it gets

the cut trace ( )

=

=−=

0

0

0

0 tt

tXfS tt ：　

of S cut by generic plane t = t0 Of which，

f (X) - t0 = 0

because it doesn't include variable t ，then it is a projective generic cylindrical surface (the concept

of projective generic cylindrical surface can be found in §2 of the eighth chapter)including

154

0ttS = , its generating line is parallel to axis t . And make equation of the projective generic

cylindrical surface simultaneous with t = 0，it gets projection

( )

=

=−′

0

00

t

tXfS ：　π

of 0ttS = on generic coordinate plane In the projection process，

0ttS = is passive set， f (X)- t0 =0 is

path set，generic coordinate plane is contain set，0ttS = is image set. This process can be descripted

as：0ttS = is through direction of axis t and be projected onto generic coordinate plane,then it

gets πS ′ .

When determining search direction，it is very simple for cored generic curved surfaces：when

πS and πS ′ are cored generic curved surfaces，of which,

πS is the cut trace of S on generic coordinate plane，then direction of connecting line between centre of πS

and centre of πS ′ can be called search direction(and length

of the connecting line can be regarded as search range，in

figure 49, “d-c-t”is the abbreviation of "directviewing

common tangent"; “ c-c-l ” is the abbreviation of

"center-connecting line";“c-t-p-c-l”is the abbreviation of

"corresponding trace point connecting line";“c-t-p”is the

abbreviation of "corresponding trace point").

But under many cases, objective generic curved

surface is coreless，it can be divided into two different kind figure 49

of methods according to different cases——translation method and inflated translation method . we

will introduce them in several：

When πS ′ is parallel and congruent the cut trace πS of S on generic coordinate plane，use

ruler to make out trace line of common generic tangent line of πS ′ and Sπon x1ox2, we call this

trace line directviewing common tangent of πS and πS ′ .Direction of this directviewing common

tangent can be regarded as search direction(figure 50).

Because this method is equivalent to translating πS to πS ′ along search direction，then this

method is called translation method (or be called simple translation method).

When πS is similar to but not congruent with πS′ ，

we can add a constant term t0 to S and make it become

S +：t + t0 = f (X) or S

+：t = f (X)- t0

then make it simultaneous with t = 0，it gets projection

( )

=

=−+

0

00

t

tXfS ：　π

of S +

on generic coordinate plane. Now two cases may

appear：one is +πS coincident with πS′ and

+πS =

πS′ (search direction can't be determined， refer to it

in figure 52)；the other is +πS not coincident with πS′ ， figure 50

but +πS ≌ πS′ ，now we can use ruler to make out their directviewing common tangent ，direction

of the common tangent can be regarded as search direction(figure 50).

This method is equivalent to inflating the directviewing figure of πS to +πS ，then translate

155

it and make it become πS′ ，so this method is called inflated translation method〔if directviewing

common tangent is not existent，we can also take direction of connecting line of corresponding

trace points of πS and πS′ (or πS′ and πS ) as search direction. Solution of corresponding trace

points：first make out tangent of trace line of +πS or πS′ through a random point of trace line of

+πS or πS′ on plane x1ox2，then make out a parallel of this tangent and make it tangent to trace line

of πS′ or +πS on plane x1ox2，the two points of tangency are the corresponding trace points of

+πS and πS′ .Solution of corresponding trace points of simple translation method is similar to this

(figure 49)〕.

But when +πS is coincident with πS′ ，search direction can't be determined，the above

discussion naturally doesn't include this kind of questions.

2.2.2 Determination of search range

So-called search range is the directviewing distances of every vertex of feasible region to the

corresponding points of Sπ(when objective generic plane is in the shape of plane,the directviewing

distance is equal to the length of line segment between trace point of generic point that the vertex is

on and foot point of perpendicular line of objective trace line). Here，we may as well look back

example one，there，the directviewing distances (can be regarded as search range) between every

vertex of convex polyhedron and objective trace line are always different in different n-dimensional

system. But if ensure that objective generic plane is in the shape of plane，then the directviewing

distances between every vertex to objective trace line are changeless. But in non-linear

programming，many objective generic curved surface can't be transformed into the shape of

hollow curved surface，this bring some difficulty to solutions of this kind of questions，so it must be

treated distinctively.

First discuss cases of objective generic curved surfaces being transformed into hollow ones. It

is divided into two cases

a.The case that search direction can be determined Assume the dimension of generic

generating line of objective generic curved surface

S：Max t = f ( xi) i =1 , 2 , … , n-1

is n-3，and it can be denoted by the following n-3 vectors

Xi1 e1 +Xi2 e2 +Xi+2 ei+2 i =1 , 2 ,… , n-3

by proper orthogonal transformation，use transform

Xi+2 ei+2 □ -Xi1 e1 - Xi2 e2 i =1 , 2 ,… , n-3

then it gets a generic coordinate plane of preferred n-dimensional system that make objective

generic curved surface in the shape of plane. Optimum point theorem of example one can be extend

to this：through trace point of generic points that every vertex of feasible region is on and make

out a parallel of search direction of objective generic curved surface ,it intersects objective trace

line(here objective trace line is a curve) at a point，then，length of line segment between the two

points is called search range from objective trace line to the vertex. When length of search range

from objective trace line to a vertex is the longest，the vertex is optimum point.

But，this method is only suitable for cases of using simple translation method to determine

search direction，but it is not suitable for cases of using inflated translation method to determine

search direction.

Under the case of inflated translation method，assume search direction has been determined，and πS′ intersects a vertex C of feasible region. Through trace of generic point that C is on and

make tangent of trace line of πS′ on plane x1ox2，then make parallel of the tangent and make it

156

tangent to trace line of +πS . The length of line segment (be called connecting line of the

corresponding trace points )between two tangent points( be called corresponding trace points) is

just the search range with relation to point C (figure 49).

Example 1.Find out optimum point of example 2.

Solution：：：：From figure 48，because when a ≠ 0 ,the figure of S is very complicated，translation is also very troublesome，so we may as well choose the figure of generic straight line

0 : =aSπ when a = 0 and translate it. The method is：first order 41

205=t (≈32.0156)，it gets

5x1 + 4x2 + 3x3 - 205 = 0

and make it simultaneous with equation 4x1-5x2=0 of trace line of generic principal axis (be called

axis trace for short, refer to it in figure 51)of Sπ,equation x3=0 and t = 0，it has

=

=

=−

=++

0

0

054

205345

3

21

321

t

x

xx

xxx

Solve the equation set , it gets x1=25 , x2 = 20 , x3 = 0 , t = 0 . This is equal to translating πS

along direction of its axis trace ，solution of the equation set is equal to coordinate of vertex Jn′(25，20，0，0) of Sπafter being translated to new position. And make the figure of πS′ (viz. when

a = 0.05 ) taking Jn′as vertex，then the translation from πS to πS′ is finished. As will be readily

seen，direction of nOJ is the search direction to be found out.

Vertex Jn′is regarded as projection trace of a generic point N of S. Choose proper direction

and unit length and make figure of upright axis t，lead Jn′to generic point N. And continuously

translate πS along direction of axis trace，then points of πS leaved in feasible region will lessen

gradually，until at last only one point left is coincident with A (continuously translate,then all points

of πS will be translated out of feasible region). From direct view，in search range,the

figure 51

157

distance between πS and vertex A is the longest，so A is the optimum point to be found.

