Elementary Vector Geometry

seymour schuster)

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ELEMENTARY

VECTOR GEOMETRY)

SEYMOUR SCHUSTER)

DOVERPUBLICATIONS,INC.Mineola, New York)))

Bibliographical Note

This Dover edition, first published in 2008, is an unabridged republication of

the work originally published by John Wiley & Sons, Inc., New York, in 1962.)

Library of Congress Cataloging-in-PublicationData

Schuster, Seymour.

Elementary vector geometry / Seymour Schuster. - Dovered.p.em.

Originally published: New York: John Wiley & Sons, Inc., 1962.Includes index.

ISBN-13: 978-0-486-46672-9ISBN-lO: 0-486-46672-81.Vector analysis. 2. Geometry. I. Title.

QA433.S38 2008515' .63-dc22)

2007050541)

Manufactured in the United States of America

Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501)))

to my parents)))

preface)

Thisshort work is the outgrowth of lectures that werepresented at a National Science Foundation Institute

held at Carletol1 College ill the summer of 1959. The

lectures were to serve the purposeof \"enriching\" the

backgrounds of geon1etry teachers. However, thepattern of events in mathematics education indicatesthat the material covered is no longerenrichment

material but, rather, essential knowledge for everyteacher.

It was just a few years ago that linear algebra'was a

course for beginning graduate students and vector analy-sis was typically taken as an upper class course by

mathematics, physics, and engineering students. Thelast decade has brought \037 revolution in undergraduate

mathematics education, and today the knowledgeof

vectors is acquired at a much earlier stage. Indeed,it isquite usual for college freshmen to study vectors andmatrices,particularly as applied to geometry. Further-

more, the studies and recommendatio11smade by the)

v)))

VI) PREFACE)

Commission on Mathematics, the School M\037thematics

Study Group, and the Committee on the UndergraduateProgramin Mathematics all point in the direction ofgetting some of these c011ceptsinto the high school

curriculum. I have 1011g felt that vector techniquesshould and ,viII find their way into the high schoolcurriculum-perhapsnotas an integral part of the mathe-matical training of all studentsbut,at the very least, as

work to excite and challenge superiorstudents.On a very elementary level, this textbook deals pri-

marily with the development of vector algebra as amathematical tool in geometry. The aim is to gain agreater insight into the theorems by attempting vector

proofs and analytic proofs in contrast to the synthetic

proofs, a knowledge of which the reader brings as aprerequisite.The elements of vector algebra are devel-oped slowly-more so than in any of the standard workson vectors. Simplegeometric explanations, as well as

numerous illustrations, are used. Beyondthis, some

analytic geometry (in two and three dimensions) isdevelopedasa natural outgrowth of the vector treatment.In addition, the vectorapproach is used to assist in otherareas of elementary mathematics:algebra,trigonometry

(plane and spherical), and higher geometry. In short,vectorshave been employed whenever it was felt thatthey would aid in gaining insight and/or in facilitating

computations and proofs. I have tried to develop verylittle machinery but to go a longway with this small

amount. Accordingly, the reader will. find himself

dealing with such topics as linear inequalities,convexity,linearprogramming, involutes, and projective theorems.

As for prerequisites, they are not listed formally

because I do not claim to have givena logical(axiomatic)

development of geometry. Loosely speaki11g, I haveassumedthat the reader is familiar \\vith the definitionsand concepts of Ellclidean geometry and with the bare)))

PREFACE) VII)

essentials of trigonometry. For example, the notionsof

angle, parallelism, and area are assumed. It is furtherassumedthat the reader knows the definitions of the

sine, cosine, and tangent functions (in the naive sense,as ratios of the sides of a right triangle). In regard toresliitsfrom geometry and trigonometry, I haye indeedtaken very little for granted. Samples of. geometricinformation that are calledupon are: formulas for the

area of a parallelogram and volume of a parallelopiped\037

the fact that two points determinea unique line, and the

result that opposite sides of a parallelogramare equal.

In Chapter 3 I use the)aw of cosinesfor motivatio11,butthe reader who has not seen it before will be consoledby

a proof given shortly thereafter.It is entirely possibleto give a vector development of

Euclidean geometry from \"scratch.\" In fact, some

people believe that a first course in geometry shouldbeginwith vectors. Others believe that the coordinatemethod should be given at the olltset, and still othershave faith in a combinatio11of approaches. The

coordinate development, in various forms, has recentlybeenadoptedby several of the current groups interestedin rewriting the high school mathematics curriculum.

For the reader interested in seeing how a strict vector

approach would do the job, I strongly recomme11d the

excellent paper \"Geometric Vector Analysis and theConceptof Vector Space\" by Professor Walter Prenowitz.This fine expositionconstitutesoneofthe chapters of the

Twenty- Third Yearboolc of the National Council ofTeachersof l\\lathematics.

Sincere thanks and appreciation are due to the teacherswho came to Carleton i11 the SlImmer of 1959 in the hopeof comfortably learning mathematics in cool Minnesotabut who, instead,laboredal1dperspiredunder the strains

of vector geometry and the 96% humidity. For reading

the n1a11uscript and for their valuable suggestiol1SI am)))

VIII) PREFACE)

especially grateful to Mr. Saul Birnbaum of the New

Lincoln School in New York City, ProfessorRoy Dubisch

of the University of Washington, Professor J. M. Sachs

of the Chicago Technical College, and my Carletoncolleague,ProfessorWilliam B. Houston, Jr. Also,

special thanks go to ProfessorDick Wick Hall, who

courageously used the text in mimeographform at aNational Science Foundation Institute in the Sllmmer of

1961. This experience enabled Professor Hall to con-tribute substantially by pointing out ID.y errors ill judg-ment and typography.)

SEYMOUR SCHUSTER)

N orthfield, MinnesotaJanuary, 1962)))

contents)

Chapter 1 ELEMENTARY OPERATIONS) 1)

1. Introduction. 2. Definition of vector. 3.Fundamentalproperties\302\2674. Linear combinations

of vectors. 5. Auxiliary point technique.6.Uniqueness of representations.)

Chapter \037 VECTORS IN COORDINATE SYSTEMS 40)

7. Rectangular systems and orientation.8.Basisvectors and applications

\302\2679. The complex

plane.)

Chapter 3 INNER PRODUCTS) 60)

10. Definition. 11. Properties of inner product \302\267

12. Components. 13. Inner product formulas.14. Work.)

IX)))

x) CONTENTS)

Chapter 4. ANALYTIC GEOMETRY

15. Ourpointofview \302\26716. The straight line \302\26717.

A,nalytic geometry of the line continued. 18.Distancefrom a point to a line. 19. Analyticmethod of proof.20.Circles.21.Spheres. 22.

Planes \302\26723. Det\037rmining a plane by points on it \302\267

24. Distance from a point to a plane \302\26725. The

straight line in three dimensions \302\26726. Angle be-

tween two lines \302\26727. Intersection of a line with aplane \302\26728. Angle between a line and a plane.)

Chapter5 CROSS PRODUCTS)

29. Cross products. 30. Triple scalar product.31. Distancefrom a point to a plane \302\26732. Dis-

tance between two lines. 33. Triple crossproducts.)

Chapter 6 TRIGONOMETRY)

34. Plane trigonometry .,35. Sphericaltrigonometry.)

Cl\037apter 7 MORE GEOMETRY)

36. Loci defined by inequalitjes. 37. A few

booby traps. 38. Segments and convexity.39.Linear programming. 40. Theorems arising in

more general geometries. 41. Applicationsof

parametric equations to locus problems. 42.Rigid motions.)

APPENDIX

ANSWERS

INDEX)

76)

135)

151)

160)

204fJ06

\03711)))

elem.entary

operations)

1. INTRODUCTION

The history of the development of mathematical ideasindicatesthat abstractconceptsarisegenerally from roots

in some \"practical\" problem. Arithmetic stemmedfromproblemsof counting, geometry arose from problems of

surveying land in Egypt, and calculusdevelopedprinci-pally from the efforts to solve the problems of motion.

Mathematics, however, goes quite beyond the point ofsolving merely the problems that initiate the particularstudy, for mathematics is concerned with building a

deductive science that is general and abstract,that may

have a wide range of application. By a deductive science

we mean, briefly, a logical development that beginswith

a basic framework consisting of a set of assumptions(calledaxioms or postulates) and a set of terms used instating the assumptions.All the logical consequences

of the assumptions are then the theorems of the deductive

science, which is concerned with abstractionsor idealiza-1)))

2) ELEMENTARY VECTOR GEOMETRY)

tions of concepts from the original problem rather than

with the original problem itself. For example, the study

of geometry is based on a set of assumptions that deal

essentially with lines and POil1tS rather than with fences

and fenceposts. Line is an abstract concept;it is an

idealization of the fence, and it admits to all sortsofotherinterpretations:a

ray.of light, the edge of a board,

the path of a molecule under some circumstances,and a host of still others. Thus geometry,with its

origins in surveying, finds application in a variety of

problems.

The pure mathematician, however, goes on-and farbeyond. Becauseoncehebegins a mathematical study,he is free to exercisehis imaginationby making logical

deductiollS (proving theorems) and developing theoriesthat are qllite apart from the realities of the motivatingproblem. li\"'or the mathematician there is a realitywithin his deductivescience.Again drawing from geo-

metry for illustration, we can pointto the developments

of fOllr dimensional geometry-even n-dimensional

geometry-in spite of the fact that our physical world is

three dimel1sional, or to the inventionof non-Euclideangeometries,that contradictEuclid'sParallelPostulate(which, for over 2000 years, was accepted as absolutemathematicaltruth). Such creations by mathematicians

were consequences of strong imagination and quitebeyond the consideration of any elementary problem inthe physicalworld.

Vector analysis is also a subject that has its rootsin physical problems. It was developed primarily tohandle problemsin physics,initially problems in me-

chanics but, later, problems in variousotherbranches of

physical science. Developments ofth\037

nineteenth and

twentieth centuries have resulted in the abstractconceptof a vector and, consequently, in a wide rangeof interpre-tationsofthisabstractconcept.The result is that vec-)))

ELEMENTA.RY OPERATIONS) 3)

tors now play a prominent role in a variety of studies,

to name just a few: physicalchemistry,fluid-flow theory,

electro-magnetic theory, economics, psychology, andelectrocardiography.

Geometrybooksare aI,vays filled with illustrations in

spite of the fact that point,line,and circle are abstract

concepts that do 110texistin physicalrealityand in spite

of the fact that beginning students are apprisedof theabstract nature of the subject at the very beginning oftheir course. The reasonis quite simple. Abstract

reasoning is difficult; students therefore need-or are atleastassistedby-thehelp of some real model (or inter-pretation) of the abstract concepts.Consequently, a

small dot marked with a sharp pencil is a convenient

model for the concept of point, and a sharppencildrawn

along the edge of a ruler leaves a n1arkthat isusedasamodel for the concept of line. Such pencil marks are agreat convenience for beginl1ers until they get to feel athome in the subject and begin to feel that there is areality in geometry itself. Later in their mathematicalstudies studentsencounter other abstract concepts, but

by this time they can, and do, use geometric models to

assist them in still more abstract reasoning. Thispattern of development is precisely what occurs in thestudy of vectors. Although the concept of vector canbe made abstract, a geometric model (directed line seg-ment) is the crutch that assists the beginner in develop-ing steadylegsinthe field that is new to him.

It is the author'sview that steady legs in abstractvector algebra are developedslowly and that reasoning

in the model (now geometry)for some extended period

should be done preliminary to engagingin the abstractstudy. Therefore this entire textbook co\037cerns itselfwith a geometricstudy of vectors (i.e., the application ofvectors to geometry),in contrastto the general abstract

study of vectors. Let us begin.)))


2. DEFINITION OF VECTOR

Earlier we pointed out that the idea of vector cameoriginally from physics. Therefore, let us consider-from the physicist'spointof view-the statement of thetelevision announcer who, beforegiving his final \"Good

night,\" states, \"The temperature is 110W 37\302\260and the wind

is' 12 miles per hour in a northeasterly direction.\" In

this simple weather announcementwe observe examples

of two distinctly different types of quantitiesin thesensethat the first (temperature) requires only a single num-ber-with units, of course-for its description,whereas

the second quantity (wind velocity) requires two facts,magnitudeand dire-ction. These examples are typical ofthe quantities encounteredin elementary physics.

Hence, the following simple classification is made:quantities that possess only magnitude are singled out and

called scalars, whereas quantities that possess both magni-

tude and direction are called vectors.In additionto temperature, examples of scalar quanti-

ties are mass, length, area, and volume; and, in addition

to velocity, examples of vectors are force,acceleration,a11delectricalintensity.

Just as the mathematician desires a geometric model

for his general concepts, so doesthe physicist.For

geometry-to the trained physicist-also has a realityby means of which he can \"visualize\" and be aidedin his

reasoning. A convenient geometric model for a vectoris a directedline segment (/) because this possessesboth magnitude (length) and direction,simultaneously.

This model, which suits the needs of physicists, is alsoquite satisfactory for our purposes, for it is our aim tostudy geometry (by mearlS of vectors). Hence, for ourmathematical development,we make the following for-

mal definition.Definition. A vector is a directed line segment. W e shall

use boldface type to indicate a vector. Thesymbol1AI)))

ELEMENTARY OPERATIONS) 5)

Q

Terminusor

endpoint)

p

Origi n) .0)

c)

(a)) (b))

FIGURE 1)

will be used to designate the length of vector A. In the

event that the vector we speakof is thedirectedsegment\037

PQ, we emphasize its vector nature by writi11gPQ,P being the origin and Q being the terminus or endpoint

of PQ (see Figure 1a). Another useful conventionwhen

\037 --7 \037

referring to several vectors OB, OC, and OD with a

common origin 0 is to call these vectorsB, C, and D,

respectively. That is, if a discussion is concernedwith

several vectors emanating from a single point, we may

designate them merely by their individual endpoints(seeFigure1b).

A vector of length one is called a unit vector. The

notion of a vector of zero length (with any direction),

although apparently peculiar, is actually a great con-

venience. We refer to such a vector as the zero vector

and shall point to its usefulnessfromtime to time. The

notation for the zero vector is o. The direction of the

zero vector is discussed further in Secti0114.Scalars,being merely magnitudes, will be real-num-

bered quantities. (In more advanced mathematics

scalars may be elements of the complexnumbers;indeed,they may be from any number field. Our needs, how-)))


ever, do not requiresuchgeneralityand will therefore be

satisfied by restricting scalars to the real numbers.)

They are designated by lower-case Latin letters: a, b, c,or by numerals.)

3. FUNDAMENTAL PROPERTmS

Our desire is to buildan algebraof vectors, and to this

end we must first present a criterion for calling two vec-

tors equal.Definition. Two vectors A and B are called equal (written

A = B) if and only if Ithe following three conditions hold:

(i) A is parallel to B;2

(ii) A and B possess the same sense of direction; and(ill) IAt

=\\BI, i.e., the length of A equals the length of B.)

It cannot be emphasizedtoostrongly that vectors maybe equal even if they do not possess the same position inspace. As a matter of fact, our definition indicates thata vector A may be relocated provided that we move itrigidly to a position parallel to its original positionandthat we do not change its length or sense of direction(seeFigure 2a). It may therefore be relocated in a posi-tion with its origin anywhere in space that we choose.When vectors are given this freedom, they are termedfree.

1The phrase \"if and only if\" points up the fact that the defini-

tion is actually a double implication, or logicalequivalence. That

is, (a) If A = B, then conditions (i), (ii) , and (iii) hold; and(b) If conditions (i), (ii), and (iii) hold, then A = B.

2 We use the word parallel in the more general sense to mean\"parallel or on the same line.\" Although this usage is not givenin high school geometry, it is quite common in analytic geometry,where two lines possessingequal slopesare calledparallel (see

Section 16). Thus a line is parallel to itself, and a vector is equalto itself. The latter would not be

tru\037if we didn't use \"parallel\"

in this generalized sense. Figure 2b exhibits two equal vectors

on the same line.)))

ELEMENTARY OPERATIONS)

Yy

Y)(a))

7)

\"\"

,

\" ,\"-

\",

\(b))

'\\,

,)

FIGURE 2)

In the study of geometry this libertyto make displace-

ments of vectors is highly advantageous. In theapplica-tionsof vectors to other sciences this freedom is notalways granted, for it is necessary to restrict vectors tosome degree. For example, in the mechanics of rigidbodiesit isoftenrequired that a vector be confined to aline;that is,it may be moved rigidly but only in the lineit lay originally. This line is referred to as its line of

action. In Figure 3a we show three vectors, F, G, and

H, which represent three forces of the same magnitude,acting on parallel lines and haviIlg the same sense of)

/////z/)

(a)) (b))

FIGURE 3)))


(a)) (b))

FIGURE 4)

direction. They would therefore be equal accordingtoour definition. F represents a pulling force at the centerofthebar and G represents a pushing force at the centerof the bar. F and G would have the samemechanical

effect and are therefore considered mechanically equal.However,H appliedat theendofthebarwould effect a

turning motion, which is quite different fromthe effect of

F = G. Thus H is not equalto the otherforces.In

such studies it would be natural to insist that two forces

having different lines of action be unequal. Thisjustifiesthe stipulationinthe theory of mechanics of rigid bodies,of permitting a vectorto bedisplaced only along its line

of action.If the fieldof applicationwere the theory of elasticity,

then it would be necessaryto restrict(force) vectors still

more. Figure 3b shows two forcesJ and K, both of the

same magnitude and directed alongthe same line of

action, acting on a soft plasticlikematerial. K has the

effect of stretching the mass, whereas J has the effect of

compressing it. This illustration indicates why, in thetheory of elasticity, two vectors applied at different)))


points are not considered equal. In this field a vectorisrestrictedto itsoriginal position; there is no freedom to

displace it. Suchvectorsare called bound.

We state, with emphasis, once again: all vectors in

this book are free, in accordance with our definition of

equality.Addition. Let A and B be any two vectors (Figure 4a).We can select a location for vector B so that itsoriginisplaced at the terminus of A. Now we constructa thirdvector, called A + B, whose origin coincides with theorigin of A and whose terminus coincides with the ter-minusof B.

The construction(Figure 4b) of B + A clearly indi-cates that B + A = A + B, for both are the same diago-

nal of the same parallelogram, and they possessthe same

sense of direction. Hence we have the followingresult.Theorem 1. (i) Addition of vectors is commutative;that is A + B = B + A.

(ii) Addition of vectors is associative; that is

(A + B) + C = A + (B + C).

Part (ii) of the theorem is easily established by usingthe definition of addition. Figure 5 illustrates the proof.)

Co>\037

\037\037

\037)

Co\037

\037\037 B+C

\037)

A) A)

FIGURE 6)))


However, the reader is advisedto phrase the proof as a

logical deduction from elementarygeometryindependent

of a figure.)

EXERCISE1. Give an elementarygeometry proof of the theorem: If equalvectorsareaddedto equal vectors, the sums are equal vectors.

As indicatedin Section1, the notion of a vector arose, his-toricallYr from an attempt to characterize physical quantities,notably force. It is interesting to note that the little-known,but nonethelessexcellent,Dutch scientist, Simon Stevin

(1548-1620), experimented with two forces in an effort toreplacethe two by a single one, called the resultant. He dis-

covered that the resultant was actually the force represented

by the diagonal of a parallelogram,of which the sides repre-sented the two original forces (Figure6). This led to his

formulation of the principlesof addition of forces, which heused extensivelyin developing a complete theory of equilib-rium-the beginningof modern statics. (It is for this reasonthat the parallelogramsof Figure 6 are sometimes referred toas parallelogramsof forces.) Among the many other accom-plishmentsof Stevin are his: (1) work on hydrostatics,which

laid plans for the reclamation of the below-sea-Ievelland of)

Force Fl)

Force Fl)

Force Fl)

FIGURE 6)))


Holland, and (2) developmentof decimal notation for numberswith the consequent methods for computation. He was thefirst to give a systematic treatment of decimals. (Seethe

chapter entitled \"Stevin on Decimal Fractions\" in A Source

Book in Mathematics, by D. E. Smith, or A History oj Mathe-

matics, by J. F. Scott,1960.)Thus our definition of vector addition is consistentwith the

desires of the physicist who is interestedin applying the tech-

niques of vector analysis to his problems. (The student of

applied science should constantly maintain a critical attitude

toward the mathematical definitions, taking care to see whetheror not they accurateiy reflect given physical situations.)Before continuing, it should be mentioned that Galileo (1564-

1642), quite independently, cameto the sameconclusionas did

Simon Stevin. Thus two scientists discovered how vectors

\"should\" add, approximately two centuries prior to the inven-

tion of vector algebra and vector analysisin the nineteenthcentury.

Our definition of addition can now be extended to find

the sum of n vectors: Al + A 2 + A g +\302\267\302\267\302\267

+ An.

Of course, this can be doneby grouping pairs (note part

(ii) of Theorem 1) and applying the definition repeatedly.

However, the geometric process might be describedsimply as follows: move A 2 so that its origin is at theterminus of AI; move Aa so that its origin is at theterminus of A 2 ; continue this process until An is placed

with its origin at the terminus of An-I. The sum Al +A 2 + A g +

\302\267\302\267\302\267+ An is then the vector whoseorigin

coincideswith the origin of Al and whose terminus coin-cideswith the terminus of An.

What would be the sum of the vectors that form aclosed polygon with arrowstakingusallthe way around?

(Try to find the answer before reading on.) Consider,for example, A + B + C + D + E + F 'of Figure 7.This is the vector whose origin coincides with the originof A and whose terminus coincides with the terminus of

F after the vectors are placed \"origin to terminus\" as)))


FIGURE 7)

described above. Hence the origil1 and terminus of the

sum vector are the same point, al1dthe vector is of zero

length. We then write

A + B + C + D + E + F = o.)This argument holds for a polygon of n sides, so theanswer to our query is: The zero vector.Multiplicationof a vector by a scalar. In arithmetic it isconvenient to introducemultiplication as an extension

of addition. For example, 3 X 4 may be thought of

4 + 4 + 4. Similarly, we can-at leastto begin with-)

\037x

\037\037

rt,\037)

FIGURE 8)))


think of multiplying a vector by a scalar as an extensionof vector addition. An illustration (Figure 8) might bethat 2A should represent A +. A. From our definitionof addition, we know vector A + A is actually a vector

parallel to A and having the same sense of direction as A;

however, the length of A + A is twice the length of A.

Therefore, if 2A = A + A, the result of multiplying A bythe scalar 2 is a vectorparalleltoA having the same sense

of direction as A but with twice the length of A.Before proceedingto the general case of multiplying a

vector by a scalar, let us considerthe questionof what

would be appropriate f\037r a definition of -A. A reasona-ble demand might be that A + (-A) = 0; so let us,for the moment, stipulate this and see where it takes us.In general, if A + X = 0, we know that (a) X must be

parallel to A, (b) IAI=

\\xl, and (c) X must have a senseof direction opposite to that of A (Figure 9). Thus-A

should have precisely the properties a, b, and c men-

tioned in the previous sentence. (Alternatively, if

A + X = 0, then A followed by X can be thought of as

a closed polygon in which the origin of X is at the ter-

minus of A.) Consequently,our definition should (and

will) stipulate that multiplying by a negativescalarhasthe effect of changing the sense of direction of a vector.

We are now ready to present a definition for the multi-

plication of a vector by a scalar.)

FIGURE 9)))


;1A)

FIGURE 10)

Definition.3 nA is a vector parallel to A with magnitude In \\

times the magnitude of A. In symbols, InAI=

Inl iAI.Further, if n > 0, nA is defined to have the same sense ofdirection as A; and if n < 0, nA is defined to have a senseof direction opposite to that of A; finally, if n = 0, nA is

defined to be the zero vector (which followsfrom the first

sentence of our definition). Figure 10 illtlstratesthe def-init.ion for n = 3, n = -3, and n = i-.Theorem 2. If m and n are scalars,then

(i) m(nA)= (mn)A.

(ii) (m + n)A = mA + nA.

(iii) meA + B) = mA + mB.

To illustrate, consider m = 5 and n = - 2. ThennA = - 2A, a vector twicethe length of A btlt directed

3 The sy-mbol Inl refers to the absolutevalue of n, which is definedas follows:If n > 0, then Inl

= n; and if n < 0, then Inl= -n.

rfhe definition asserts that the absolute value of a number is

always non-negative; e.g., 131= 3, 1-31 = 3, and 101

= O. We

shall need the fact that \\mnl= Imllnl, the truth of which should

he clear from the definition.)))


oppositely to A. (i) states:

5(-2A) = (5)(-2)A= -lOA.

(ii) states: (5 + (-2))A = 5A + (-2)A or

3A = 5A + (- 2)A.

(iii) states: 5(A + B) = 5A + 5B.

Proof. (i) By examining the length of the left mem-

ber of (i), our definitionof multiplication of a vector bya scalar yields)

1m(nA) I=

ImllnAI= ImllnllAI = ImnllAl

=I (mn)AI. (1)

Equation 1 proves that the vectors of (i) are equal in

length. That they are parallel follows from the fact

that both are multiplies of A. The reader is left to checkthat the directionshave the same sense. (Hint. Usethe definition of nA..)

(ii) If m + n = 0, both sidesof (ii) represent vectors

that point in the same direction \037s A. If m + n < 0,both sidesof (ii) represent vectors that point in the direc-tion oppositeto that of A. The comparison of their

lengths is left to the reader. (Hint. Usethe definition

of nA.)

(iii) Let us suppose that A and Bare nOllparallel and

nonzero. We consider the trianglePQR,which defines-? -?

A + B (see Figure 11), by having A = PQ, B = QR,--7then P R = A + B . We construct triangle P' Q' R' ,-? -? \037

where mA = P'Q', mB = Q'R', then P'R' = mA + mB.

SincemAilAandmBilB.triangle PQR is similar to tri--?angle P'Q'R'. Thus P'R' is meA + B), and we have the

result that meA + B) = mA + mB.

The reader should consider two questions cOllcerning

the proof of (iii). The first is: What of the direction of)))


p)A) mA)

p')

FIGURE 11)

m.(A + B) as comparedto that of mA + mB? The

proof explicitly concerned itself, with the lengths and

parallelism of the two vectors,but it didnotdiscusssenseof direction. The second question is: Does the proofbreakdown if A\\.IB?

NOTE. Since part (i) of Theorem 2 states that a

change of parenthesis is legitimate,we now know there

would be no confusion if we eliminate the parenthesis

entirely and write mnA. For example,)

3(2A) = (3 \302\2672)A

= 3 \302\2672A = 6A.)

Subtraction. As il1 elementary arithmetic, where sub-tractionis an operation that is the inverse of addition,we definesubtractionof vectors as the inverse of' vectoraddition. More precisely,if a is a real number, we writea - a = a + (-a) = O. This equation expresses thefact that subtractioll is defined in terms of addition andthat subtraction is the inverse of addition. Mathe-

maticians say that the realilumber -a is the inverse of

a relative to the operation of additioll or simply that - ais the additive inverse of a (similarly, a is the additiveinverseof -a). We carry these ideas over to vectors in

defining subtraction.Definition. A - B = A + (-B) where -B is written

to mean -lB.)))


Geometrically the operation of subtraction can takeany of the forms preeented in Figure 12. Note that thediagonal A of the parallelogram is equal to the sumB + (A

- B) and also to the sum (A- B) + B. Such

algebraic checkingis advised for the beginner who is

having difficulty in finding the correctorientationfor the

difference vector A-B. With these few tools of addi-

tion and subtraction we can begin applying vectorstoelementary geometry.

EXAMPLE 1. We shall use our vector operationsto work an

elementary exercise, one equivalent to the theorem which states

that the diagonals of a _parallelogram bisect each other. Let

0, B, and C be threepoints not on one line. Call M the mid-point of segment BC (see Figure 13a). We shall prove that--+ --+--+

OM =\037(OB + OC). In accordance with the convention dis-

\037 --7 \037

cussed on page 5, weshallwrite B = OB, C = OCand M = OM.Then)

\037

M = B + BM)

\037 \037 \037 \037

and M = C - MC = C - BM (since Bill = MC).)

B)

B) -B)

FIGURE 12)))


B) B)

D)

o)o)

c) c)

(a)) (b))

FIGURE 13)

Adding, we get)

or)

2M=B+C

M =\037(B + C).)

If we considerour figure to be part of a parallelogram OBDC(Figure13b),then the result states that the line joining vertexo to the midpoint of diagonal BC is one half the diagonal OD,

\037

for IODI=

\\B + CI. In equivalent (and more usual) languagewe may conclude that the diagonals of a parallelogram bisecteachother.EXAMPLE 2. Prove that the line joining the midpointsof

any t\\VO sides of a triangle is parallelto the third and equal toone half of it.

In trianglePQR (seeFigure 14) let 111 and N be midpointsof PQ and PR, respectively. Call)

\037 \037 \037 \037 \037 \037

A = PN = NR, B = PkI =J.lfQ, C = MN, and D = QR.

Then .C = A - Band C + A - D - B = o.(Note that these equations follow from summing vectors about

triangle MNP and quadrilateral JfNRQ, respectively.)Adding vector D to both sides of the last equation, we get

C + A - B = D.Since C = A - B, we have 2C = D, which (by the definition

of equality of vectors) proves simultaneously thatNM\\lRQ and

that the segment NM equals one half of the base RQ.)))


p)

R:)

FIGURE 14)

EXERCISES

1. Establish the easily rememberedSHORTCUTLEMMA:

\037 \037 \037

AB + BC = AC. (This lemma has also been appropriately

named the Bypass Lemma by Professor D. W. Hall.)

2. Reproduce Figure 15 on another sheet. Then constructand labelthe vectorsC -

A, B - C, B + C, and -B - C.\037 \037 \037

3. Show that PQ + TlS=2MN, where P, Q, R, and S are

four arbitrarily chosen points and where \037f and N are the)

FIGURE 15)))

20) ELEMENTARY. VECTOR GEOMETRY)

midpoints of PR and QS, respectively. (Hint. Sum the

vectors around the polygon NMRS and NMPQ.)

4. Draw triangle ABC, with P the midpoint of A.B, Q the mid-point of BC, and R the midpoint of CA. If X is any point,

\037 \037 \037 \037 \037 \037

show that XA + XB + XC = XP + X Q + XR.

5. A sailboat is acted upon by the wind and current. Thewind velocity is 8 mph E and the current velocity is 5 mph N.

Compute the magnitude .of the resulting force and give ageometricconstructionthat shows the direction of the resultant.

6. Usingthe fact that the sum of the forces.(vectors)actingon a body in equilibrium is the zero vector, solve the following

problem.

A weight of 100 lb hangs by a wire and is pushed by a hori-zontal force until the wire makes an angle of 1r/4 (or 45\302\260)

with the vertical. Find the magnitude of the horizontal force

and the tension in the wire.

7. Show that the midpoints of consecutive sides of a quadri-

lateral are vertices of a parallelogram.8. If A, B, C, and D are any four points (not necessarily in a

\037 \037 \037 \037 \037

plane), prove that AB + AD + CB+ CD = 4PQ where Pand Q are the midpoints of A.C and BD. (How does thisrelateto Exercise2?)

9. Using the definition of subtraction, the commutative and

associative properties of addition, show that the sums B +(A

- B) and (A- B) + B, do actually reduce to equal A.

10. Establish the uniquenessof 0, thus justifying the termi-nology: the zero vector. (Hint. Consider the possibility of a

vector 0' that has the properties of O. Then prove that0' = 0.))