Coordinate of a vertex Jm′of πS in the new position is

x1≈29.698, x2≈23.759 , x3 = 0 , t = 0.

Through Jm′and make parallel of axis t ,it intersects the extension line of OJn ( Jn is on

generic point N，it is opposite trace of N ) at Jm，then Jm is the opposite trace of generic point M，so the intersection point is marked as M. Because directviewingly measure | Jm′Jm |≈38.032 ,

then it gets t ≈38.032，this is in accord with the result of making directview AA′∥Ot through

A and it intersecting S at point A′.

Then it gets the optimum solution

x1 = 0 , x2 = 0 , x3 = 23 , t ≈38.032.

Viz. we shouled plan production of A 0，production of B 0，production of C 23 pieces，then we

can get the maximum profit about 38.032 thousand yuan.

b. The case that search direction can't be determined

For objective generic curved surface whose search direction can't be determined，every vertex

quality of convex polyhedron can be judged by determining expansivity. So-called expansivity is

length ratio of length of connecting line between a vertex and centre of πS to length of line

segment between intersection point of the connecting line and πS and centre of πS .

Expansivity is greater，equality is better.

When πS is in the shape of hollow，this expansivity is easy to find out in the figure. But

when πS is solid, becasuse the difficulty is very great,then it is difficult to adopt.

Example 2: Assume constraint condition is the same with the example，and objective function

is 125

541

3

41

4

41

5

400

41

13

41

5

41

4

Max

2

321

2

321

−

+++

+

+−

=

xxxxxx

t

find out optimum solution.

Solution：：：：Substitute t = 3 into objective function and make it simultaneous with t = 0，it gets the

directview figure of

=

=−

+++

+

+−′

0

0425

541

3

41

4

41

5

400

41

13

41

5

41

4

S

2

321

2

321

t

xxxxxx：π

it is coincident with

=

=−−

+++

+

+−′+

0

0425

541

3

41

4

41

5

400

41

13

41

5

41

4

S

2

321

2

321

t

t

xxxxxx：π

(figure 52)，so this is an example that search direction can't be determined. The centre of πS is

located at generic E ，trace of the generic point is

. 0,0,41

20,

41

25

−−eJ

Connecting lines EA，EB，EC，ED，ER，EQ between E and every vertex separately intersect

158

πS at points A′，B′，C′，D′，R′，Q′，and from direct view it has

, AE

EA

QE

EQ

BE

EB

RE

ER

DE

ED

CE

EC

′>

′>

′>

′>

′>

′

So，C(10，18，0，0) is the optimum point to be found out. Substitute coordinate of C into equation

of S，it gets t ≈ 22.29.

figure 52

In the above examples，we determine search range or expansivity by comparison ，linear

programming of example one also adopt this method. We call this method comparison method by

a joint name.

2.3 The case that objective generic curved surface is solid

Let's introduce solution process of this kind of questions by a concrete example，then extend it

to a generalized case.

Because objective function t is really regarded as variable，then case that objective function

is implicit function has come into consideration. Then ，we design the following objective

function into implicit function，and call it objective implicit function.

Example 3:Assume constraint condition still the same with that of example one，then

objective implicit function is

04003

32096128

3

64

15

184

15

262

25

360

3

115716

25

160

25

265

2132

121

22

3

2

2

2

1

=+−++−−

−−+−++

txxtxtx

txxxtxxxS：　

try to find out optimum solution.

Solution：：：：arrange objective implicit function into quadratic function of t

159

04009612825

36016

25

160

25

265

3

320

3

64

15

184

15

262

3

1157

2121

2

3

2

2

2

1

321

2

=

++++++−

−

++++

xxxxxxx

txxxt　

from root expression

a

c

a

b

a

b

a

acbbt −±−=

−±−=

2

22

422

4

of quadratic equation with one unknown at2 + bt + c = 0 find out

1414000102403391044526725

5888

5

8384

25

127366456560

25

563824

25

936976

1157

1

1157

160

1157

32

5785

92

5785

131

321

32

31212

3

2

2

2

1

321

+++++

++++

±

+−−−−=

xxxxx

xxxxx

xx

xxxt

Here，objective function t has two roots， it represents that πS has two intersection

opportunities with every vertex of convex polyhedron in translation. But because of solving

maximum value，then we choose

1414000102403391044526725

5888

5

8384

25

127366456560

25

563824

25

936976

1157

1

1157

160

1157

32

5785

92

5785

131

321

32

31212

3

2

2

2

1

321

+++++

++++

+

+−−−−=

xxxxx

xxxxx

xx

xxxt

Here，the directview figure of S is a solid generic paraboloid，so the above method looks

helpless. Then we borrow ideas of traditional nonlinear programming problem using limit method

(note:not extremum method) to solve to solvethis kind of questions.

Take t11= 0.89, t21= 2.19, t31= 3.47 in turn，substitute them into S successively，choose two

among six inequalities of constraint condition each time and change them into equalities (the other

four don't change ，but bracket them and put them under all equalities) then make them separately

simultaneous with S，choose the following equation sets according with constraint condition (all

solution are feasible solutions)，divide them into groups of three each according to t ，they are

called feasible groups：

[ ]

≥≤++≤++≤++

=

===

+++−−

=

=

　；；　；

：

06622346224622

0

47.3,19.2,89.0

141400010240565601157

1

1157

160

1157

32

0

0

group

feasible

row

th-

1 the

3321321321

3

2

3

3

2

1

1

xxxxxxxxxx

t

xxx

x

x

160

[ ]

≥≤++≤++≤++

=

===

+++−−

=

=

06622346224622

0

47.3,19.2,89.0

141400045267225

936976

1157

1

1157

160

5785

131

0

0

group

feasible

row

th-

2 the

1321321321

1

2

11

3

2

1

xxxxxxxxxx

t

xxx

x

x

；；　；

：

[ ]

≥≤++≤++≤++

=

===

+++−−

=

=

06622346224622

0

47.3,19.2,89.0

14140003391045638241157

1

1157

160

5785

92

0

0

group

feasible

row

th-

3 the

2321321321

2

2

22

3

1

1

xxxxxxxxxx

t

xxx

x

x

；；　；

：

these three feasible groups represent intersection points of three axes x1，x2，x3 of three edges of

convex hexahedron and gradually translatory πS′ ，feasible solutions (refer to it in table 1)found out

separately represent coordinates of these intersection points .

The same variable's value of the same t of different feasible groups is usually different，we call

their non-negative difference value absolute difference of corresponding solutions of the feasible

groups. But，there are only three non-negative differences of the same variable of the same t ，we

only choose the maximum and call it maximum absolute difference.

Table 1 gives maximum absolute difference of corresponding solutions of the above feasible

groups：

table 1(accurate to 0.01)

Value

of t

the orre-

sponding

solution

the 11

-th feasible

group

the 21

-th feasible

group

the 31

-th feasible

group

maximum absolute

difference

x1 0 0 0 0

x2 0 0 0 0

0.89

x3 0 0 0 0

x1 0 9.1 0 9.1

x2 0 0 11.6 11.6

2.19

x3 11.8 0 0 11.8

x1 0 17.9 0 17.9

x2 0 0 23 23

3.47

x3 19.5 0 0 19.5

It can be observed from table 1，with the increase of value of t ，absolute differences of the

corresponding solutions will increase gradually. But we expect these absolute differences not only

decrease gradually but also tend to zero quickly.

161

And choose t12= 4.1 , t22= 4.2 , t32= 4.22 , t42= 4.24 , t52= 4.2588

in tur, and substitute them into objective function S successively，choose two among six inequalities

of constraint condition each time and make them simultaneous with S which is substituted into

new t value ，and choose solutions found out which are all equation sets of feasible solutions，divide them into three groups of five each according to t：

[ ]

≥≥≤++≤++

=

=====

+++

++++

+−−−

=

=++

0066223 4622

0

2588.4,24.4,22.4,2.4,1.4

1414000339104452672

25

1273664

25

563824

25

936976

1157

1

1157

160

5785

92

5785

131

0

4622

group

feasible

row

th-1 the

21321321

21

21

2

2

2

1

21

3

321

2

xxxxxxxx

t

xx

xxxx

xx

x

xxx

，，，　

：

[ ]

≥≥≤++≤++

=

=====

+++

++++

+−−−

=

=++

004622 4622

0

2588.4,24.4,22.4,2.4,1.4

1414000339104452672

25

1273664

25

563824

25

936976

1157

1

1157

160

5785

92

5785

131

0

66223

group

feasible

row

th-2 the

21321321

21

21

2

2

2

1

21

3

321

2

xxxxxxxx

t

xx

xxxx

xx

x

xxx

，，，　

：

[ ]

≥≥≥≤++

=

=====

+++++

+++++

+−−−−

=++

=++

000 4622

0

2588.4,24.4,22.4,2.4,1.4

14140003391044526725

5888

5

8384

25

127366456560

25

563824

25

936976

1157

1

1157

160

1157

32

5785

92

5785

131

0223

6622

group

feasible

row

th-3 the

321321

213231

212

3

2

2

2

1

321

321

321

2

xxxxxx

t

xxxxxx

xxx

xx

xxx

xxx

xxx

，，，　

：

these three feasible groups represent intersection points of three edges QC，RC，DC of convex

162

hexahedron separately intersecting πS′ five times，coordinates of these intersection points are just

solutions of these three feasible groups. Table 2 give all feasible solutions of these three feasible

groups.

table 2(accurate to 0.0001)

Value

of t

the orre-

sponding

solution

the 12

-th feasible

group

the 22

-th feasible

group

the 32

-th feasible

group

maximum

absolute

difference

x1 8.078 17.77 10 9.692

x2 18.961 6.345 16.7656 12.616

4.1

x3 0 0 1.22 1.22

x1 9.2926 12.342 10 3.0494

x2 18.3537 14.487 17.5554 3.8667

4.2

x3 0 0 0.442 0.442

x1 9.5336 11.504 10 1.9704

x2 18.2332 15.744 17.702 2.4892

4.22

x3 0 0 0.2597 0.2597

x1 9.7742 10.71 10 0.9358

x2 18.1129 16.935 17.8591 1.1779

4.24

x3 0 0 0.14 0.14

x1 10 10 10 0

x2 18 18 18 0

4.2588

x3 0 0 0 0

It can be observed from table 2，absolute differences of the corresponding solutions decrease

gradually，it represents that the distances between intersection points of πS′ and three edges QC，RC，DC also decrease gradually. At last， πS′ intersects these three edges at optimum point C.

This method is also applicable for the case that objective generic curved surface is "hollow"，it is also applicable for some linear programmings.For example in example one，when t separately

choose 570、590 and 610 , S separately intersect DC1, Q2C1 and R1C1 at three points，tabel 3

lists coordinates of these intersection points：

table 3

Value

of t

the orre-

sponding

solution

DC1

Q2C1

R1C1

maximum absolute

difference

x1

10

3

22

18

3

32

x2

10

3

58

6

3

40

570

x3 8 0 0 8

x1

10

3

26

14

3

16

x2

14

3

56

12

3

20

590

x3 4 0 0 4

x1 10 10 10 0

x2 18 18 18 0

610

x3 0 0 0 0

163

As will be readily seen from table 3，when t is far from maximum value ，the maximum

absolute difference is also much more than zero；When t takes maximum value 610，maximum

absolute differences of the corresponding solutions are all zero.

In order to differentiate it from comparison method introduced before，we call this method

absolute difference method.

In non-linear programming field，because of lacking of ideal tools in the past ，so a kind of

all-purpose common algorithm hasn't been formed. But，after preferred n-dimensional system

appeared ，this situation will change much. Here “absolute difference method” introduced will be

general all-purpose algorithm of non-linear programming. aithough this method is lack of use value

because of its complexity，but by people's studious exploration，this method will become

mainstream algorithm of linear programming and non-linear programming questions ultimately.

Now，we will extend example 3 to a general case：In general quadratic programming，constraint conditions are most linear inequalities. In order to reduce discussion difficulty ，we also

assume constraint conditions as linear inequalities.

Assume a generic curved surface

S：t = f (X) ( X = xj；j = 1 , 2 , …, n-1)

of n-dimensional space is regarded as figure of objective generic curved surface，each equation of

system of linear equations

gi ( X )≤ai ( i =1 , 2 ,…, r )

X≥0

is regarded as each generic plane forming simplex(viz. convex polyhedron). And assume cut trace

of S on generic coordinate plane is ( )

=

=

0

0

t

XfS ：　π

Choose a proper value of t as initial value(this is similar to interior point method determining initial

point，but they are also quite different two things)and substitute it into S then make it simultaneous

with t = t0，then it gets the cut trace

( )

=

==

0

0

0 tt

tXfS tt ：　

of S cut by generic plane t = t0 Of which，f (X) = t0 is a projective generic cylinder surface including

0ttS = and its generating line parallel to axis t .Make equation of the projective generic cylinder

surface simultaneous with t = 0 ，it gets projection

( )

=

=′

0

0

t

tXfS ：　π

In each feasible group， πS′ intersects one edge of convex polyhedron. Because under normal

conditions，the number of constraint equations is not more than n－1，so， πS′ will at most separately

intersect k edges(k≤n－1) of convex polyhedron at one point

( xk1 , xk2 , … , xk,n-1 , 0 ) ( k = 1 , 2 , … , r ；r≤n－1 ).

Then make t take t1 , t2 , … , tδ in turn , make t0＜t1＜t2＜…＜tδ ，and make

πS′ change position constantly and separately intersect the k edges of convex polyhedron at one

point ( xl k1 , xl k2 , … , xl k,n-1 , 0 ) ( l =1 , 2 , … ,δ； k = 1 , 2 , … , r ；r≤n－1 ) one after

another. Coordinate of the point can be found out by using the following method：

In all r+n-1 inequalities of system of linear equations

gi (X)≤ai ( i = 1 , 2 , … , r )

164

X≥0，

take those inequalities which satisfy the given tm ( m =0 , 1, 2, …, δ；δis positive integer) and

change them into equalities, every n-2 of them can be a group and separately be simultaneous

with

( ) ( )

=

==

0

,,1,0

t

mtXf m δL

（the other r+1 inequalities keep changeless，but bracket them and put them under all equalities）and form feasible group（can form n－1 feasible groups at most）.