4. LINEAR COMBINATIONS OF VECTORS

Now that we have learnedto addand subtract vectors,

and also to multiply vectors by scalars, we can combine

these operations to generate new vectors and to enrich

our algebra of vectors. For example,if we are given A)))


and B, we can perform our various operationsto getA + B, A - B, 2A -

3B, 5A + 6B, etc. Such com-binationsof A and B are called linear combinations of

A and B. The set of aillillear combinationsof A and B

could be written {xA + yB Ix and y real}.4

The defil1i-

tion of linear combinations is now extended in the fol-lowing way: If AI, A 2, A 3 , . . . , An are n vectors and

Xl, X2, X3, . . . ,X n are n scalars, the vector

xiA I + X2A 2 + X3 A g + \302\267\302\267\302\267+ xnAn

is said to be a linearcombination of AI, A 2, A 3 , . . . ,An.

The present section is devoted to the study of linear

combinatiolls of certain sets of vectors, and the ideas

contained herein are perhaps the most difficult in the

el1tire book. Thus we shall proceedslowly. The reader

is cautioned to study the definitionsand to take them

literally!

We Iloted earlier that the equationA + X = 0 impliesthat X = -A, that is, X is the vector obtained by mul-

tiplying A by -1. SinceX can be derived from A, wecould, in a sense,say that X is depe11dent upon A; or,conversely,that A is dependent UpOll X. Similarly, if,ve began with A + 3Y = 0, we could write

Y = --lA

A = -3Y.)and)

Thus Y is shown to depel1d 011A and A to depel1d onY. It might be preferable to state simply that A a11dYare dependent. It is almost trivial to observe that

such dependency would be impossibleto sho,v if the scalar

coefficients of A and Y were both zero. Hellce we excludethis case from consideratiol1 in making the followingdefinItion.

4 The synlbolism { I }, borrowed from set theory, is useful in

defining sets in the following \"ray: {zIS(z)} represents the set of allz satisfying the condition or sentence S (z).)))


Definition. Two vectors A and B are called linearlydependent if and only if there exists two scalars a and b,not both zero, so that aA + bB = o.

Remark:Thestudentshouldrecognize the fact that

definitions of words are statements of logical equivalence

(see footnote 1, page 6); i.e., boththe statement and its

converse hold. Using our present definitionto illustratethis explicitly, we would say that the definition states:(1) A and B being linearly dependent implies thatthere existscalarsa and b, not both of them zero, so thataA + bB = 0; and (2) if a relation aA + bB = 0 holds,with not both a and b equal to zero (i.e., at leastonebeingnonzero), then A and B are linearly dependent.

Algebraicallyour definition is equivalent to saying thattwo vectors are linearly dependentif and only if (seefootnote 1 for explanation of the phrase \"if and only if\one of them is a scalar multiple of the other (show this!).

A geometric interpretation would be the following:Twovectorsare linearly dependent if and only if they areparallel. The reader can verify these interpretations

formally by constructing a generalargument of the form

that preceded our definition. One fine point, however,

must be mentioned; this concerns the presenceof 0 asoneof the vectors under consideratio11. First, we notethat 0 and any vector A are linearly dependent, for10 + OA = 0, which is the definingconditionfor 0 and

A to be linearly depe11de11t(1 is the required nonzero

scalar). Our geometric i11terpretation would say thato isparalleltoA, where A may be any vector. This may

appear strange to the beginning student of vectors; how-

ever, it is a matter of great convenience to retain the con-vention that the zero vector is parallel to every vector!Instead of regarding the zero vector as having no direc-tion (as would appeal to some), we regard 0 as havingany direction. Vectors specify magnitude and direction,so we chooseto say that 0 has any or all directio11ssimul-)))


taneously. Of course, the zerovectoristhe only vector

with such a property, for any other vector(line segment

of nonzero length) has a u11iquedirection. Later,when

we work with perpendicular vectors, we shall have occa-sionto regard the zero vector as perpendicular to everyvector. Again, this is consistent with the idea that thezerovector has any direction.

In the event that two vectorsarenot linearly depend-

ent, we call them linearly independent. Summarizing,we say that:)

(1) A pair of vectors, of which one is a zero vector, is alinearlydependentsetofvectors.

(2) A pair of parallel nonzero vectors is a linearlydependentset.

(3) A pair of nonzero, nonparallel vectors is a linearlyindependentset.)

Confining our attention merely to pairs of vectors

would contradict the generalizing spirit of mathematics

and, fllrthermore, would leave us with a rather meager

theory of linear dependence. We therefore proceed toextendOllr notions with the following definition.Definition. A set AI, A 2, . . . , An of n vectors is called

linearly dependentif there exist a set of scalars Xl, X2, . . . ,

X n , so that)

xlAI + X2A 2 + \302\267\302\267\302\267+ XnAn = 0,

where not all the x's are zero (i.e. when at least onedoesnot

equal zero). If, on the other hand, the equation

xlAl + X2A 2 + \302\267\302\267\302\267+ xnAn = 0

implies that Xl= X2 = \302\267\302\267. = Xn = 0, . then the set

AI, A 2, . . . , An is said to be linearly independent. For\037 \037 \037

example, if three vectors, AB, CD, and EF, are given,

one of the following cases must occur: (a) It is possible)))


to find three scalars a, b, and c, not all zero, such that)

\037 \037 \037

aAB + bC D + cEF = o.) (*))

(b) It is not possible to find three such scalars. In

case (a) the three vectorsare said to be linearly depend-ent. In case (b) they are said to be linearly independent.In bothcases(*)will hold if a = 0, b = 0, and c = o.

The jump from two vectors to n vectors may be rathersteep,sowe shall provide an intermediate step ill the formof a geometric theorem.

Theorem 3. If A and B are linearly independent, then

any third vector C; which is parallel to (or in) the plane

determined by A and B, can be expressed as a linear com-bination of A and B.

Proof. It should be recalled that vectors may be

arranged so that they possess a commonorigin,that is,we are dealing with free vectors. Therefore, even whenthe segments representing A, B, and C might be in space,they can be moved to positions in the same plane (seeFigure 16). For A and B with a commonorigindeter-mille a pla11e; alld C (being parallel to this plane) maythen be displacedsothat it is actually in the plalle of

A and B.)

'B)

\037)B

\037A)

))

A)

FIGURE 16)))


A)A) I \037A

A /x >o and y>O

(b))

I CIyBI B

IxA

\037----

/0

oX < 0 and y > a

(a))

FIGURE 17)

Now, if C is a nOllzerovector, thereisa parallelogram

with diagonal C alld '\\vith edges along A and B. An

explicit constructioll of this parallelogramis given in

Figure 17 and call be described as follows:Call0 thecommon origin of the vectors A, B, and C, and call A,

B, and C the respective endpoints of the three vectors.Constructa line \302\243through C parallel to B, alld call Dtheintersectionof\302\243with the line of actiol1 of A (when the

\037

origin of A is 0). Then OD, beingparallelto A, is some\037 \037

multiple of A; let us say OD = xA. l\037\"urthermore, DC,

being parallel to B, is some multipleof B; let us say\037

DC = yB. Then)

C = xA + yB,) (2))

which shows C to be a linear combinationof A and B.

In the event that C is the zero vector, the theorem is

trivially true, for)

o = OA + OB,) (3))

and our proofis complete.Toassociatemore strollgly the idea of Theorem 3 '\\vith

the concept of lillear dependence, we state theCorollary. Any three vectors in the same plane. arelinearly depertdent.)))


Proof. Equati<?ns 2 and 3 imply that there exist

scalars Xl, X2, Xg, (not all zero), so that

xlA + X2B + xgC = O.

For, when C \037 0, at least one of X and y in (2) is non-zero. When C = 0, we can set Xg equal to any realnumber,say Xg

= 1, and Xl= X2 = o.

EXAMPLE 3. We shall use the concept of linear combinationto achieve another view of Example 1. Suppose we wish toprove that the diagonalsof a parallelogram bisect each other.Let the parallelogrambe OA.BC (see Figure 18), with P theintersectionof the diagonals. Again, using the convention of

\037 \037 \037

writing P = OP, A = OA, and B =OB, we have P = mB

\037\"'hd=

meA + C) and PA\037

n(A- C), where m and n are scalars

\037

to be determined. Since A = P + P A, we may write

A = meA + C) + n(A- C),)

and grouping the A-terms and C-termsgives

(m + n -l)A + (m - n)C = o.

Because A and C are linearly independent, their scalar coeffi.-

cients in the last equation must both be zero. Hencem + n = 1 and m - n = o.

Thus m = n =\037, which proves that P bisects both diagonals

simultaneously.)

B)

o)

FIGURE 18)))


Continuing with the development on linear combina-tionsofvectors in the plane, we next establish an impor-tant theorem that serves as a very strong instrumentforwork with geometric problems.

Theorem 4. Let A, Band C be located 80 that they havea common origin O. Then:)

(i) If C has its endpoint on the linejoiningthe end-pointsof A and B,

C = lA + mB, where l + m = 1.

(ii) Conversely, if C has a representation in the form

C = lA + mB, where l + m = 1,

C has its endpointon the line joiningthe end-pointsofA and B.

Before proceeding with the proof, the readershouldobserve that the theorem is concerned with three vectorsin the same plane (often called coplanar vectors),and if

A.and B are nonzero and nonparallel, then every vector

in the plane of A and B (C being one such) is a linear

combination of A and B, by Theorem 3. However, thetheorem singlesout particular linear combinations by

means of a condition on the scalar coefficients.

For the converse point of view, we would state (and

the reader should verify) that any linear combinationof

two linearly independent vectors, A and B, is equal to avector in the plane of A and B when the two are situated

to possess a common origin. Once again,the theorem

states that certain of these li\037\037ar combinations have

interesting geometric implications.Proof: (i) Here we have as our hypothesis the fact

that A, B, and C all emanate from the same point 0,and that C has its endpoint on the linejoiningthe end-pointsof A and B. We follow the convention of calling)))

28) ELEMENTARY 'VECTOR GEOMETRY)

o)

FIGURE 19)

A, B, and e the endpoints of A, B, and C, respectively

(see Figure 19).Let e divide segment BA in the ratio l:m, where

l + m = 1. (The reader should convince himself thata given ratio can always be transformed \037lgebraicallyso that the two parts sum to Ullity.) TheIl)

\037 \037

C = B + Be = B + lBA = B + leA - B)= lA + (1

- l)B.

Thus C = lA + mB.

(ii) For the converse, our hypothesisstates that A, B,

and C emanate from the samepoint and C = lA + mB,where l + m = 1. We must show that C has its end-point on the line jOillillg the endpoints of A and B.

Now C = lA + mB = lA + (1 - l)B;or C = B + leA

- B).)

A geometric examination of this last equation com-

pletes the proof, for the equationstatesthat e may be

reached by traveling from 0 to B, and then alonglineBA.)))


EXERCISES

1. What happens if A or B is the zero vector?

2. What are land m if C = A?

3. What can be stated regarding the location of C if

(a) both land m are positive?

(b) l is negative?(c) 1 = O?)

4. Give constructions for cases where(a) l = m =

\037;

(b) 1 = ! and m = j;

(c) l = i and m = -i.

5. (a) If the ratio of division is 2:3 = l:m,find land m so that 1 + m = 1.

(b) Do the same for the ratio 4:3.(c) Do the same for the ratio 5: -3.

EXAMPLE 4. Theorem 4 reduces Example. 1 (p. 17) to atriviality. For, let 0, B, and C be threepoints not on one line,with M the midpoint of BC (see Figure20). Then JJI divides

BC in the ratio I: I(=i-:\037), and Theorem 4 allows us to write

M = i-B + i-C.)

o)

c)

FIGURE 20)))


A)

c)

FIGURE 21)

5. AUXILIARY POINT TECHNIQUEThe introduction of an auxiliary point to play the role

of commonoriginfor severalvectorsunder consideration

often facilitates the use of Theorem 4. The following

three examples are devoted to an exploration of thistechnique.EXAMPLE 5. We shall now provide a new approach to the

problem of Example 2. Although more difficult perhaps, thenew approach may be instructive.

Let triangle ABC be given, with jJ{ and N midpoints ofsides AB and AC, respectively. Let 0 be a point in generalposition(not coincidingwith any of the already named pointsor lines). Applying Theorem 4 to the three vectors A, C, and

N that emanate from 0 (seeFigure21),we get

N =\037A + \037C;)

and similarly,)M =

\037A + \037B.

Since we desire to compare MN with BC, we must examinevector N - M. Thus

N - M =\037C

-\037B

=\037(C

- B),)\037

which proves both desired results: that N - M (= ]jfN) IS\037 \037 \037

parallel to C - B (= BC) and that MN =\037BC.)))


EXAMPLE 6. We shall once again prove that the diagonalsof a parallelogram bisect each other, but this time by using

the auxiliary point technique, Theorem 4,. and still anothertool that has heretoforebeenunemployed.

Let the parallelogram be ABCD, calling P the intersectionof the diagonals. Furthermore, let 0 be a point in generalposition (see Figure 22). Once again, according to our con-vention, we write

\037 \037 \037 \037 \037

A = OA, B = OB, C =OC, D = OD, and P = OPe)

Establishing our hypothesis in vector language, we write

\037 \037

AD = BC or)D - A = C - B.) (4))

Before proceeding further, let us pause to discussthe approach.We are interested in finding the precise ratio in whichP divides

AC and also the ratio in which P dividesBD. Thus, if we can

get a representation of P, say

P = nA + mC, where n + m = 1,

then we would know, by Theorem 4, that)

\037 \037

IPCI: IAPI= n:m.

(Note how the coefficients in the statement of Theorem 4 arerelated to the vectors in Figure19.) Consequently, we seek

P as a linear combination of A and C so that the sum of the)

A)

c)

o)

FIGURE 22)))


coefficients be unity. _1\\ fu\037ther observation, which is the basisfor the new tool we promised, is that P has two possible repre-sentations as a linear combination of two vectors emanatingfrom 0; namely, there must be n, m, r, and 8 so that)

where)

P = nA + mC = rB + sD,n + m = 1 and r + s = 1.)

(5))

If we can succeed in producing theserelationssothat n, m,r,and 8 are determined, we will have achieved our end.

OUf first step toward (5) is to add A + B to both membersof (4), getting)

B + D = A + C.) (6))

Since the sum of coefficientson eachside of equation (6) is 2,we divide both membersby 2, obtaining)

\037B + \037D=

\037A + tC.) (7))

Both members of (7) comply with the conditions of Theorem4. Thus the left member represents a vector emanating from

0, and whose endpoint must be on BD; and the right memberrepresentsa vector emanating from 0, and whose endpoint

\037

is on AC. Therefore the vector (on each side) must be OP,for P is the only pointonbothBDand AC. ....L\\.pplying Theorem

4 once again, we concludethat P divides both BD and AC inthe ratio \037: \037, which is the desired result.

EXAMPLE7. Employing the same techniques, we attemptto prove the familiar result: The medians of a triangle meetina point two-thirds the way from a vertex to the oppositeside.

Let triangle ABC have M and N midpoints of sides BC and

AC, respectively (see Figure 23). CallP the intersection of

medians AM and BN, and let 0 be a point in general position.Applying Theorem 4 to the fact that M is the midpoint of BC,we have)

\037

OM = M. = tB + \037C.

Similarly, N being the midpoint of AC yields)

(8))

\037

ON = N = tA + tC.) (9))

Attempting to achieve P as a linear combination of A and M,and also as a linear combination of Band N, \\ve subtract (9))))

ELEMENTARY OPERATIONS

from (8) to eliminate C:

M - N - IB - lA-

2\"\" 2\"\".

Then, adding \037A + N to both members, we get

iA + M =\037B + N. (10))

33)

The sum of the coefficients on each side of (10) is \037, so we

multiply both members by i. Thus

l A +2M - IB + 2NT -g-

--g- \037.

The left member of (11) representsa vector whose origin is

o and whose endpoint is on AM, and the right member of(11) representsa vector whose origin is 0 and whose endpointis onBN. But the left and right members of (11)aredifferent

representations of the same vector. This forcesus to concludethat this vector is P, that is,)

(11))

P = j-A + iM = j-B+ iN. (12)Equation 12 tells us that P divides AM in the ratio i:j-.(Noticeonceagain how the coefficients in the equation allow

\037 \037 \037 \037

us to deduce that lAP!: IPMI = !BPI:!PN!= t:t.

Query: Why is it now clear that all three medians intersect

in a point?

Remark. In the examples presented,the point 0of common origin of the vectors ,vas always selected in)

A)

B) M) c)

FIGURE 23)))


general position. This is, of course,not necessary.

Moreover, placing the point of common origin in somespecial,judiciouslychosellpositionoften leads to con-

siderable simplification. Exercises 1, 2, and 3, which

follow, have been provided to illustrate this point.)

EXERCISES1. Carry out Example 5 when point A is the point of emanationof the severalvectorsin the proof.2. Dothe same for Example 6.

3. Do the same for Example 7.

4. Prove that the midpointsof consecutive sides of a quadri-lateral areverticesof a parallelogram by making use ofTheorem4 (compare with Exercise 7 on p. 20).5. Given triangle ABC with N on AB and ilf on AC, such thatMN is parallelto BC,prove that AN:NB = AM:MC. Whatis the relationshipbetweenthe fractions AN

If>!Band NM/BC?

6. Prove that the line joining a vertex of a parallelogram to

the midpoint of an oppositeside trisectsthe diagonal crossed

by it.

7. Generalization of 6. Let ABCD be a parallelogram wit.h

P on AD such that AP = (l/n)AD. Prove that BP inter-

sects diagonal AC in a point Q, whose distance from A is[l/(n + l)]AC.8. (a) Prove the analogue of Theorem 3, namely: If A, B,

and C are three nonzero, noncoplanar (cannot be placedin oneplane) vectors, any vector in space can be expressed as alinear combination of A, B, and C. (Hint. A parallelepiped

is the space analogue of the parallelogramin Theorem3.)(b) Using the result of part a, prove that any four vectors

in space form a linearly dependentset.)

6. UNIQUENESS OF REPRESENTATIONS

The importance of Theorem 3 liesin the statement that

any vector in a plane call be expressedas a linear com-

bi11ation of two given linearly independent vectors in)))


that plane. The question we now pose is: Is such a

representation unique? More precisely, let A and B be

the given linearly independent vectors and C somearbi-trarily chosen vector in the plane of A and B. Then, byTheorem 3, we know that there exist scalars m and n sothat)

C = mA + nB.) (13))

But is it possiblethat there is another representation,

perhaps different from (13), of C as a linear combination

of A and B?

Suppose, then, that there are scalars rand s so that)

Then

and)

C = rA + sB.

mA + nB = rA + sB

(m -r)A + (n

- s)B = O.)

(14))

However, A and B being linearly independent impliesthat m - r = 0 and n - s = 0;hence,m = rand n = s.Thus (14)is precisely the same represelltation of C as is(13). As a result, we say that the representation of C as a

linear combination of A and B is unique.

Similarly, we could prove the general statement that

the representation of a vector as a linear combination of

linearly independent vectors is unique. SupposeC =

alAI + a2 A 2 + \302\267\302\267\302\267+ anAn = blAl + b2 1\\2

+ \302\267\302\267\302\267+ bnAn,

where {AI, A 2 , . . . ,An}5

is a linearly independent setof vectors. Then)

(al -bl)A I + (a2 - b 2 )A 2 + \302\267\302\267\302\267

+ (an-

bn)An= O.

Again, the linear indepelldence of the A's \037mplies that

al = bl, a2 = b 2 , . . ., and an = bn . The reader)

5Again, the brace symbolism represents a set, with a listing orroster of the elements of the set.)))


should check the geometric construction in Theorem3 toobserve the uniqueness of the representation of C as alinear combination of A and B.

We now employthesefacts to attack some problems of

geometry.

EXAMPLE8. We return once more to the parallelogramproblemin orderto prove that the diagonals bisect each other.Referring to Figure 24, we write)

\037

PT = meA + B).) (15))\037

We can achieve another representation of PT by considering\037 \037 \037

it as one side of triangle PQT, that is, PT = PQ + QT.\037 \037 \037

But since QT is part of diagonal QS, we may write QT =\037

nQS = nCB - A). Thus)

\037

PT = A + neB -A).

From (15) and (16) we conclude that)

(16))

\037

PT = mA + mB = (1 -n)A + nB.)

Since A and B are linearly independent, the representation of\037

PT as a linear combination of A and B Inust be unique. That)

R)

FIGURE 24)))

ELEMENTARY OPERATIONS) 37).IS,)

m=1-n) and) m =n,)

which imply that m = n =\037, proving the desired result.

The reader shouldcomparethis approach with that of

Example 3 to see how closelyalliedthey are (as are the

concepts of linear independence and uniqueness of

representation).

EXAMPLE 9. We turn to the median problem, using thenew approach. Referring to Figure 23 again, we note that

\037 \037 \037 \037

AP is a part of median AM, so we write AP = mAM. Simi-\037 \037

larly, PN = nBN. We seek to determinern and n.)

\037 \037 \037 \037

AP = mAM = m( \037AB + \037AC).) (17))

\037 \037

AB and AC constitute a pair of linearly independent vectors.\037

If we succeed in finding another representation of AP as a\037 \037

linear combination of AB and AC, we may then employ ournew technique. To this end we write)

\037 \037 ---4AP =, AN + NP)

\037 \037

=\037AC - nBN)

\037 \037 \037

= tAC- n(tBA + tBC))

\037 n\037 n\037 \037

=\037AC -

2BA -

2 (AC- AB).)

Finally,)\037 I-n\037 \037

AP =2

AC + nAB.

Comparing(17)with (18) gives)

(18))

m\037 m\037 \037 1-n\037

\"2AB +

2\"AC = nAB + 2 AC,)

which-because of the uniqueness of representation-allows)))


us to state that)

m-=n2)

and)m 1- n- =

2 2)

The solution to this pair of linear equations is)

2m=-

3)and)

1n =-.

3)

The reader should observe that there was a persistent\037

effort to get two representations of AP as a linear com-\037 \037

bination of the two specific vectors AB and A C. A usual

difficulty encountered by the beginningstudent revolvesaroundthe problem of getting the \"right\" relationships.He knows that hewill often get two independent relation-

ships by summingvectorsaround two different polygons,

e.g. ,)

\037 \037 \037

AP = AB + BP) and)

\037 \037 \037

AP = AN + NP.)

But where to go from here? If he bears in mind the---?

general aim to get different representations of AP (for

instance) as a linear combination of the same set of

linearly independent vectors, he may employ the techniqueof equating coefficients.)

EXERCISES1. Prove that the medians of a triangle meet in a point bythe method of Example 9, but now by getting different repre...

\037

sentations of NP.

2. Re-do Exercises 6 and 7 on page 34, making use of themethod of Example9.3. Show that the line join.ing the midpoints of a median to avertex of a triangle trisects the sideoppositethe vertex.4. Given quadrilateral ABCD, with the condition that BDbisectsAC, call F the intersection of AB and CD; call G the

intersection of AD and Be. Prove that AC is parallel to FG.)))


(This is a difficult problem, so we provide the following hint.\037 \037 \037 \037 \037 \037

Let DG = xDA and DF =yDC. Note that DB =.uDA +

\037 \037 \037 \037

uDC. Then call CG = mBG and expressBG as a linear com-\037 \037

bination of DA and DC, in two different ways, to determinex in terms of u. Follow a similar procedure to determiney.

\037

in terms of u. Finally, if you expressGF as a linear combina-\037 \037

tion of DA and DC, you will have the desired result.)

5. Prove the following Theorem: Let the points B' and A'divide the sides CA and CB of triangle ABC in the ratiosn: (1-

n) and m: (1 - m). Let the point P be chosen sothat A'CB'P is a parallelogram with AC' and CB' as two ofits sides. CallD the intersection of CP and AB. Then Ddivides AB in the ratio m: n. (Howwould you choose nand

m to be sure that CD would bisect angle C?))))

vectors in

coordinate

systems)

'1.RECTANGULAR SYSTEMS AND ORIENTATION

The reader is undoubtedly familiar with rectangular

coordinate systems in the plane, where two perpendicularlinesare chosen as axes. One is called the x-axis, theother the y-axis, and their point of intersection the

origin. A positive direction is chosen arbitrarily on eachaxis,and the customary correspondence is made between

points of each axis and the real numbers, where the

positive real numbers are on that side of the origin

arbitrarily designated as the positive part of the axis (see

Figure 25). The standard convention (by no means

binding) is to have the horizontalaxiscalledthe x-axis,

with its positive side to the right of the origino. Thevertical axis is then the y-axis, with its positive sideabove the origin (see Figure 25b). Now, consider any

point P in the. plane. From P we drop perpendiculars

to the axes. Call Px a11dPy the feet of these per-40)))

VECTORS IN COORDINATE SYSTEMS) 41)

y) y)

PyP (x, y)

----1 tI y

(-3,2) . . (3,2) II

x\037x

0X

P%)

(a)) (b))

FIGURE 26)

pendiculars on the x- and y-axes,respectively.The real

number associated with P x on the x-axis is calledthex-coordinate (or abscissa) of P and the real numberassoci-ated with Py on the y-axis is called the y-coordinate(orordinate) of P. P is then designated by a11orderedpair(x, y) of real numbers: the abscissa occupying the first

position in the pair and the ordinate occupyingthesecondposition.If the point P is on the x-axis, we give it ay-coordinate of 0; and if P is on the y-axis,we give it

an x-coordinate of o. Thus the origin0 has coordinates.

(0, 0). This method of associating pairs of real numbers

with points has the advantage of associating exactly one

pair with each point, and exactlyonepointof the plane

with each ordered pair of real numbers.Rectangularcoordi11atesystems in space follow the

same general pattern as do such systems in the plane.Three mutually perpendicular (intersecting) lines are

selected as axes: the x-axis, the y-axis, and the z-axis,

with the point of intersection of the three axes being

called the origin. Again, a positive direction is chosen)))

42) ELEMENTARY' VECTOR GEOMETRY)

arbitrarily on each axis, and the usual correspondence is

made between the points of each axis and the real num-bers. The zeropointoneach axis is taken at the origin,and the positivereal numbers are on that side of theoriginwhich has been designated as the positive side of

the axis.

Following the same pattern of development as in the

plane, we consider any point P in space. From P wedrop perpendicularsto the axes, and we call P x, P y and

P z the feet of these perpendicularson the x-, y-, and

z-axes, respectively (see Figure 26a). The real numberassociatedwith P x on the x-axis is called the x-coordi11ateof P; the real number associated with Py on the y-axis

becomes the y-coordinate of P; and the real numberassociated with P z on the z-axis becomes the z-coordinateof P. If P isonany of the axes, not all the perpendicu-lars may be drawn. Thus we state further that if P is

on the x-axis, its y-coordinateand its x-coordinate are

both zero. Similarly, if P is on the y-axis,its x- and

z-coordinates are both zero. Finally, the coordinatesof the origin are all zero.

It is quite convenientto visualize P (when P is not onan axis) at the corner of a rectangular solid, with theorigin 0 at the oppositecorner, as shown in Figure 26b.This may suggest to the reader other approaches he

might prefer in finding the coordinatesof P, or in locatingP if its coordinates are given. For example, suppose weknow the coordinatesof P. We can then locate P byfinding the point on the x-axis that corresponds to its

x-coordinate, then movingparallelto the y-axis a dis-

tance that corresponds to the y-coordinateof P, and then

parallel to the z-axisa distancethat correspondsto the

z-coordina te of P.

When we designate a pointby its coordinates in space,the order of the triple of numbersfollows the alphabetic

order, that is: (x, y, z). We stated that, in dealing with

plane coordinates, it is customary to take the x-axis as)))

VECTORS IN COORDINATE SYSTEMS) 43)z)

z)

y)

/)

y)

A)

x) x)

(a)) (b))

FIGURE 26)

horizontal, etc. In space, the conventionis to have the

positive z-axis pointing upward and the xy-planehori-zontal. It isalsocustomary to have one positive axis

pointing toward us and the otherpointingto our right.

However, which axis points toward us and which to our

right seriously influences our vector development.Consequently,before we finally decide on the orientationof the axes, it would be well to explain the notions of

right-handed and left-handed triples of vectors.Let {A, B, C} be an ordered set of three linearlyinde-

pende11tvectors (which can always be considered asemanating from the same point 0). Since the vectorsare linearly independent,they do not all lie in one plane.Moreover, no two are on the same or parallel lines. Thusthe first a11d second vectors, A and B, form an a11gle(J (0 < 8 < r).l Remember! It's the order of first,

1Here, and throughout the entire book,angle measurement is in

terms of radians. The use of degrees is due to an. unfortunate

historical accident and serves to confusestudents,especially when

they reach the study of trigonometric functions in calculus.

Because radian measure serves the mathematician in good stead

throughout the whole field of mathematics, we prefer to propa-gandize by making exclusiveuseof it.)))


c)

o)

{A, B, C} right-handed triple

(a))

0\\\\\\

\\

\\)

C

{A, B, C} left-handed triple

(b))

FIGURE 27)

second, tllird il1 the triple that counts. Now, consideran observerstatio11edonthesideof the plane of A and Bthat allowshim to walk from a point on A through theangle 8 to a pointonB,with his outstretched right arm

always pointing away from 0 (seeFigure 27). That is,

he walks about, keeping point 0 always on his left. If

the observer's head is on the same side of the plane of)))


A)

FIGURE 28)

A and B as the vector C, we say {A, B, C} is a right-ha'nded or positive triple. It should be clear, then, that{B,C,Ai and {C, A, B} are also right-handed triples,whereas{B,A, C}, {A, C, B} and {C, B, A} are left-

handed or negative triples. If you think of the vectors asseats at a circulartable,all clockwise readings (starti11g)

{A,B, C} right-handed) {A,B, C} left-handed)

(a)) (b)')

FIGURE 29

(a) ()rdinary screwgoesin. (b) Ordinary screw comes out..)))


at any letter) yield triplesof the same orientation and all

counterclockwise readings yield triples of the opposite

orientations. This situation is described succinctly by

stating that a cyclic permutation of the vectorsof a triple

does not change its orientation as a left-handed or right-

handed triple, as the case may be.A useful mnemonic is found by studying a screw.

Think of the first two vectors of the triple as beingon the head of an ordinary (right-handed) screw. Turnthe first vector into the second through the angle lessthan 1r. If thishasthe effect of driving the screw in

the general direction of the third vector, we say thetriple is right-handed.)

8. BASIS VECTORS AND APPLICATIONSLet i, j, and k be three unit vectors in the respective

positive x-, y-, and z-directions. We take {i, j, k}to be a positivetriple,and this establishes the right-handedness of tbe (x, y, z)-coordinate system. That is,if three vectors in the positivex-,y-, and z-directions form

a right-handed triple, we call the (x, y, z)-coordinate

system right-handed; otherwise we say it is left-handed

(see Figure 30.).

Since i, j, and k are linearly independent, we know

(by Exercise 8, p. 34) that any vector in space can be

expressed as a linear combination of them. In par-\037

ticular, let P = OP where 0 = (0,0,0)and P = (x, y, z);then P = xi + yj + zk. P is referred to as the positionvector of point P (see Figure 30c).

We wish to extend our algebra of vectors so that we

may work with the vectors i, j, and k. To explore

addition, subtraction, and multiplication by scalars inthis form, let)

and)

A = a1i + a2i + a3k

B = b1i+ b 2j + bgk.)))

VECTORS IN COORDINATE SYSTEMS 47)

z z)

k)

j):y) :y)

:JC)(a))

x)

(x,y, z)-systemright-handed

(b))

{i,j,k} right-handed)

z)

y)

x)

(c))

FIGURE 30)

Then by Theorem 1,A + B = (al + b1)i+ (a2+ b2)j + (a3 + b3)k,

and by Theorem2 (iii)

mA = mali + ma.2j + ma3k,)

and)

mA + nB = (mal + nb1)i+ (ma2 + nb 2)j+ (ma3 + nb 3 )k.)))