( )

( ) ( )

( )

−≤≤−=≥

≤≤−=≤

=

==

−≤≤−−==

≤≤−−==

11,,2,1 0

1,,2,1

0

,,2,1

112,,2,1,0 0

12,,2,1,0

group

feasible

row

th-

the

222)(

222)()(

111)(

111)()(

22

2222

1 1

1111

njpnqx

riqnpaXg

t

mtXf

njpnqx

riqnpaXg

k

qj

pipi

m

qj

pipi

l

；　

；　

；　

；　

：

L

L

L

L

L

δ

(in the formula，r is the rank of coefficient matrix of system of linear equations gi (X)≤ai) and then

separately solve them .

Assume xημ and xξμ are the same variable belonging to different feasible group，then xημ and

xξμ are called corresponding solutions of πS′ ，and 0≥− ξµηµ xx is called absolute difference of

the corresponding solution. When πS′ approximates to optimum point，it has 0→− ξµηµ xx .

But，when 0=− ξµηµ xx ， S will not always has optimum solution. Near some

non-optimum points of convex polyhedron，it also has the case 0→− ξµηµ xx . Assume when

0=− ξµηµ xx ,correspondingly it has t = tμ，if we want to check whether tμ has maximum

value or not，if only choose a minimum value Δ(Δ＞0，Δ should correspondent with the exact

value required.For example，if we demand it accurate to 0.0001, then we can order Δ=0.0001) and

substitute t = tμ+Δ into objective function S and make it simultaneous with other equations of

( )( ) ( )! 1! 2

!1

+−

−+

rn

nr 　　equations，find out whether feasible solutions exist or not . If all solutions are

unfeasible solutions ，then t = µt is the maximum value；If feasible solutions still exist ，then

t = µt is not the maximum value，computation should be carried on.

Considering equation sets which have unfeasible solutions ( be called unfeasible sets) can also

be determined by computation actually，then we will compute ( )

( ) ( )! 1! 2

!1

+−

−+

rn

nr 　　 equation sets

each time，the workload is very great. So，we must manage to simplify computation steps . In the

following we will introduce two methods of simplifying computation steps：graded search

methodand unconstrained search method.

First，graded search method. The concrete practice is，divide constraint condition into two

sections：

The first section is called preferred constraint equation ，it is composed of principal part

g (X)≥0 of constraint condition. According to what we introduced in §1，constraint condition can

be classified by magnitude of constraint force. So this section of constraint conditions all have great

constraint force；

165

The second section is called prevailing(or be called recognized) constraint equation，it is

composed of subsidiary section X≥0 of constraint condition . Constraint force of this section of

constraint condition is usually very weak.

Because in the model we created, πS′ usually translates from near the origin along search

direction ，its trace line moves to the first quartile of plane x1ox2 gradually. So in feasible groups(or

unfeasible groups)，constraint conditions all composed of prevailing constraint equations are called

initial group；Constraint conditions all composed of preferred constraint equations are called

terminative group；Constraint conditions composed of both preferred and prevailing equations are

called intermediate group. Of which，according to the amount of prevailing equations included in

intermediate group, it can also be divided into first grade intermediate group，second grade

intermediate group，…，etc.

In example 3，the11-th, 21-th, 31-th feasible groups are all initial groups，the 12-th, the

22-th feasible groups are intermediate groups(because in the example there is only one prevailing

constraint equation in the intermediate groups，so they are called intermediate group a joint

name)，the 32-th feasible group is terminative group.

Considering optimum solution is usually near terminative group，so in the computation of the

11-th , the 21-th , the 31-th three feasible groups，the value of t increases very fast，computational

accuracy is also very low(in fact， the value of t may increase much faster，accuracy may be even

lower. Even，we can consider giving up the computation of initial group and directly carry the

computation of intermediate group or terminative group)；And in the 12-th , the 22-th , the 32-th

feasible groups，the increase speed of the value of t slow down，computational accuracy increased.

Second，unconstrained search method.

In spite it is feasible group or unfeasible group，we choose one among each of them and

compute together. For example in the equation set

( )

( ) ( )

=

==

−≤≤−−==

≤≤−−==

0

,,2,1

112,,2,1,0 0

12,,2,1,0

111)(

111)()(

1 1

1111

t

mtXf

njpnqx

riqnpaXg

m

qj

pipi

δL

L

L

；　

；　

( )

−≤≤−=≥

≤≤−=≤

11,,2,1 0

1,,2,1

group

positive

-contra

222)(

222)()(

22

2222

njpnqx

riqnpaXg

qj

pipi

；　

；　

L

L

inequalities bracketed are deleted，it is equivalent to being deleted constraint condition，so it is

called unconstrained search method. But constraint condition can't be wholly ignored，or else we

can't determine solutions found out are feasible solutions or not. So we put constraint condition

behind equation set as contrapositive group.

This equation set represents an edge and its extension line of convex polyhedron.

In concrete compution，we usually choose two adjacent edges' equations of the same vertex

of convex polyhedron.

Determine whether two equation sets represent two adjacent edges of the same vertex of

convex polyhedron or not，the method is very simple：according to interleaving theorem (refer it in

theorem 1——theorem 4 of the sixth chapter) between two linear figures of the sixth chapter，each

equation set has n-1 equalities，of which, n-2 equalities are the same，only one is different. When

the two equation sets are simultaneous，simultaneous group has the unique solution ，and the

166

unique solution is feasible solution.

But，solving the unique solution of simultaneous group of two eqution sets need increase

amount of calculation invisibly. So，in the two equation sets，we choose one having feasible

solution，the other has no feasible solution. Of which，equation set having feasible solutions (can be

called the original equation set for convenience) represents an edge of convex polyhedron，equation

set having no feasible solution represents extension line of the edge's adjacent edge. When

πS′ translates along search direction，absolute difference of corresponding solutions of two equation

sets should decrease gradually.If absolute difference of corresponding solutions of two equation

sets doesn't decrease but increase gradually，it represents that vertex of intersection position of two

edges has fallen behind πS′ ，now we should give up the adjacent edge's equation set(be called old

equation set for the convenience of description)having unfeasible solution，and choose the other

equation set(be called new equation set) having unfeasible solution. In order to ensure that absolute

difference of corresponding solutions between new equation set and the original equation set

decreases gradually，adjacent edge(its extension line) denoted by the new

equation set should intersect edge denoted by the original equation set at another vertex. This

demand that new equation is secondary intermediate group of the original equation set, To be

specific，the new equation set should have one prevailing constraint equation less than that of the

original equation set but one preferred constraint equation more than that of the original equation

set.