We see, therefore, that sums, scalar multiples, and, in

general, linear com1?inations of vectors expressed in termsof i, j, and k can also be expressedas linear combinations

of vectors i, j, and k. That is, every vector in space

can be expressed as a linearcombination of these three

unit vectors. Mathematicians describe this situationby

saying the vectQrs i, j, and k span or generate the space

under consideration. Furthermore, the set of vectors

{i, j, k} is said to serveasa basisfor the space. A basis

is a linearly independentset that generates the space. The

need for a basis to generate the space is quite clear, butwhat is the purposeof the stipulation of linear inde-

pendence? The \"answer lies in the desire to have a vectorexpressedin a unique manner relative to a given basis,that is (in our present discussion), if

A = a1i + a2j + agk= c1i + C2j + cgk,

we may conclude,as we did earlier, that)

al = CI, a2 =C2, a3

=Cg.)

Suppose, on the other hand, that we used a linearly

dependellt set {i, j, k, h} to generate the space. Then

any vector V could be written in the form)

v = vIi + v2j + vgk + V4 h .) (19))

rfhe li11ear dependence of the set {i,j, k, h} implies the

existence of scalars CI,C2, C3, and C4 (not all zero) such that)

cli + C2j + cgk + C4 h = o.) (20))

Addi11g (19) alld (20)yields

V = (VI + cl)i + (V2 + C2)j + (V3 + cg)k + (V4 + c4)h.

Thus, if h \037 0, we have demonstrated the existen:ceofdistinct representationsfor V as a linear combination of

i, j, k, and h. Actually there are an infinite number of)))

VECTORS IN COORDINATE SYSTEMS 49

such representations, which can be demonstrated by

multiplying (20) by an arbitrary scalar beforeadding itto (19).

Remark. The reader has no doubt observed that we

have avoided any reference to the plane as being two

dimensional and space being three dimensional. Ourreasoningis in accordal1ce with the thinking of mathe-maticians who deal with so-called \"vector spaces.\"

They do not choose to assumeany knowledge of the C011-

cept of dimension. They seeit as naturally related to

other concepts involving vectors. In fact, they define

the dimensiol1 of a space-and do so in one of two com-

pletely equivalent ways:)

(1) Thedimension of a space is taken as the numberofvectorsin a basis for the space. (Of course, this numberis shown to be independent of the choice of the basis.)

(2) The dimension of a spaceis the maximum number

of linearly independent vectors in the space.)

Consequently,we see that the line would be one dimen-sional. Theorem 3 and its corollary imply that the planeis two dimensional. Finally, Exercise 8 on p. 34 estab-lishesthe three-dimensional character of what we havecalled \"space.\" The foregoingdiscussion,which notes

that i, j, and k are linearlyindependent,and that adding

any other vector to this set of three would result in a

linearly dependent set, further establishesthat \"space\"

is three dimensional.)

EXAMPLE 10. Let pointsbe given as follows: 0 = (0, 0, 0),A = (1, 3, 4), B = (1,5,2),C = (-2, 1,6), D = (-2, 5, -2),and E = (0,2, -2).

(1)The position vector of point A is given by A. = i + 3j +4k and the position vector of point E is E = 2j - 2k.

\037

(2) How should we write the vector AB? Noting Figure\037

31, we see that AB = B - A. Hence we write)))


\037

AB =. (i + 5j + 2k) -(i + 3j + 4k)

= (1 - l)i+ (5- 3)j + (2 -

4)k

= 2j- 2k = E.

(3) How do \\ve write a vector emanating from 0, pointing\037

toward C but half the length of C ? We seek! OC. This ismerely)

\037C= i( -2i + j + 6k)

\037C= -i + ij + 3k.

(4) How do we write a vector from 0 to the midpoint ofsegmentBC? Calling it! the midpoint of segment BC, weemploy Theorem 4 to ,,,,rite)

\037 \037 \037

OM = M = iOB + \037OC (21)

=\037(i + 5j + 2k) + \037(

- 2i + j + 6k)= -

\037i + 3j + 4k.

Observing that M is the position vector of M, we can state thecoordinatesof the midpoint of BC: M = (-t, 3, 4). This)

z)

y)

x)

FIGURE 31)))


c)

o) D)

FIGURE 32)

procedure can be generalized to find the midpoint of any seg-ment. Call

PI =(Xl, YI, Zl) and P2 =

(X2, Y2, Z2).

Then, if ill is the midpoint of P IP 2, we can write)

\037 \037 \037 \037 \037

0111 =\037OPI + \037OP2 =

\037(OPI + OP2)

=\037[(xli + Ylj + zlk) + (x2 i + Y2j + z 2k)]

=\037(Xl + x2)i + l(YI + Y2)j + \037(Zl + z2)k.

Thus the midpoint is

M =

(

Xl + X2 YI + Y2 Zl + Z2

)

.2

'2 ' 2

(5) What are the coordinates of the point of intersection

of the medians of triangle BCD? Referring to Figure 32, weseek the coordinatesof P. Recalling that P divides DM inthe ratio 2:1, and using Theorem 4 once again, we write\037 \037 \037 \037

OP = jOM + -lODe Employing equation 21, we have OP =\037 \037 \037

f( \037OB + lOC) + iOD,or)

\037 \037 \037 \037

OP = (t)OB + (t)OC + (i)OD.) (22))))


Consequently,)\037

OP = (j.)(i + 5j + 2k) + (j.)(-2i + j + 6k)

+ (j.)( -2i + 5j - 2k).Then P = (-1, 11/3, 2). It is hardly necessary to state thatthese methods are valid no matter what the relative positionsof the points are. That is, even if 0, B, C, and D are in oneplane,the formula (22) still holds true.

EXAMPLE 11. Prove that the (radial) vectors drawn from

the center of a regular polygon to its verticessum to the zero

vector.The reader should-before reading the next paragraph-

attempt a solution to this problem when the polygon is atriangle, square, and pentagon. He may choose to set thepolygon within a coordinate system and perhaps, if necessary,resortto somehelp from trigonometry or elementary geometry.Before proceeding, the reader should give some thought tothe original problem, which does not specify any particularregular polygon but concernsthe general case.

Consider the origin of a two-dimensional coordinatesystem

to be at the center of the regular polygon of n sides (see Figure33) . Let S be the sum of the radial vectors. If S is not thezerovector,it has a unique inclination of a radius with respect

to the x-axis. Rotate the polygon 21r/n radians about the

origin. The sum vector S is now inclined a + 21r/n radiansto the x-axis. Sincethe rotated figure has precisely the same)

y)y)

\037x) %)

FIGURE 33)))


appearance on the coordinatesystem as the unrotated one, thenew sum 8' of the radial vectors must again be inclinedaradians to the x-axis. Thus we have two vectors Sand 8'inclined differently while both represent the single sum in

question. Since we know that the sum vector in unique, i.e.,S = S',the only compatible solution is the vector 8 = 5' = O.)

It should be emphasized that it was the symmetry of

the regular polygon that enabled us to constructthisproof. The student of natural science should always beon thelookoutfor such intrinsic properties that may helpto simplify the physical as well as the geometric con-siderationsof his problems. Scientists encounter-and

make use of-symmetry in such diversefields as geome-

try, algebra, botany, zoology, electrical circuit theory,mechanics,and optics. The reader who wishes to k110W

more about the prominent role played by symmetry in

art and science is referred to the fascinatinglectures of

Hermann Weyl, which are contai11ed i11his book, Sym-

metry, Princeton University Press. A portion of Weyl'sbook ,is reproducedin The World of Mathematics byJames R. Newman.)

EXERCISES

1. (a) Locate on a sheet of graph paper the points A =

(6,4,10), B = (-6,4, -10), C = (4, -6, -10), and D =(0, 10, 4).

(b) Write the position vectors A, B, C, and D in terms of theIi, j, k}-basis.

(c) Find the sum A + B + C + D graphically, and check

this against the computation in the Ii, j, k}-system.\037 \037

(d) Compute A - B, D - C,BD, and AC.

(e) Find the midpoint of segnlent AB.

(f) Find the coordinates of the point X that divides segmentAB in the ratio 2: -1. ' .

(g) Find the coordinates of the point Y so that B is the mid-point of segment AY.

.

(h) Find the median point of the triangle ABD.)))


2. Let A = 2i - 4j, B = -i -2j, C = i + j + 3k, and

D = 2i - 3j+ k.(a) Determine 2A as a linear combination of i, j, and k..

(b) Determine -3B as a linear combination of i, j, and k.

(c) Determine3B- 2A as a linear combination of i, j, and k.(d) Find A + B - C as a linear combination of i, j, and k.

(e) Is {A, B, C} a linearly dependent set?(f) Is {A, B, C, D} a linearly dependent set?(g) Let OAXC be a parallelogram, with 0 = (0, 0, 0), A =

(2, -4, 0), and C = (1,1,3). Find the fourth vertex. (Hint:How is the sum of two vectors related to a paralielogram?)

3. Do the position vectors A, B, and C of Exercise1 form a

linearly dependent set?

4. Let A = i - j, B = i + j, and C = j- k.

(a) Is {A, B, C} a linearly dependent set?(b) Expressthe vectorV = 2i + 4j - k as a linearcOlnbina-

tion of A, B, and C.

5. (a) Provethat the sum of the eight vectors from the centerof a cube to the vertices is the zero vector. Do this by assign-

ing coordinates to the vertices, writing the eight vectors

explicitly, and then summing.(b) Let \037 be a regular dodecahedron (12 faces) and sits

circumscribedspllere. Prove that the sum of the vectors from

the center of S to the verticesof \037 is the zero vector. Howmany such radial vectors are there ? (You might look this

up in a solid geometry text.))

9. THE COMPLEX PLANE

A two-dimensio11al space that naturally admits toanalysis il1 terms of vectors is the complex plal1e,which

may be familiar to the reader from his studies in algebra

and trigonometry. 111 order to see the complexplane

from the vector point of view, consider a rectangularcoordinatesystem with a basis consisting of two unitvectors:1and i. 1 is taken il1 the positive x-directiol1arIdi il1the positivey-direction (see Figure 34). The unitvector i is actually the imaginary Ul1it i = VI - 1.)))

VECTORS IN COORDINATE SYSTEMS

Y)

55)

i)

1)

o)%)

FIGURE 34)

That is, the role of basis vector i (of Section 8) is now

played by 1,and the roleof j isnow played by i.)

Old

i)

New

- 1- i)

.J)

\037

If P = (x, y), the vector OP = xl + yi, or, more\037

simply, the vector OP may be thought of as the complexnumber x + iy. Thus every complex number is actuallya vector il1 the plane. If scalarsare taken to be real

numbers, the multiplication of vectors by scalars cor-

responds precisely to multiplicationof complex numbers

by real numbers (see Exercise 2, page57). Is it also

true that vector addition corresponds to addition of com-

plex numbers? To answer this question we considertwo

complex numbers: CI =Xl + iYI and C2 =

X2 + iY2,their related vectors being Cl

= xII + y1i and C2=)))

56) ELEMENTARY VECTOR GEOMETRY

y)

(Xl + X2,Yl + Y2)- -----------1

r C2 I i

Yl + Y2 I I

I I

\"1------ __.J

ICl:)

o)Xl + X2)

X2>1)

\037X)

FIGURE 36)

X21 + Y 2 i. Addition of complex numbers allows us tosay)

CI + C2 =(Xl + X2) + i(Yl + Y2);

and addition of vectors permits us to write

Cl + C2= (Xl + x2)1 + (Yl + Y2)i.

Thus we see that Cl + C2 corresponds to the vectorCl + C2.

EXAMPLE 12. Considering Example 11 in the light of

this vector approach to the complex plane, we can reach an

extremely simple and elegantsolution.Let the regular polygon be centered at the origin, with one

of its radial vectors lying on the x-axis. Figure36 illustrates

the approach for the pentagon. If the radial vectors arechosen of unit length, we know from algebra that the n vectors,as complexnumbers, are simply the solutions of the equation)))


xn = 1. But the sum of the roots of this equation is zero,for the coefficient of the Xn-l term is zero. This completesthe proof.)

EXERCISES

1. Sketch the vectors representing the complexnumbers 2 - iand 3 + 2i. Give a construction for their addition and sub-traction, and compare this geometric view of the operationswith the algebraic.

2. Exhibit the following scalar multiplesof 2 - i:

2(2 - i), j(2 - i), \037(2- i), -(2 - i), -2(2 - i).

What geometric fact can one deduce about the set of (real)

scalar multiples of a complexnumber?

Remark on Length and Absolute Value. From ourbrief discussionof complex numbers as vectors we cangain someinsight regarding the symbolism I \\,

which is

used to denote the absolute value of a real or complex)

y)

x)

FIGURE 36)))


number as well as the length of a vector. The two

notions, absolute value and length, are actually inter-

related, or, more accurately stated, the notion of absolute

value is a special caseof the length of a vector. This is thebasisfor the same choice of symbolism.

The absolute value or modulus of a complex n umber

a + bi, written la + biLis defined as V a 2 + b2

. But

if we view the complexnumberasa vector in the complex

plane (see Figure 37), then the length of V = al + bican be determinedby the Pythag orean theorem

lal + bit= V a 2 + b2 =

fa + bil.

In the event that we are dealing with real numbers, wemay treat themas a subsetR if

of the complex numbers.

That is,

R# = {a + bitb = O}.

Thus the length of r = al + Oi isIrl

= v? = lal.In both caseswe see that the absolute value of a number)

iy)

a + bi)

b)

x)

FIGURE 37)))


-2)>, , I I I I 1'1'1 I I

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

131=3=1311-21=2=1-21)

<3)

::.-)

FIGURE 38)

is merely the length of the vector associated with thatnumber. One may also considerthe real numbers as

constituting a one-dimensional space, Le., as a line (seeFigure 38), with the same result: absolute value of the

real number equals length of the vector.Fromthis geometric view of absolute value one can see

the plausibility of the following properties of absolute

value:)

(i) I al=

1- al > 0

(ii) la + bl< lal + Ibl

(iii) lal-

\\b(< \\a + bl

(iv) \\lal-

Ibll< \\a

-bl.)

EXERCISE

1. Considering a and b to be complexnumbers, give geometric

interpretations of the algebraic properties (i) through (iv)

above.)))

\

.Inner

products)

10.DEFINITION

In this chapter we begin to introduce quantitativeaspectsintoour vector algebra. Of particular interest

are the notions of distance and angle. In orderto seehow best to introduce such concepts, it might be advisa-bletohave a look at them in the framework of coordinategeometry.

Sincethe planeis more easily dealt with than three

dimensions, we let A = (aI, a2) and B = (bl, b 2 ) be

points in the (x, y)-plane (Figure39). The distance d,

between A and B, \\vhich w\037 denote IABI, can be found

(by the Pythagorean theorem)from the formula)

or)

d2 = (al -

b l )2 + (a2 -b 2 )2,

d = vi (al -b 1)2 + (a2 - b 2 )2.)

(23)

(23a))

Note that it makes no differencewhether A or B

is considered first, for d is alsoequal to

60)))

INNER PRODUCTS) 61)

V (b1 - al)2+ (b 2- a2)2.

Expanding (23) results in

d2 = a12 + a22

+ b 12

+ b22 - 2(alb 1 + a 2b2). (24)

If we introduce--for conveniencein the present dis-

cussion-the notation A *B = alb l + a2b2, we can rewrite(24) as

IABI2= A *A + B*B - 2A *B. (25)

Hence we have distance between two points describedcompletely in terms of the symbol

* (al1d, of course, +

and -).Turning to anglemeasurement, we examine Figure 39a,

seeking an expression for the angle (J. By the law ofcosines we write)

IABI2=

IOAI2 + IOBI2- 21oAIIoBIcos(J;)

y)

x)tal

-bII

B--------,F

I

lI a 2 -b21

I

A)

(4)) (b))

FIGURE 39

N ate. In (b) above, it is possible that a - b is positive, nega-

tive, or zero. We therefore designate the distance (non-negative)between A and F (also B and F) by absolute value in order to

assure the non-negative character of distance. Consequently

(23a) can be written d = Vial -bll

2+ la2 -

b212

.)))


or, using our * notation,)

IABI2= A *A + B*B - 2 V A

* A v' B*B cos 8. (26)

Uponsubtracting (26) from (25) and simplifying, wefind)

A*Bcos 8 = ,

V A*A VB*B)(27))

so we seethat angles can also be expressed in terms of the* notation.