When absolute difference is zero， πS′ translates to intersection point of two edges of convex

polyhedron denoted by two equation sets. Further translate it，then absolute difference again

increases gradually， πS′ changes from intersecting extension line of one of two edges to intersecting

this edge，and originally it changes from intersecting the other edge to intersecting its extension

line .So，as long as absolute difference gradually decreases at the beginning，and suddenly increases

gradually in the sequel，it shows that πS′ has translated intersection point of the two edges.Now we

can choose another extension line of an edge having common point with the new edge and make

itcontinuously intersect the translating πS′ .……，until the optimum point is found. For example，in example 3，

≥

≤++

≤++

≤++

=

=

=

0

66223

4622

4622

ositive

-contrap

0

0

0

2

321

321

321

3

1

x

xxx

xxx

xxx

t

x

x

and

≥

≥≤++

≤++

=

=

=++

0

0

66223

4622

ositive

-contrap

0

0

4622

2

1

321

321

3

321

x

x

xxx

xxx

t

x

xxx

167

are separately equations of OQ and CQ . Take 2.418, 2.937 and 3.47 of t successively，and order

x3=0，substitute it into S ,it gets

3.472.937,2.418,

1414000339104452672

25

1273664

25

563824

25

936976

1157

1

1157

160

5785

92

5785

131

21

21

2

2

2

121

===

=

+++

++++−−−

xx

xxxxxx

and make them separately simultaneous with equations of OQ and CQ，the results are listed in table

4：

table 4( accurate to 0.001)

Value

of t

the orre-

sponding

solution

OQ

CQ

maximum absolute

difference

x1 0 -23.4 23.4

x2 13.71 34.7 20.99

2.418

x3 0 0 0

x1 0 -8 8

x2 18.3 27 8.7

2.937

x3 0 0 0

x1 0 0 0

x2 27 27 0

3.47

x3 0 0 0

It is observed from data of the table，when t＜3.47 of objective function S，solution of

equation of CQ is unfeasible solution .So now πS′ intersects CQ at extension line. When t≥3.47，πS′ intersects CQ in feasible region，but intersects extension line of OQ in the outside of feasible

region. Now we can change equation of OQ into equation of CR and carry on the computation.……

In this method， πS′ continuously intersects another and yet another edge of convex

polyhedron，so，we call this method “one edge method”. Traditional method(such as interiorpoint method) using extreme value theory to solve

non-linear programming tries to include objective function and constraint condition these very

complex equations all in an equation. Although this idea is very new，it is unavoidable to be too

naive. Because such a result will always change property of equation ，its accuracy and reliability

are doubtable. And because the method is too complex，it is not only very inconvenient in

application，but also greatly limits people's study course in nonlinear programming field.

Preferred n-dimensional system method is an example of successfully applying High

Dimensional Euclidean Geometry .

Because of the appearance of preferred n-dimensional system method，we may observe

problems and consider problems at a higher initial point，so，many concepts which are very

ambiguous in the past become transparent. For example constant distance theorem and optimum

point theorem of §1 in this chaper，only on the basis of oblique axes transform and oblique axes

draughting ,it can be found and successfully solved.

Preferred n-dimensional system method regard objective function t as a variable，it is one

dimension more than traditional sloution in spacial thinking manner. The change of thinking

manner brought by preferred n-dimensional system graphical method help us realize the transform

from planar thinking manner to spatial thinking manner. What does this imply? It implies that we

put objects to be observed and studied all in a very spacious planar region (viz. generic coordinate

168

plane)，but we stand in in the air to observe and study them i. Apparently，it has a more wide-field

than we standing in the original planar region to observe and study them，and more

easily to grasp the overall situation.

References

1．Edited by Peking University Higher Algebra Beijing：People's Education Press，1978

2．Hongfan Basis of Discrete MathematicsWuchang：Huazhong Institute of Science and

Technology Press，1983

3．Edited by School of Mathematics of Nankai University Introduction of Analytical Geometry

of Three Dimensions Beijing：People's Education Press，1978

4．NI Guoxi Common Matrix Theory and Method Shanghai：Shanghai Scientific and

Technical Publishers，1984

5．Zhang Rijin，Li Shenglin Graphics Set of Stereoscopic Projection Beijing：National

Defence Industry Press，1979 6 ［USSR］Qialei et al. wrote Zhangyan et al. translatedDescriptive Geometry Beijing：Higher Education Press，1955

7．Edited by Tsinghua University Higher Mathematics Beijing：People's Education Press，1964

8．［USSR］Felibf wrote Xie Shenjian，Zhou Jiyi translated Hyperspace Descriptive

Geometry and Its Application Beijing：Tsinghua University Press，1983

9．［USA］Einstein wrote Lihao translated Relativistic Significance Beijing：Science

Press，1961

10．Edited by School of Mathematics of Wuhan University Linear AlgebraBeijing：People's

Education Press，1977

11．Fan Yingchuan et al. wrote Teaching Materials of High Mathematics Beijing：Higher

Education Press，1964

12．《Mathematics Handbook》Compile GroupMathematics Handbook Beijing：People's

Education Press，1979

13．Jian Zhaoquan Solving Included Angle Between Two Planes of E n By Unit Circle

Method Engineering Graphics Archieves，1986(1)

14．Edited by Beijing Institute of Education Higher Algebra Beijing：Beijing Press，1979

15．Qian Songdi et al. wrote Operations Research Beijing：Tsinghua University Press，1990

169

Postscript

Owing to help and guidance of many experts and scholars and warm-hearted people of all

circles，then the author can dedicate its self taught and studied achievement of 17 years to readers

of wide scope. In initial stages of study，many scholars and experts gave the author illuminative guidance. Of

which，the most outstanding ones are Mr Li Shenglin of BeiHang University、Mr You Chengye of

Peking University、Mr Mei Xiangming of Capital Normal University ,Mr Fu Ruonan of Beijing

Normal University and Lu Daxiong of Beijing Institute of Education. When the study made some progress，Mr Zhou Jiyi of Tsinghua University、Mr Zhang Yunhe of

Beijing Institute of technology 、Mr Zhang Rijin of College of Building Materials and Light

Industry of Beijing Union University、Mr Liu Guangxu of Tianjin Nankai University and Mr Zhu

Xinmin of 《Guangming Daily》Press gave acknowledge and supporting in time，and made some

basic ideas and methods be published Beijing《Engineering Graphics Archieves》and《Potential

Science Magazine》，and be lucky to be published in annual meeting of 1987 in Beijing Engineering

Graphics Society.

Section of making a point of this book (the first chaper)is finished with the cooperation of

professor Hou Bingtao of professor Hou Bingtao of Beijing Institute of armored forces of Chinese

People's Liberation Army. Mr Hou thoroughly corrected the contents of some theorems and gave

strict and scientific proof to most theorems in person after reading over the manuscript of the first

chapter for many times，then section of making a point of this book kept its legs steady in science at

last. He also called section of making a point and methods extended by it "relation " method.

At the beginning this book is given to Professor Shen Yidan of Beijing Institute of Technology

in the form of series of manuscripts .Mr Shen carefully checked and approved series of manuscripts

and gave high appraise for them. He suggested that the author arrange series of manuscripts into

book manuscripts and directly publish it. After the book manuscripts are finished，Mr Shen

reviewed and embellished it for many times patiently，and worked out name for this book in person.

He also rushed about in many ways，appealed to concerned departments and experts to give

attention to ideas and methods of this book. And contacted issuing office in person using his

reputation and relation, did his best to make this book be published.Herein the author gave cordial

thanks and heartfelt admiration to him .

Besides，researcher Mr Pan Jianzhong of Geometry and Topology Research Office of

Graduate School of Chinese Academy of Sciences gave enthusiastic guidence to revision and

modification of this book，and strictly checked it on，hereon I especially give thanks to him.

During publication of this book，Mr Li Yingan of Atomic Energy Press 、Mr Ma Bohua of

Population Press、Mr Liu Wenchao of Military Medica Science Chinese、Mr Lan Chaowen of

Institute of Automation of Chinese Academy of Sciences and Mr Chen Linfeng of Beijing No.606

post box also gave great support，hereon I also give thanks to them!

Auto CAD making of iconographs in this book are finished under the guidance of Mr Bu

Yanwu，I also especially give thanks to him!

Limited to qualification of the author，there must be much error in this book，readers are

expected to point out them so that they can be corrected.

Contents of the ninth chapter aren't agreed with or acknowledged by any experts，the author is

responsible for the responsibility .

170

Compiler

July,1996 Beijing

Solutions to part exercises The first chapter

1.1 be changed into 15e1-14e2.

1.2 be changed into 7

33

7

1321 ee ′−′ .

1.3 -5e1+e2.

1.4 59e1-34e2.

1.5 ( ) ( ) 21 2472812 ee −−− .

The third chapter

3.4 Because generic points are divided into sufficient-order、difficient-order and zero-order，

then accordingly,there are three graphic methods of general graphic method、indirect graphic

method、and direct method.

3.8 36x1+126x2-198x3-132x4-88x5-99x6-72=0.

3.10 x1=8, x2= 4, x3=6, x4=10, x5=12, x6=18, x7=4 .

The sixth chapter

6.1 L1、L2 are mutually interleaving planes，the distance between them is about 5.5149.

The seven chapter

7.6 φ≈71°45′.

7.7 φ≈47°48′.

7.8 φ≈89°53′.

7.9 φ≈73°34′.

The eighth chapter

8.4 φ3≈74°12′(the others are all zero angles).

8.5 φ1≈25°31′， φ2≈64°30′.

8.6 φ1≈64°04′， φ2≈33°46′.

8.7 The characteristic root found out is λ1=1,

, 6764.1328659 , 3236.116328659 32 ≈−=≈+= λλ

so the nonzero included angle between two figures are φ2≈84°41′, φ3≈39°26′.

171

Indexes of special terms or symbols

vocable or symbol the ubiety of pages

"relation" method-----------------------------------------------------------------------------------------------1

has relation σ：σ ------------------------------------------------------------------------------------------2

has no relation σ： --------------------------------------------------------------------------------------2

overlapping direction-----------------------------------------------------------------------------------------14

oblique axes draughting--------------------------------------------------------------------------------preface

preferred n-dimensional system-----------------------------------------------------------------------------17

image relation sign“□”------------------------------------------------------------------------------------19

shadow relation sign“○”-----------------------------------------------------------------------------------19

oblique axes transform-----------------------------------------------------------------------------------------19

principal overlapping direction-------------------------------------------------------------------------------22

principal overlapping space-----------------------------------------------------------------------------------22

generic point---------------------------------------------------------------------------------------------------23

sufficient-order generic point--------------------------------------------------------------------------------30

difficient-order generic point--------------------------------------------------------------------------------23

zero-order generic point--------------------------------------------------------------------------------------23

projective trace------------------------------------------------------------------------------------------------24

opposite trace--------------------------------------------------------------------------------------------------24

generic curved surface----------------------------------------------------------------------------------------28

generic curve---------------------------------------------------------------------------------------------------28

generic plane---------------------------------------------------------------------------------------------------30

generic straight line-------------------------------------------------------------------------------------------30

generic coordinate plane--------------------------------------------------------------------------------------30

generic sphere--------------------------------------------------------------------------------------------------32

generic circle---------------------------------------------------------------------------------------------------42

generic cylinder surface---------------------------------------------------------------------------------------42

solid generic curved surface----------------------------------------------------------------------------------56

hollow generic curved surface-------------------------------------------------------------------------------56

generic circular conical surface------------------------------------------------------------------------------59

generic paraboloid---------------------------------------------------------------------------------------------61

projective generic cylinder surface-------------------------------------------------------------------------113

dimension theorem--------------------------------------------------------------------------------------------29

direct graphic method-----------------------------------------------------------------------------------------30

simplex principal overlapping direction---------------------------------------------------------------------36

oblique number“S(xs)” ----------------------------------------------------------------------------------36-37

oblique sign“A(xb)”-----------------------------------------------------------------------------------------37

indirect graphic method--------------------------------------------------------------------------------------30

general graphic method--------------------------------------------------------------------------------------30

points having common generic------------------------------------------------------------------------------33

points having common generic sign“ ”--------------------------------------------------------------33

points having common generic theory-----------------------------------------------------------------33-34

leading-axis method-------------------------------------------------------------------------------------------59

cutting-trace method------------------------------------------------------------------------------------------59

172

synthetical graphic method-----------------------------------------------------------------------------------61

singly through--------------------------------------------------------------------------------------------------62

mutually through-------------------------------------------- -------------------------------------------------73

through point--------------------------------------------------------------------------------------------------73

mutually interleaving-----------------------------------------------------------------------------------------74

orthodromic vector-------------------------------------------------------------------------------------------77

orthodromic space----------------------------------------------------------------------------------------76-77

common orthodromic space---------------------------------------------------------------------------------78

normal space--------------------------------------------------------------------------------------------------78

common-orthodromic normal space-----------------------------------------------------------------------78

common normal space---------------------------------------------------------------------------------------79

exterior direct sum -------------------------------------------------------------------------------------------81

exterior sum---------------------------------------------------------------------------------------------------81

passive set------------------------------------------------------------------------------------------------------87

path set----------------------------------------------------------------------------------------------------------87

accepting set---------------------------------------------------------------------------------------------------87

Jianshi solution theorem-------------------------------------------------------------------------------------90

angle vector----------------------------------------------------------------------------------------------------92

orthogonal-unit vector----------------------------------------------------------------------------------------92

common vector------------------------------------------------------------------------------------------------93

uncommon vector---------------------------------------------------------------------------------------------93

Jianshi solution solution-------------------------------------------------------------------------------------94

Jianshi method------------------------------------------------------------------------------------------------94

exterior product method and inner product method----------------------------------------------------106

exterior product----------------------------------------------------------------------------------------------106

Jianshisolution principle------------------------------------------------------------------------------------118

coefficient theorem------------------------------------------------------------------------------------------120

exterior product method and projective method---------------------------------------------------------132

objective trace line-------------------------------------------------------------------------------------------141

constant distance theorem----------------------------------------------------------------------------------142

optimum point theorem-------------------------------------------------------------------------------------144

translation method--------------------------------------------------------------------------------------------154

inflated translation method----------------------------------------------------------------------------------154

axis trace------------------------------------------------------------------------------------------------------156

expansivity----------------------------------------------------------------------------------------------------157

comparison method-------------------------------------------------------------------------------------------158

objective implicit function-----------------------------------------------------------------------------------158

feasible group--------------------------------------------------------------------------------------------------159

absolute difference--------------------------------------------------------------------------------------------160

absolute difference method----------------------------------------------------------------------------------163

unfeasible group-----------------------------------------------------------------------------------------------165

graded search method----------------------------------------------------------------------------------------164

initial group----------------------------------------------------------------------------------------------------165

terminative group---------------------------------------------------------------------------------------------165

intermediate group--------------------------------------------------------------------------------------------165

unconstrained search method--------------------------------------------------------------------------------164

173

Appendix：：：：solution process of part exercises The First Chapter

1.1 solution：

under this relaton expression，arrange and transpose the right end of vector

3e1+5e2-4e3σ 3 (3e1-5e2)+5 (-2e1+e2)-4 (-4e1+e2 ) ，

it has 15e1-14e2

viz. the vector become 15e1-14e2 after the transform.