Drawing motivation from this discussion-primarilyfrom (27)-we start afreshwith vectors (not restricted to

two dimensions) and definethe inner product of A and Bas)

A \302\267B =IAIIBI cos 8,) (28))

where 8 is the angle between the two vectors when they

are arranged to emanate from the samepoint. Accord-

ingly, the notions of distaIlce (length) and anglearebothincorporatedin our definition of inner product. Notethat it is immaterial whether 8,

- 8, or 2r - 8ischosen,

for cos 8 = cos ( -8)

= cos (2r - 8). (SeeFigure 40.)

Because of the notation employed, the inner productis alsocalledthe dot product. Still another name is the)

FIGURE 40)))

INNER PRODUCTS) 63)

scalar product, for this method of \"multiplying\" two

vectors yields a scalar (examine equations 28). All

these terminologies enjoy popularity in mathematicsand physics textbooks, so we shall employ them all in

order that the reader may feel at home with anyone.

Two immediate corollariesto the defillitiollare)

A \302\267B = B \302\267A (commutativity of dot product)

and A \302\267A =IAI2 (since cos 0 = 1).

If A is perpendicular to B, A. B = o. (Why?)

However, if A \302\267B = 0, there are three possibilities:)

(1) A = 0,

(2) B = 0,

(3) A is perpendicular to B.

If we agreethat the zero vector is perpendicular to everyvector (see page 22 for justification of this convention),we can combine these conclusionsin a single statement:

Theorem 5. If A is perpendicular to B, then A \302\267B = ();

and conversely, if A \302\267B = 0, then A is perpendicular to B.

Word of Caution. Many beginning students havedifficulty in believing that vectors in space are perpel1-dicular if they do 110t intersect. It must therefore beemphasizedonceagain that the definition of equality ofvectors permitsus to move a free vector as long as it iskept parallel to its original position; this enables us tothink of any two vectors as intersecting. It is the sin-cere hope of the author that the reader has long sinceunderstoodthispointand hence is thoroughly bored withthe word of caution.)

11. PROPERTIES OF INNER PRODUCTAn examination of Figure 41 leads to an interesting

geometric interpretation of the inner product. From)))


B)

o) c) A)

FIGURE 41)

triangle OCB we see that cos () =

:\037I

or)

\037 \037

loci=

lOBI cos (J.)

\037 \037

Since OB = Band OC is the \"vectorprojectioll\"of

B on A, we write

projection of BOll A =IBI cos 8,1 (29)

which smacks of the inner product A. B. In fact,multiplying by IAI results ill the right member of (29)

becoming A.B. That is,

IAI (projection of B on A) =lAllBI cos 8 = A \302\267

B, (30)

which gives us a geolnetric associationwith the inner

product concept.Before proceeding, we must clarify the ideas contained

in (30). The reader may feel-and rightly so-that per-haps too much has been iIlferred from a rather special

picture. We therefore provide the following definition.)

1 Note that (29) gives the projection of one vector upon anotheras a scalar, which may be positive, negative, or zero.)))

INNER PRODUCTS) 65)

Definition. By the projection of B on A, written pr AB,we mean the orthogonal projection of B on the line of actio\"n

of A. The pr AB is obtained by dropping perpendicularsfrom the origin and endpoint of B to the line of act\037.on of A

(see Figure 42). The distance between the feet of these

perp\037ndiculars is the magnitude of the pr AB. If the angle (J

between A and B (when A and B are arranged to emanatefrom the same point) is acute, then prAB is positive; if (J

is obtuse then pr AB is negative.Now that the notion of projecting one vector upon

another has been made precise,we restate (30) as

Theorem 6. A. B =(prBA)IB\\

= (prAB)\\A\\.

The completion of the proof is left as an exercise for

the reader.From this result we prove that the inner product is

distributivewith respect to addition, which we state asTheorem7. A. (B + C) = A. B + A. C.)

\037-

rvA

)

\\ /y

prA B> 0)

y

prA B < 0)

I

I

I

I

I

I

I

-CL__/)

A< II

\\)

\037)I

I

I

_ _ -..t:L _\\

y

prA B> 0)

I

I

I

ri A> __

I)

I

I

_J:J__/

y

prA B> 0)

FIGURE 42)))


I

II I

AI I

11._____b...____\037

prA B prAC)

II

I

II

I

___-1_____'----v----/'--v---'

prA (B -I: C) pr A B)

A)

\\.)v

prA(B + C))

J)\\) \037

pr A C < 0)

1)

FIGURE 43)

Proof. (See Figure 43.)

A \302\267(B + C)

= prA(B + C)IAI (by Theorem 6)= (prAB + prA C) IAI (justify!)= (prAB)IAI + (prAC)IAI

= A \302\267B + A \302\267C (by using Theorem 6 once again).

Corollary.)

(A + B) \302\267(C + D) = A \302\267C + A \302\267D + B \302\267C + B.D.)

Proof. By Theorem 7 we write)

(A + B) \302\267(C + D) =

(A + B) \302\267C + (A + B) \302\267D,)

and leave the remainder of the prooffor the reader.

EXAMPLE 13. vVe prove by vector methods that the per-pendicular bisectors of a triangle meet in a point.

Becausevectors may be \"moved,\" we adopt what may seemto be a strangeapproach. The reader should become familiarwith it, for such an approach is extre\037ely useful when severallines meeting ill a point must be sho\\\\yn.

Let the triangle be ABC, with the perpendicular bisectorsof AB and BCmeetingin point 0 (see Figure 44). Let M, N,and P be the midpoints of sides AB,BC,and AC, respectively.

1\"he approach. \\ve adopt is to show that OP is actually per-

pendicular to A C.)))

67)INNER PRODUC.TS

We begin by stating the hypothesis in vector language.)

\037 \037 \037 \037

OM. AB = 0 (by Theorem 5, for OM ..L AB). (31)

Rewriting (31), using the fact that M is the midpoint of AB,we get.)

(}A + j-B) \302\267(B

- A) =0,)

which, expanded according to the corollary of Theorem 7,becomes)

i B \302\267B - iA \302\267A = o.

Hence A. A = B \302\267B, which expresses the fact that the

length of A equals the length of B. We leave it to the readerto show in a similar manrier that)

B \302\267B = C \302\267c.)

\037 -+Now what must be shown is that OP is perpendicularto AC,

which expressed vectorially is)

P \302\267(C

- A) = o.)

We therefore expand p. (C -A), using relations from our

hypothesis in the hopes of showing this dot product to be)

A)M)

(a))

c)

B) B)

(b))

FIGURE 44)))

68 ELEMENTARY VECTOR GEOMETRY

zero. Thus)

P \302\267(C

- A) = (\037A + !C)\302\267

(C- A)

:c:\037C

\302\267C - tA \302\267A.

Thus P \302\267(C

- A) :c:0 (for A \302\267A = C \302\267C),

from which we conclude the desiredresult.)

This example illustrates once again the fact that ajudiciousselectionof the point from which the severalvectors'are consideredto emanate often leads to a simplesolution to an otherwisedifficult problem.)

12. COMPONENTS

In the study of mechanics it is frequently useful-and often necessary-to considera singlevector as the

sum or resultant of two other vectors. For example, in

pushing a lawn mower (see Figure 45a), the force F

is exerted along the bar of the mower, but the questionsasked in physics are: What is the force Fh that contributes)

Fh)

(a)) (b))

FIGURE 45)))

INNER PRODUCTS) 69)

to the horizontal motion of the mower? And, what

force Fv is \"wasted\" by being applied vertically down-ward? Thesequestionsare answered by considering F

as the sum of Fn and Fv (Figure 45b). Fh is called thehorizontalcomponent of F, and Fv is called the verticalcomponent of F. 1'he problem of determining thecomponents is a problem in dot products as is easilyseen by noting that the magnitude of a component isthe magnitude of the projection of F. For example,IFhl

=IFllcos 81, and this relation can easilybe trans-

formed into a dot product by making use of a un\037t vector

U along the horizontal. ThenIFni

=IF

\302\267ul, and we

may further write)

F n= (F. U) U) (32))

The unit vector U serves as a \"gimmick\" in two capaci-ties. First, to assistus in writing the magnitude of a

component in the language of dot product; and second,it enables us to write Fh explicitly as a vector becauseU itself, in (32),imparts a direction to the right memberwhile it doesnot distortthemagnitude.

2

EXAMPLE 14. In plotting \\tvind forces on plane graph paper(Figure 46) it is found that the ,vind force, at present, is vectorF = 3i+ 4j (i and j are basis vectors in the xy-plane). We

shall express, in terms of scalar products: (1) the magnitude

of F; (2) the component of F in the direction of the x-axis;(3)the component of F along the y-axis; and (4)the componentof F along vect or A = -i + 5j.

(1) IFI= yF. F =- Y (3i + 4j) \302\267

(3i + 4j)

(2) Fx = (F. i)i = [(3i+ 4j)\302\267

i]i)

2 Some authors prefer to define components as scalars ratherthan as vectors. This is purely a matter of taste,. which is ofteI;l

colored by pragmatic considerations. Readers of mathematicailiterature would do well to heed the advice given to Alice byHumpty Dumpty: \"When I use a word, it means just what I

choose it to mean-neither more nor less.\)


(3) Fy= (F. j)j = [(3i+ 4j)

\302\267jJj

(4) The unit vector along A is

11 1

, Thus the component

of F that we seek is)

(

A

)

A A AF.. = F \"

'AI IAI

= (F\" A)IAI2

= (F\" A)A\" A

=\302\2533i+ 4j) \" (-i + 5j))

(-i + \037\037:( \037i + 5j)\

13. INNER PRODUCT FORMULAS

The form of the answers in Example 14 is qllite cumber-someand of such a nature that it is difficult for anyone to

gain any insight from the answers. We therefore turn)

y)

x)

-3)

-4)

-5)

FIGURE 46)))

INNER PRODUCTS) 71)

attention to simplifying those expressiollSthat involve

the inner product of vectors ill terms of their rectangular

components (in the directions of the rectangular axes).

We shall take advantage of the fact that vectors in spaceare generally no moredifficult to handle than vectors in

a pla\037e. Thus our computations will deal directly withvectors in three dimensions, admitting the possibility of

i, j, and k components.First, we note the fact that the length of the basis

vectors is one impliesV i. i =.Vj\302\267

j= Vk. k = 1.

Hence)i \302\267i = j \302\267

j= k \302\267k = 1.

Secondly, we observe that

i \302\267j

= j \302\267k = k \302\267i =0,)

(33))

(34))

which follows from the mutual perpendicularityof i, j,

and k. Now, let)

A = ali + a2j + a3k) and) B = b1i + b.Gj + b 3k.)

Then, makillg use of Theorem 7 (and its corollary),

(33) and (34), we compute as follows:

A \302\267B = (ali + a2j + a3k) \302\267(b1i + b2j + b3 k)

= a1i \302\267b1i + a1i \302\267b 2j + a1i \302\267

b3 k

+ a2j \302\267b1i + a2j \302\267b 2j + a2j \302\267

b3 k

+ a3k \302\267b1i + a3k \302\267b 2j + a3k \302\267b 3k

= a1i \302\267b1i + azj \302\267b 2j + a3k \302\267b 3k.

Finally, we have a -Simple formula for the inner productof two vectors:)

A \302\267B = a1b 1 + a 2 b 2 + a3b3-) (35))

The length of A ca n ll0 \\V be found by the formula

IAI= V A \302\267A = Va12 + a2

2 + a32. (36))))


From the definition of A \302\267B, (35) and (36), we solve for

cosine of the anglebetween A and B, getting

a1b1 + a2 b 2 + a 3b3cos 8 = \302\267

(37)Y a 1

2 + a22+ a3

2 Y b12 + b2

2+ b3

2

(Note that (35) and (36) are precisely the formulas

used in our heuristic reasoningon pages60-62,where we

were informally exploring a method for introducing quan-titativeaspectsintoour vector algebra.)

EXAMPLE 15. We return to Example14 to make explicit

computations of the answers given there in unsimplified form.)

(1) IF\\= y 3 \302\267

3\"+ 4 .4 = 5

(2) Fz = [(3.

1) + (4 \302\267O)]i

= 3i.

(3) F y= [(3

#0) + (4 . 1)]j = 4j.

-i + 5j(4) Fa = (3(-1) + 4, 5) (-1)(-1) + 5.5

=

\037\037( -i + 5j).)

We now ask the question:What is the angle a between Fand A of Example 14?

Using (37), we V\\Trite

F.Acos ex = \037

yF \302\267F Y A \302\267A)

-3 + 20

5y 26)

(3i + 4j) \302\267(-i + 5j)

y3 .3 + 4 .4 Y (-1)( -1) + 5 \302\2675

17

5 y 26 ')

or)

17 V26cos ex =

130.

If we were interested in a precisevalue for a in terms of degreesor radians,it would now be a simple matter to find ex by con-

sulting a table for the values of the cosine function.)))

INNER PRODUCTS) 73)

p\\)

v

prSF)

J) s)Q)

FIGURE 47)

14. WORK

The work done in applying a force of magnitude f

through a distance 8 is definedby physicists to be the

product fs. Thus work is done if and only if motion

occurs. Furthermore, it s\037ould be emphasized that

only that force which produces the motion is used to com-

pute the work done. If we have a force vector F appliedto an object (see Figure 47) with the effect of moving the

object along a straight line from P to Q,the force used to

compute the work done is that of the component of F\037

(along PQ). That is, calling S = PQ,)

W =(prsF)ISI.) (38))

(We write (38) in terms of the projection of F to allowthe possibility of work being negative. Although

negative work may sound strangeto the uninitiated,

the concept is one of great practical vallIeto the physicist

and engineer, who need it for an adequatemathematicalformulation of the fundamental laws of mechanics andelectricity.)

But formula (38) can be rewritten

W =IFI cos 8\\sl;)

or, finally,)

W=F.S ,)))


which expresses the idea that dot product can be viewed

as work done by one vector in the direction of the other;or perhaps the readermight prefer to view work as aphysical interpretation of the dot product.)

EXERCISES1. Using the approachof Example 13, prove that the mediansof a triangle meet in a point. That is, let A11f and BN be

two medians of triangle ABC, and call P their point of inter-section;then show that the extension of CP actually bisectssideAB.

2. Let ABC be a right triangle with hypotenuse BC, and Don BC be the foot of the perpendicular from A. Consider

one force of magnitude -.: acting in the directionof .iE,

IABI

and another of magnitude \037 acting in the direction AG.

IAcl

Prove that the resultant is a force of magnitude \037 acting

IADI\037

in the direction of AD. (Hint. Usecomponents.)3. Find a vector perpendicular to V = 2i - 3j and whoselengthis four times the length of V.

4. The coordinatesof two points are (3, 1, 2) and (2, -2, 4).Find the cosine of the angle between the vectorsjoiningtheoriginto thesepoints.5. Prove that the sum of the squares of the diagonals of a

parallelogram is equalto the sum of the squares of the sides.

6. Prove that the sum of the squares of the sides of any (not

necessarily plane) quadrilateral exceedsthe sum of the squares

of the diagonals by four times the square of the line segmentthat joins the midpoints of the diagonals.

.

7. By means of dot products prove that an angle inscribedin asemicircleisa right angle. (If AB is a diameter, 0 the center,)))

INNER PRODUCTS) 75)

--+ --+and P any point on the circle,then OB = - OA. Evaluate

--+ --+the dot product AP \302\267

BP.)

8. Prove that the altitudes of a triangle meet in a point.(Hint. Use the approach of Example 13, and rely heavily onTheorem 5.)9. Prove that if two circles intersect, the line joining theircenters is perpendicularto the line joining their points of

in tersection.)

10. Let) A = 2i - 3j + 4k

B = - 2i + j - k.)

(a) Find A-B.

(b) Find prB A and prA \"B.

(e) Find the component of A along B.

(d) Find the work done by force vector A in moving a par-ticle from the origin to (2,0, 0) along the x-axis.

(e) Find the work done by A in moving a particle from theorigin to (1, 2, -1).

11. Let F be the sum of n forces F 1, F 2 , . . . , Fn, all actingat the point O. Then

W=F-S=F l .S+F 2 .S+... +Fn-S. (Why?))

In addition, if F results in n consecutive displacements repre-sentedby 8 1, 8 2 , . . . , Sn, then)

F \302\2678 = F \302\2678 1 + F \302\2678 2 + \302\267\302\267\302\267+ F \302\267

8n. (Why?))))

analytic

geometry)

15. OUR POINT OF VIEW

The reader should note that the definitions of addition,

subtraction, and inner product of vectors werenotmadeintermsof-coordinates.At the outset we treated ve.ctors ina coordinate-freefashion. (When such is the case, it isoften stated that the conceptsare \"independent of a

coordinate system.\") And yet, manyapplications-par-ticularly to geometry-werepossible.As we proceeded

in our development, new techniques wereemployed;it isin this light that one should view coordinate systems.That is, coordinatesystems should be looked UpOll as'another instrument rather than as allother branch of

mathematical study. The philosophy that the authorissuggesting takes the following form.

The branch of mathematics that we call geometry

deals with, among other things, the propertiesof lines,planes,circles,and spheres. The solutions to problemsin this field may be reachedby various approaches,

76)))

ANALYTIC GEOMETRY) 77)

some being more natural or stronger than others-depending on the problem at hand. The first approachlearned is usually the synthetic method, which hassome approximation to that exhibited in Euclid's Ele-

ments. The present work is devotedto a study of the

vector and analytic approach. Hence givena geometric

problem, one could give it a vector interpretation, or onecouldimpose a coordinate system on the problem and useanalytic techniques. Of course, any combination of thethree methods is alsopossible.

In summary, coordinate systems should not be viewed asintrinsic to geometry but., rather, as another (very powerful)mathematicaltool with which one attacks geometric problems.

Combining our notionsof vectors-particularlythoseof the scalar product-with those of coordinate sys-tems, we shall develop some of the elements of analyticgeometry.)

16.THE STRAIGHT LINE

We begin with the problem of finding the equatioll of a

straight line in the plane. More precisely, if P = (x, y)is an arbitrary pointona givel1 line \302\243,we seek a mathe-

matical relation that distinguishesP =(x, y) from those

points in the plane that are not 011\302\243.

There are, of course, many ways to specify a unique

line. We begin by specifying two points P1(XI,Yl) and

P 2 (X2, Y2),1 and seek the equation of the line determined

by these points. The aim is to arrive at an equation in

terms of the coordinatesof the specified points.Since P, PI, and P 2 are all one line (see Figure 48),

we may employ Theorem 4 to write)\037 --+--+OP = (1 - t)OPl + tOP2 -)

1 This appears to be the most natural way to specify a uniqueline. It was, in fact, included as Euclid's first axiom, which,

freely translated, states that one and only one line can be drawn

joining two points.)))


y)

(a))

y)

x) %)

(b))

FIGURE 48)

Or, in accordance with our convention of designating

vectors with a common origin by their endpointsalone,we rewrite this equation in terms of the positionvectorsof the points)

P = (1 - t)P1+ tP 2 .) (39))

Equation 39 is often referred to as a vector equation of

line PIP 2. As t takes on different real number values,we get different vectorsfor P. For example, if t =

j-,)

P = (1 - j-)P1+ j-P2=

\037P l + j-P 2)

is the position vector of the midpoint of segment Pt P 2.The auxiliary variable t is called a parameter and (39)

is also referredto as a parametric (vector) representation

of line \302\243.

Rewriting (39) in terms of the basis vectors,we have)

xi + yj = (1 - t)x1i+ (1-t)Ylj + tX2i + tY2j

= [Xl + t(X2 - xI)]i+ [YI + t(Y2- YI)]j.)

However, the representationofa vector in terms of basis)))


vectors is unique. Therefore

{

X = Xl + t(X2- Xl)

Y= YI + t(Y2

- Yl)

are the parametric equations of line \302\243in analytic or

coordinate (in contrast to vector)form.EXAMPLE 16. Find parametric representations of (a) thex-axisand (b) the line joining (3, 1) and (-2, 3).

(a) The reader knows, of course, that the equation of the

x-axis is y= 0, but we shall now attempt to verify that this

is the case by using (40). Taking two points, say (0, 0) and(0, 1), on the x-axis,we write)

(40))

{

X = 0 + t(1 - 0)y

= 0 + t(O - 0).)Thus the parametric equations of the x-axis are)

{

X = t

Y= O.)

No matter what the value of t, the y-coordinate is always zero.Therefore y

= 0 is a sufficient description of the line.

(b) Calling PI = (3, 1) and P 2 = (-2, 3), and applying(40), yields)

{

X = 3 + t( - 2 -3)

y= 1 + t(3 - 1).)

In simplified form:)

x = 3 - 5t

Y= 1 + 2t.

If, instead,we had begun with PI = (-2,3)andP2 = (3,1),(40) would yield:)

{

X = - 2 + 5t

Y= 3 -

2t,)

a different parametric representation of the same line. Thus

we see that the parametric repre8entationof a line i8 not unique.)))


It depends on the choiceof points used to derive the parametricequations.

In many instances it is desirable to eliminate theparameter and to write a line in some form without anauxiliaryvariable. Inorderto accomplish such elimina-

tion we rewrite (40) as

x -Xl

= t(X2- Xl)

Y-

YI = t(Y2 -YI).)

and divide the second equation by the first, getting)

Y-

YI Y2-

YI- ,X

- Xl X2 -Xl)

(41))

which is the equation of the line,2in terms of the coordi-

nates of PI and P 2 alone. For this reason (41) is calledthe two-point form for the equation of a line. This formstatesthattheratioof the difference in the y-coordinatesto the differenceof the x-coordinates (taken in the sameorder) of any two points on the line is the same,namelyY2

-YI

(see Fl.gure 49). N t th t th I t to e a isisequiva en 0

X2 - Xlthe stateme11tthat all the triangles in Figure 49 are

similar. It should alsobenotedthatthisratio,Y2-

YI,

X2- Xl

is precisely the ratio of the coefficientsof t in the para-metric form of the line. Thus at least one feature of the

parametric form is independent of the points used to)

2Strictly speaking, (41) is not the equation of the line, for it is

not satisfied by (Xl, Yl). The equation satisfied by all points of

the line is Y- Yl = Y2

- Yl(x - Xl), while (41) describes this

X2- Xl

line with the point (Xl, Yl) deleted. We chooseto use (41),inspite of the slight inaccuracy, because of its easily remembered

symmetric form.)))


y)

Y2-

Yl)

X-Xl \037)

X2 -Xl) :>1)

o)X)

FIGURE 49)

derive the equations. (See Example 16b where 2

5 is

the ratio of the coefficients of t.)

S.h

\302\267Y2-

Yl \302\267 \302\267..h I b

.Incet e ratIo IS SO IntrInSIC to tea ge ralCX2

- Xl

representation of the line, we discuss it further in the

light of it being the tangent of the angle labeled a in

Figure 49. We proceed by first giving angle a a name,the angle of inclination of line \302\243. The angle of inclination

of a line is the angle formed by the given line and thepositivex-axis. Thatis,a ismeasured counterclockwise

from the positive side of the x-axisto the portionof line

\302\243that lies above the x-axis. If \302\243is parallel to the

x-axis, we say that the angle of inclination of \302\243is zero.

The tangent of the angle of inclination is termed the

slope of the line. Thus)

Y2-

YI

slope of \302\243= tan a = \302\267(42)

X2-

Xl)))


Consequently, (42) states the formula for the slopeofa line in terms of two given points; and (41) says, ineffect,that any two points used in the computationyield the very same slope.

A difficulty occurs when X2-

Xl = 0; then the divisionthat produced(41)is not legitimate, for dividing byzero is undefined in arithmetic. Analyzing this case

separately (see Figure 50), we see that the linemust be

vertical in order that Xl= X2. The angle of il1clination

is 1r /2. The reader is undoubtedly familiar with thefact that the tangent of 1r/2 is undefined; i.e. there is noreal number that equals tan 1r/2. Thus vertical lineshave no slopes; the concept is merely undefined for suchlines. However, if a \037 1r/2 and 0 < a < 1r, thentan a is defined, hence vertical lines are the only ones that

have no slope.)

Y)

(Xl, Y2))

(Xl, Y1))

x)

X=X1)

FIGURE 60)))


The equation of the vertical line under discussion is

x =Xl which, if confusing to the reader, should be

thought of as stating: for every value of Y, X is always

equal to Xl. V erticallines all have the form)

X = constal1t;

and the reader can easily show that horizontal lines allhave the form)

y= constant.

Continuing our analysis of the equation of the straightline, we solveequatio1141for y:)

Y2-

Yl

+Y2

-Yl

+Y = X Xl YI.X2 - Xl X2

-Xl)

Simplifying this expression by making the replacements)

Y2-

Ylm=

X2-

Xl)

and)Y2

-Yl

b = Xl + YI,X2

-Xl)

we get)

y=mx+b) (43))

as the equation of the line. Note that m, the coefficientof x, is the slope of the line. Is therea geometric inter-

pretation of the constant b? The answeris easily deter-

mined by observing the Y= b when x = o. Thus the

point(0,b) is on the line. This is the point at which the

line (43) crosses or intercepts the y-axis,and it 'isthere-fore called the y-intercept. Equation 43 is termed theslope-interceptform for the equation of a line, for, fromthis equationwe can immediately read off the slope mand the y-intercept,(0,b) of the line.

EXAMPLE 17. We return to the line (of the last example)determined by PI = (3,1) and P 2 = (-2, 3) and ask several

pertinent questions regarding it.

(1) What is its equation in two-point form?)))

84

By (41) we have)

ELEMENTARY VECTOR GEOMETRY)

y - 1x-3)

3 - 1-2 - 3)

or)

y- 1

x-3)2- ---5) (

Observe that its slope is52

.))(2) What is the equation of the line in slope-interceptfornl?

Solving for y, we get y= -

\037x + 151-, so that m = -

\037 and

b = 115 \302\267

(3) We graph the line (Figure 51). A simple procedure that

applies the slope-intercept form is first to locate the y-intercept(0, 15

1) as one point of the graph. Then,making use of the

slope idea, we proceedfive units to the right and two unitsdownward to locatea secondpoint of the graph.

(4) What is the x-intercept?In the process of answering this rather simple questionwe

shall digress a bit and reflect on the nature of \"the equation ofthe line.\" Logically, the equation y

= - ; x + 151- is a sen-

tence, which is true for only certain choices of x and y. Thatis, some orderedpairs (x, y) render the sentence true; thesepairs are the coordinatesof points that are said to lie on theline y

= -\037x + 15

1-. All the ordered pairs that render thestatement false represent points not on the line y

= -\037x + 15

1-.We therefore check to see \\vhether a given point, say (3, 1),)

y)

x)

FIGURE 61)))


is on the line by substituting the coordinates in the sentenceand determining whether the sentenceis renderedtrue. Here

we have 1 = -\037(3) + 15\037' which is true. Hence (3, 1) is

actually on the line.In the language of set theory it would be said that a line is

a set of points that, expressed analytically, is a set of ordered

pairs. Thus a line \302\243,whose slope is m and whose y-interceptis (0,b), is defined by the statement:

\302\243= {(x, y)ly = mx + b}.)

The line in Example17is then the following set:

{(x, y)ly = -\037x + 151_ }.

Continuing the line of reasoning, we considerthe problem

of finding the point of intersection of two lines, \302\2431and \302\2432,

where)

\302\2431:y= m1 x + b1

\302\2432:y= m2x + b 2 .)

(44))

The point of intersection (there is, at most,oneunless\302\2431=\302\2432)

is the ordered pair rendering both statements \302\2431and \302\2432true,

simultaneously.3 That is, we seek the orderedpair that sat-isfies both equations simultaneously. Finding the pair, if it

exists, is then a matter of elementary algebra-solving for thesolution to a set of two simultaneous linear equations.

Thus finding the x-intercept of our line is the problem of

solving for the solution of the pair of equations:)

y = -\037x + J5\037

y= 0 (x axis).

The x-interceptis thereforeC\302\245-,0).)

3 The use of the symbols\302\2431and \302\2432to mean both lines and sen-tencesin one paragraph might be frowned upon by some logicians,but this economy of sYlnbols should not cause the reader any

confusion. In the rigorous axiomatic treatment. of analytic

geometry a straight line may be defined as a set whose defining

sentence is of the form ax + by + c = 0 (see equation (49)).

This definition includes vertical lines as well as those that haveslope defined.)))

8'6.) ELEMENTARY VECTOR GEOMETRY)

A Euclidean theorem states: If two lines are cut by atransversal so that a pair of corresponding angles are

equal, the two lines are parallel;and conversely, if two

parallel lines are cut by a transversal, corresponding

angles are equal. Applying this to our treatment of

analytic geometry, we may say that two lines are parallel

if and only if they possess the same angle of inclination.For nonverticallines, it may be stated that two lines are

parallel if and only if they have the same slope.

EXAMPLE18. To find the equation of the line parallel toy

= -\037x + 1;- and passing through (10, -2).

Since the line we seek must have slope m = -\037, we may

\\vrite it in the form)

y= - ; x + b,) (45))

,vith b remaining to be determined. In accordancewith the

foregoing discussion, the substitution of (10, -2) in (45)

should render it true. Thus -2 = -\037(10) + b. Conse-

quently, b = 2, and the equation is completely determined:y

= -\037x + 2. (For another approach, see (equation46) in

Exercise8 below.))

EXERCISES

1. Derive the equation of the line determinedby (5, 4) and

(3, 1) in:(a) the vector form as expressed in (39);(b) Parametric form;

(c) two-point form.

2. (a) What is the slope of the line determinedin ExerciseI?(b) What are its x- and y-intercepts?

3. Derivethe equation of the line:

(a) through (-5, 4) and having y-intercept -5; that is,through (0, -5).

(b) whose slope is 2 and x-intercept is.-3.(c) parallel to 2x -

y= 3 and through (1, 1).

4. What is the point of intersection of the line whoseequationis 2x -

3y= 12 and the line whoseequationis 2x -

5y= 121)))


5. Sketch (a) 2x -3y

= 12.

(b) 2x -5y

= 12.

(c) 3y + 7 = o.(d) y

- 2x = 1.(e) y

= 4.

(f) x = ?r.

(g){

X = 2 - 2t

Y= -1 + t

(h){: \037_ t.

6. (a) Find the point of intersection of the medians of thetriangle determinedby A = (-2, 1), B = (5, -2), and C =

(3, - 2). (Don'tuse the equations of the medians!)(b) What is the equation of the median emanating from B?

(c) Find the equation of the interior angle bisectorat A.(Hint.A vector emanating from A, which bisects angle A,

\037 \037

is V =AB

+AC .

\037 \037

IABI IACI

7. (a) Find the equation of the line with an inclination ofr/6 and passingthrough (1, -2).

(b) What is the equation of the line through (1, -2) whoseslopeis undefined?

(c) What is the equation of the line through (1, -2) whose

slope is zero?