1.2 solution：

because submatrix composed of the front rows of coefficients matrix is

−

−

12

53

we found out inverse matrix of the submatrix is

−−

−−

7

3

7

27

5

7

1

so

=

−−

−−•

−

−

−

=

7

17

7

2

1

1

7

3

7

27

5

7

1

14

12

53

M

now，the relation of exercise 1.1 becomes

′+′

′

′

213

22

11

7

17

7

2eee

ee

ee

σ

σσ

substitute the new relation expression into vector

3e1+5e2-4e3，

it has 2121217

33

7

13

7

17

7

2453 eeeeee ′−′=

′+′−′+′

viz.the vector becomes 217

33

7

13ee ′−′ .

1.3 Solution：

substitute the transform into vector 4e1-5e2+3e4，the vector becomes

4e1-5e2+3 (-3e1+2e2 )= -5e1+e2 . 1.4 solution： 5e1-4e2+6 (9e1-5e2 )=59e1-34e2 ，

the vector becomes 59e1-34e2. 1.5 Solution：

174

( ) ( ) , 24728122

1

2

28712 212121 eeeeee −−−=

+−+−

viz. the vector becomes ( ) ( ) 2472812 21 ee −−− .

The Second Chapter

2.7 Solution：

abbreviated formula of transform in exercise 2.6 is

−

−

+−

+−

214

213

214

213

456

23

456

23

eee

eee

eee

eee ○○□□

2.11 Solution：

the corresponding oblique axis transform is

.

489

485

483

489

485

483

213

213

213

213

213

213

　○○○□□□

+

+

+

−−

−−

−−

eee

eee

eee

eee

eee

eee

The sixth chapter

6.1 solution：

from theorem 4，L1 and L2 are two mutually interleaving planes，and they have a

one-dimensional common-orthodromic space. The common-orthodromic space can be denoted by

vector {11，-13，-4，1}. Separately find out another orthodromic vector

{0，10，9，-16}， {-6，0，1，16}

on L1 and L2 ，then the common normal space is

=−+

=++−

=+−−

016910

0166

041311

432

431

4321

xxx

xxx

xxxx

T：　

one solution vector of T is {715,715,-286,286}.

through one point

−− 0,0

5

3,

5

4A

of L1 , and make generic plane

S：715x1+715x2-286x3+286x4+1001=0

taking this vector as normal vector This is the equation of exterior sum

120 LLS +=

175

(L20 is a new plane gainded by translating L2 and making it intersect L1 at point A).order

,22

⊥′LL I

it has

=++−

=+−−

=−+++

=−+−+

0166

041311

05223

030432

131

1321

1321

1321

xxx

xxxx

xxxx

xxxx

M：　

it gets intersection point M1 (2,3,-4,1) .

translate T and make it through point M1 and intersect S at

=+−+

=++−

=+−−

=++−+

02216910

0166

041311

01001286286715715

432

431

4321

4321

2

xxx

xxx

xxxx

xxxx

M ：　

it gets the intersection point

−−−−

4147

1859,

4147

10582,

4147

2574,

4147

67212M .

Then ，distance

5149.54147

1666.22870

6006600615015150154147

1 2222

21

≈≈

+++=MM

between the two points is just the distance between L1 and L2 .

The Seventh Chapter

7.6 solution：

from formula(1)，

,3131.09999

31

942536916144251649

3410301228cos ≈=

++++++++++

−−−−−=ϕ

so it gets φ≈71°45′.

7.7 solution：

from formula(2),direction vectors of the two straight lines separately are

H1={2,-44,-87,-74}, H2={6,-16,-8,-2},

from formula(3) ,6717.036014985

1560cos

21

21 ≈=•

•=

HH

HHϕ

so φ≈47°48′.

7.8 solution：

two planes intersect at the straight line

176

=++−−

=−+++

=+−−+

054353

0233

035432

4321

4321

4321

xxxx

xxxx

xxxx

H：　 ，

then the orthodromic vetor of H is H={79,-82,93,-92}.

taking the vector as normal vector and make generic plane separately intersecting I1 , I2 ,it gets

straight line

=−+−

=−+++

=+−−+

092938279

0233

035432

4321

4321

4321

1

xxxx

xxxx

xxxx

H ：　

=−+−

=++−−

=++−−

092938279

06246

054353

4321

4321

4321

2

xxxx

xxxx

xxxx

H ：　

separately found out H1={1372,-853,-1050,877}, H2={913,956,2217,2173}.

so， , 0021.0113845234481622

15039cos ≈=ϕ φ≈89°53′.

7.9 solution：

because it found out H1={2,-44,-87,-74},

then it is known that ⊥′

1H is a generic plane

F：2x1-44x2-87x3-74x4= 0

use F∩I1 ,it gets the straight line

=−−−

=−+++

=+−−+

07487442

0233

035432

4321

4321

4321

xxxx

xxxx

xxxx

H：　

H={124,-223,586,-553}

H⊥′：124x1-223x2+586x3-553x4= 0

=−+−

=−+++

=+−−+

0553586223124

0233

035432

4321

4321

4321

2

xxxx

xxxx

xxxx

H ：　

H2={8031,-5802,-2371,1628}

is found out,so , 2828.010643219014985

357155cos ≈=ϕ φ≈73°34′.

The Eighth Chapter

8.4 solution：

arrange the equation of generic plane into

177

,2

4

2

5

2

3 3214

xxxx −+=

substitute it into the equation of unit generic sphere

, 12

4

2

3

2

2

2

1 =+++ xxxx

it gets ,4402430202913 323121

2

3

2

2

2

1 =−−+++ xxxxxxxxx

its characteristic matrix is

−

−−

−−

202012

202915

121513

λ

λ

λ

：　G

make elementary transformation

( ) ( )( )

⇒

+−−−

−−−

−−

⇒

−

−−

−−

116334150

45430

131512

122020

152920

131512

2 λλλ

λλ

λ

λ

λ

λ

( ) ( )( )( )

−−

−⇒

+−

−−−

54400

040

001

2165800

45430

001

2 λλ

λ

λλ

λλ

for G then the equation of the generic sphere becomes

,14

54 2

3

2

2

2

1 =++ xxx

so， cosφ1=cosφ2=1,

,2722.054

2cos 3 ≈=ϕ

becauseφ1=φ2=0, then choose φ3≈74°12′to define the included angle between the two

generic planes.

8.5 solution：

determinant of coefficient

, 1145

14−=

−−=D

of the two posterior terms of the plane,

( )( )

( )( )

, 21935

234

, 51343

123 but

21

21

21

2

21

21

21

1

xxxx

xxD

xxxx

xxD

−=+−

−−−=

−−=+−

−−−=

so it gets , 11

2

11

19 ,

11

5

11

13214213 xxxxxx +−=+=

substitute it into the equation of unit generic sphere,it gets

178

, 121154150651 2

2

2

2

1 =++ xxxx

from coefficient theorem，it has

( )

( ) ,3922.5651150541506511212

1

,2277.1651150541506511212

1

22

22

≈ −+++×

=

≈ −+−+×

=

B

A

so， cosφ1≈0.9025, cosφ2≈0.4306， φ1≈25°31′，φ2≈64°30′.

8.6 solution：

determinant of coefficient , 1625

42=

−−=D

( )( )

( )( ) 21

21

21

2

21

21

21

1

1924325

542

216232

454

xxxx

xxD

xxxx

xxD

+=+−

−−−=

+−=+−

−−−=

of the two posterior terms of I1 it gets

. 16

19

16

24 ,

16

2

16

16214213 xxxxxx +=+−=

substitute it into the equation of unit generic sphere ,it gets

, 2568486211088 21

2

2

2

1 =++ xxxx

from coefficient theorem，it gets

( ) ( )

( )

, 8313.04564.370

16cos , 4373.0

5436.1338

16cos

, 4564.37093719317092

1

, 5436.133893719317092

110886218486211088

2

1

21

22

≈=≈=

≈−=

≈+= −+++=

ϕϕ

B

A

φ1≈64°04′，φ2≈33°46′。

8.7 solution：

determinant of coefficient , 4

212

423

652

−=

−

−−

−−

=D

( )( )

( ), 12820

21322

4232

65326

321

321

321

321

1 xxx

xxx

xxx

xxx

D +−=

−−+−

−+−−

−+−−

=

( )( )

( )

( )( )

( ), 12816

22322

4332

62326

23222

4323

63262

321

321

321

321

321

321

321

2

xxx

xxx

xxx

xxx

xxx

xxx

xxx

D

−+−=

=

−−+−

−+−−

−+−−

=

−+−

−+−−−

−+−−−

=

179

( )( )

( ). 12824

32212

3223

32652

321

321

321

321

3 xxx

xxx

xxx

xxx

D −+−=

−+−−

+−−−

+−−−

=

of the three posterior terms of G1.then it gets

+−=

+−=

−+−=

,

326

324

325

3216

3215

3214

xxxx

xxxx

xxxx

substitute it into the equation of unit generic sphere,it gets

. 1369060281378 323121

2

3

2

2

2

1 =−+−++ xxxxxxxxx

make elementary transformation

( )( )

+−−−

−⇒

+−

−

⇒

+−−

−−⇒

+−−

−−

−−

⇒

−−

−

−−

⇒

−−

−

−−

32865932865900

010

001

19511800

010

001

159106660

6210

001

15910630300

62550

783045

451828

301318

783045

281845

181330

453078

2

22

λλ

λ

λλ

λ

λλλ

λλ

λλλ

λλ

λ

λ

λ

λ

λ

λ

λ

for characteristic matrixso，the characteristic root of the characteristic matrix is λ1=1,

, 328659 , 328659 32 −=+= λλ

then， cosφ1=1,

, 7724.0328659

1cos , 0927.0

328659

1cos 32 ≈

−=≈

+= ϕϕ φ1=0°，φ2≈84°41′, φ3≈39°26′.

choose nonzero angleφ2, φ3 to define the included angle between G1, G2.

8.10 solution：

first found out two solution vectors

{ }

+

−

−−

−

−

−

−−

=

−

−

−−=

−=−−=−

−−=

322

252

424

320

253

425

3202

2532

4254

3,2,0,2

252

4248

1

0010

2532

4254

8

1

21

4321

4

431

4321

3

ee

eeee

eeeeeee

β

β

{ }82,70,136,53

202

532

254

312

232

454

43 −−−=

−

−

−

−−

−

+ ee

180

of I1 and find out two new normal vectors

{ }, 0,8,1,8

7013653

2234

1

1000

827013653

3202

34

1421

4321

1 −=

−−

−−=−−−

−−=

eeeeeee

β

{ }, 258,173,328,214

7013653

818

202

178213653

018

302

17

827053

088

322

178270136

081

320

17

827013653

0818

3202

17

1

43

21

4321

2

−−=

−−

−

−

−

−−

−+

+

−−

−

−

−−−

−

−

=

−−−

−

−=

ee

ee

eeee

β

perpendicular to each other of I1.assume

{ }

{ }

{ }

{ },258,173,328,214249873

1

0,8,1,8129

1

82,70,136,5332929

1

3,2,0,217

1

2

24

1

13

4

42

3

31

−−==′

−==′

−−−==′

−==′

β

β

β

β

β

β

β

β

e

,e

,e

,e

then it gets orthogonal matrix

−−

−

−−−

−

=

249873

258

249873

173

249873

328

249873

214

0129

8

129

1

129

832929

82

32929

70

32929

136

32929

5317

3

17

20

17

2

Q

from formual (3) X=Q′Y

of the eighth chapter it gets

−−

−−−

−−

−

4

3

2

1

4

3

2

1

1

249873

2580

32929

82

17

3

249873

173

129

8

32929

70

17

2

249873

328

129

1

32929

1360

249873

214

129

8

32929

53

17

2

y

y

y

y

x

x

x

x

＝：　σ

choose β1 ,β2 as new normal vector of I1，then it gets

181

=−−+

=+−

0258173328214

088

4321

321

xxxx

xxx

substitute σ-1 into the new equation of I1 , it gets

=

=′

0

0

4

3

1y

yI ：　

and substituteσ-1 into the equation of I2，then it gets

=−−+

=++−

0249873

166

129

5

32929

683

17

20

0249873

992

129

19

32929

931

17

4

4321

4321

2

yyyy

yyyy

I ：　

determinant of coefficient

,193712917

8322

12917

400

193717

683

17

20

129

5

193717

931

17

4

129

19

,12917

270

129193717

20504

1937129

166

193717

683

17

20

1937129

992

193717

931

17

4

,1937

14

1937129

166

129

51937129

992

129

19

21

21

21

2

21

21

21

1

yy

yy

yy

D

yy

yy

yy

D

D

+=

+−

−

=

−−=

−

+

−

=

=−−

=

of the two posterior terms of I2′, so

,21937

4161

21937

1937200

,21937

1937135

21937

10252

214

213

yyy

yyy

+=

−−=

substitute the equation of y3 , y4 into the equation of unit generic sphere

12

4

2

3

2

2

2

1 =+++ yyyy

it gets ,1074575272320319374432440182690961 2

221

2

1 =++ yyyy

( )

, 4621.03.503268

21937cos , 0214.0

7.234910895

21937cos

, 3.5032687.117203813117707082

, 7.2349108957.117203813117707082

106891618.1108055317.3235414164

21

1616

≈≈≈≈

≈−≈

≈+≈

≈×+×+≈

ϕϕ

B

A

φ1≈88°46′, φ2≈62°28′.