8. (a) The equation for the line through (Xl, Yl), with slope

m, is given by)

Y-

Yl- = m. (point-slope form)x -Xl)

(46))

Justify (46). (See footnote 2, page80.)(b) The equation for the line with x-intercept of (0, a) and

y-intercept of (0, b) is given by)

X+

Y = 1. (intercept-form)a b)

(47))

Justify (47).)))


9. Booby traps:(a) Sketch x + y

= x + y.

(b) Sketch{

X = 1

y = t.

(c) Find an equation for a line whose inclination is 1r/2and.whose x-intercept is o.)

17. ANALYTIC GEOMETRY OF THE LINE CONTINUEDReturning to the useofvectors,we pose the question:

What is the equation of the line \302\243through the fixed

point Po = (xo,Yo) and perpendicular to the vectorN = ai + bj?

If P =(x, y) \"is the general point of the line \302\243(see

\037 ---?

Figure 52), we know that PoP 1..N or pop. N = o.Therefore)

or)

[(x- xo)i + (y -

yo)j]\302\267

(ai + bj) = 0

a(x -xo) + bey

-Yo)

= 0, (48)

which is the equationof line\302\243. We can rewrite (48) as)

ax + by- (axo + byo) = 0)

y)

%)

FIGURE 52)))


or, more simply, as)

ax + by + c = 0,) where c = -(axo + byo), (49))

which is the general equation for a straight line in the (x, y)-

plane. Equation 49 can be solved for y (providedb \037 0), in which case we will have the line in slope-

intercept form. In the event that b = 0, (49) yields thevertical linex = -cia (a and b cannot both be zero, forthen (49)would become the trivial sentence 0 = 0, whichis true for all points in the plane). Thus (49) includesallpossiblestraight lines, vertical and nonvertical.

If the concept of slope is carried over to vectors, we

find that the slopeof N'isbla (if a \037 0) and the slope of

(49), which can be found from its slope-intercept form

y = -(alb)x - cia, is -alb. We therefore conclude

that:

two lines (neither of them vertical) are perpendicular ifand only if their slopesare negative reciprocals of one

another.

That is, if m1 is the slope of line \302\2431,and m2 the slope ofline \302\2432,)

\302\2431.1 \302\2432) if and only if)

1)

ml=) --.)mt) 4J)

(It should be observed that there are two parts to this

result but only one has beenjustifiedhere. The reader

should therefore complete the justification.)If \302\2431.1 \302\2432and \302\2431is vertical (no slope), then \302\2432must

be horizo11t.al, i.e., m'], = O.

EXAMPLE19. Find the equation of the line through (2, -1)and perpendicular to the vector 3i - 4j.

Following the scheme of reasoning that climaxed with (48),

\\ve write)

or)[(x - 2)i+ (y

- (-l))j]. (3i -4j)

= 0;

3(x - 2) -4(11 + 1) =

0,)))


which, simplified and put in the form of (49), reads:

3x -4y

- 10 = O.

EXAMPLE20. Find the equation of the line through (2, -1)and perpendicular to 3x -

4y- 10 = o.

Writing the given line in slope-intercept form)

Y_ 3,.. _ 5- 4N 2')

we see that its slope is I. Therefore the slope of the desiredline is -1 /: =

-t. (Check this with slope of N in Example19.) Therefore, by using the point-slopeform (46), the line

we seek can be written)

y- (-1). 4= --,x-2 3

EXERCISES

1. If a line \302\243is given by ax + by + c = 0, find an expression

for a vector N perpendicular to \302\243,in terms of the coefficientsin the equationof \302\243.

2. Find the equation of the line perpendicularto 2x -y

- 1 =

o and through the origin.

3. Find the equation of the perpendicular bisector of the seg-ment joining (1, -3) and (3, 5).

4. Find two vectors of unit length perpendicularto 2x -y

- 1= O.)

or) 4x - 3y - 5 = O.)

18. DISTANCE FROM A POINT TO A LINE

The problem of determining the (minimum) distancebetweena pointPo=

(xo, Yo) and a line)

\302\243:ax + by + c = 0

is easilyhandled by vectors. We shall actually derivea formula for this distance,but it is the author's sug-

gestion that the reader not memorizethis formula. One

of the stro'ng advantages of thinking in terms of vectors-

as opposed to pure analytics-is that fewer formulas

need to be remembered. A thorough familiarity with the)))


Po)

\302\243)

FIGURE 53)

basic vector tools often enables one to solve problems by

following a basic pattern of reasoning from first prin-

ciples-just as easily as applying a complicated formula.

The present problem is a caseillpoint.Let P =

(x, y) be a general point on \302\243. Then the

minimum distance d from Po to \302\243could be thought of\037

as the magnitude of the projection of PPo 011the per-\037

pendicular PF to \302\243(see Figure 53). Thlls the dot prod-uct will serve as an aid to finding d. Usingthe result of

the last section, we write a vector perpendicularto \302\243ag

N = ai + bj. Applying Theorem6, we have)

\037\037

t.PPo. NId = IprNPPol =

INI

[ ( ).

( )e

J

a.i + bj- x -Xo 1 + y

-Yo J

e

_ /V a 2

+ b2

_ taxo + byo- ax -

byl .-vi a 2

+ b2)))


But, - ax -by

= c, so we finally have

d = laxo+ byo + c\\,

Va2

+ b2

which is a formulafor the distance d in terms of thecoordinatesof the given point Po and the constants inthe given line \302\243.

EXAMPLE 21. Find the distance from Po = (1, 3) to the

line \302\243:y= jx

- 1.Although formula (50) for the distance has been derived in

this very section, we shall illustratethe point of the suggestionmade above by abandoning the use of (50) in favor of workingfrom first principles.

We begin by writing \302\243in the form x -2y

- 2 = 0, sothat a vector perpendicular to \302\243can immediately be writtenas i - 2j. In the computation used in arriving at d, thereader doubtlessly observed that we were forced to divide byINI, which meant that we were actually computing the pro-

jectionby projecting on the unit vectorI:'

\" It might, there-

fore, be more convenient if we convert the perpendicular vector)

(50))

y)

J\\(l,3))

1) \302\243)

%)

-2 -1 0)

FIGURE 64)))


to a unit vector at the outset of our present computation.Sucha unit vector is)

U =i - 2j.o

The distance d is then the magnitude of the projection of-+

PoP I on U, where PI is any point on \302\243. A rather simplechoice for PI is the y-intercept of \302\243,namely (0, -1). Hence,)

-+ \037 \037

d = IP\037I. UI =I (pru PoPI)IUII

= Ipru PoP II

_ [(0- l)i + (-1 _ 3)j].i

-2j

o

_ (-i _4j)

.i - 2j _ 1-1 + 81

o 0

= 7 V5 .5)

Any other choice of PI, say PI = (2,0), should give the same

result:)

d=) [(2 - l)i + (0_ 3)j].i-

2j

o

705)

1+6

o)

Even a generalpoint PI = (Xl, YI) of \302\243would yield the explicitcomputation:)

d =[(Xl

- l)i + (Yl- 3)j] . i

\037\037j I

IXI-

2YI + 51

V 5)

But, since (Xl, YI) is on \302\243,we know that Xl-

2YI\"= 2. There-

fore)

d = !2+ 51 = \037 =7 0 .

o 0 5)))


. - i + 2j .Suppose we had begun With U =

V5as the umt

vector perpendicular to \302\243. Then the sign of the pro-jection would be changed; but since we are taking themagnitude,orabsolutevalue of the projection, the valuefor d wouldbethe same as in our previous computations.

As soon as the method of projection is grasped, the

computations can be madesimply and directly, without

any reference to a formula. However,a thorough under-

standing of. vector thinking comes only with the practiceof such thinking. And it is precisely to promote suchpracticethat exercises are included. Be sure to do agoodshareof them!EXERCISES

1. Find the distance from (1, 2) to x -2y

= 5.

2. Find the distance from (1, -2) to x -2y

= 5.

3. Find the distance from (1, -2) to -x + 2y + 5 = O.

4. Find the magnitude of the projection of the segment ABon the line x -

2y= 5, where A = (1, 1) and B = (2, -1).

5. Find the magnitude of the projection of segment AB on thevector 3i -

4j where A and B are the sameas in 4.)

19. ANALYTIC METHOD OF PROOFIn accordance with the philosophy described in Section

15, we shall now illustrate the applicationof analytic

methods to proving results of geometry. The spirit of

these illustrations will be to divorce our work from theideasof synthetic geometry (e.g., congruence, similarity)in favor of working with coordinates, slope, and othernotions that are analytic in character.

EXAMPLE 22. Prove that the line joining the midpointsof

two sides of a triangle is parallelto the third side and equal toone half of it.

Again, in accordance with the philosophy put forth in Sec-tion 15, we view this as a geometry problem and impose a)))


y)

B(b, C))

0(0,0))%)

A (a, 0))

FIGURE 66)

coordinate system upon it. vVe therefore impose the coordi-nate system in such a manner as to facilitate our work. Forexample,we may choose the origin as one vertex of the tri-

angle and the x-axis to be alongone side of the triangle (see

Figure 55). If we do this, and if we desire our proof to holdfor all triangles, there is no further choice available. Calling

the triangle OAB, we assign coordinates as follows:)

o = (0, 0), A = (a, 0), B = (b, c).)

Next, we compute the coordinates of M, the midpoint of AB,

and N, the midpoint of OB,by means of the formula derivedin Example lOde)

M =

(

a\037

b

,\037))

and) N =(

\037C

)

.2' 2)

c/2 - c/2 .The slope of MN =

bj2 _ (a + b)/2

= 0, whIch proves that)

M N is parallel to 0 A.)))


As for the secondpart of the result, we compute the lengthof segment MN.)

IMNI =\037 (;

-\037r

+(\037

- a

;b

Y

=\037 \0372

=1;1

,

which shows that 11IINI=\037IAOI.

EXAMPLE 23. Prove that the diagonals of a parallelogrambisect each other.

Again we elect to place the coordinate axesin a convenient

position. Let the parallelogram be OABC, where 0 = (0,0),A = (a, 0), and C = (b,c),asnotedin Figure56. Ifwe ascribe

coordinates, say (d, e), to point B, we find that d and e aredependenton the coordinateschosenfor the other three ver-tices of the paralleiogram. That is, the condition that the

figure is a parallelogram, namely, that its oppositesides areparallel,forcescertaincoordinates upon B.

Since CB is parallel to OA, the slope of CB is zero; that is,(e -

c)/(d- b) = 0, which implies that e = c. And, since

OC is parallel to AB, (c -O)/(b

- 0) = (c -O)/(d

- a); ord = a + b. Thus, B = (a + b, c).

Now that we have imposed the parallelogram condition, weare free to attack our problem-findingthe mid-points of the

diagonals. The midpoint of OB =\302\253a+ b)/2, c/2) and the

midpoint of AC =\302\253a+ b)/2, c/2), which establishes the result.)

y)

C(b,c) B(d,e) = (a+b,c))

x)

0(0,0)) A (a, 0))

FIGURE 66)))


y)

B(b, C))

0(0,0))M(\037,O))

A (a, 0))

%)

FIGURE 67)

EXAMPLE 24. Prove analytically that the mediansof a tri-

angle meet in a point.Let the triangle be OAB, as sho,vn in Figure 57, where

o =(0, 0), A = (a, 0) and B =

(b, c). CaHill the midpointof OA, N the midpoint of AB, and Q the midpoint of OB.

(

a

) (a + b C

) (

b C

)

Then M =2'

0, N =2 ' 2 and Q =

2' 2.

The equationsof the median lines are found to be:)

cON: y

=

a+bx

BM: y =b _

\037a/2) (x

-;))

or)2cx ac

y = -2b-a 2b-a)

AQc/2

: y =b/2 _ a (x -

a)) or)c

11=

b _ 2a(x - a).)

For the point of intersection of ON and B]1,l,,ve solve the first

t\\VO equations as a simultaneous pair, getting)

a+bx=

3)and)

cy

= -.3)))


For the point of intersection of ON and AQ, we solve thefirst and third equations simultaneously, arriving at the samepoint \302\253a+ b)/3, c/3). Thus, the mediansof triangle OAB

all meet at \302\253a+ b)/3, cj3).

For the purposes of further illustration, we use thevector approach to checkthe coordinatesof the point P

of intersection of the medians. Referring to Example

lOe, the position vector of P is written

P =}-O + }-A + lB

= }(Oi+ OJ) + -i(ai) + }(bi + cj)

= j(a + b)i+ ; j

Thus P = (a + b)j3, cj3),which checks with the above

com pu ta tion.It seems clear that the median problem is more easily

handledby use of vectors than by pure analytics.

EXERCISES1. Prove:In any triangle the sum of the squares of the medians

is equal to three-fourths the sum of the squares of the threesides.

2. Prove:The sum of the squares of two sides of a triangle isequal to one-half the square of the third side, increased bytwice the square of the median on that side.3. Prove:The sum of the squares of the four sides of a paral-lelogram is equal to the sum of the squares of the diagonals.(Comparewith vector proof.)

4. Prove: The sum of the squares of the four sides of anyquadrilateral is equalto the sum of the squares of the diagonalsincreasedby four times the square of the line joining the mid-points of the diagonals.

5. Prove analytically:(a) the medianof a trapezoid.is parallel to the bases;(b) the lines joiningthe midpoints of the sides of a quad-

rilateral form a parallelogram;)))

ANALYTIC GEO.METRY) 99)

(c) in any quadrilateral, the segments joining the midpointsof opposite sides intersect in a point that is the midpoint

of the segment joining the midpoints of the diagonals.

6. If line \302\2431has a slope of m1 and an angle of inclination 8 1,

and if line \302\2432has a slope of m2 and angle of inclination 8 2,

use the formula for tan (82- (J1) to determine tan (J, where (J

is the' angle between \302\2431and \302\2432(measuring counterclockwise

from \302\2431to \302\2432). See Figure 58.

7. Discuss the difficulty of using the result of Exercise 6 when

the lines \302\2431and \302\2432are perpendicular.

8. Find the angles of the triangles

(a) (5, 0), (8, 4), (-4, 13).(b) (2, 6), (6, 0), (-3; 8).

9. Theslopesof two lines are given as -2 and 3. Find theslopesof the lines that bisect the angles betweenthem.

10.If0 =(0, 0), A = (V3, 1), and B =

(2 V3, 1), find theequation of the line bisecting angle AOB.)

y)

\302\2432)

\302\2431)

%)

FIGURE 68)))


11. If \302\2431is given by a1X + b 1y + C1 = 0 and \302\2432is given by

alX + b1y + Cl = 0, we define a linear combination of \302\2431and

\302\2432as an equation of the form

m(alx + b1Y + C1) + n(a2x + b2Y + C2)= 0 (51)

(a) Why is the l\037cus described by a linear combinationof two lines also a line, if m and n are not both zero?

(b) If P is the point of intersection of \302\2431and \302\2432,prove

that any linear combination of \302\2431and \302\2432also passes through

P; and conversely, every line through P is a linear combination

of \302\2431and \302\2432. The consequence of this result is that the set

of all lines through P, often called the pencil of lines with vertex

P, can be written\037y

the single equation 51, where m and n

range over all the real numbers(excludingthe casem = n = 0).(c) If \302\2431is parallel to \302\2432,prove that any linear combina-

tion of them is also parallel to each of \302\2431and \302\2432. In this case

(51) represents the set of all lines parallel to \302\2431(or \302\2432),often

called a pencil of parallel lines.12.Using the results of Exercise 11, find the line through the

point of intersection of x - 2y = 3 and 4x -2y

= 15, and

passing through the origin, without finding the point of inter-section of the given lines. (Be sure to follow (51) carefully!)

13. (a) Write an equation that representsthe pencil of lines

through (1, -2).(b) Write an equation that representsthe pencil of lines

\\vith inclination 7r /6.

14. (a) Find the line through the point of intersection of thelineswhose equations are x -

2y + 4 = 0 and 2x + 3y- 3 =

0, and through the point (0, \037).

(b) Find the line through the point of intersection of the

lines whose equations are 3x+ 2y + 8 = 0 and x + 8y + 7 =

0, and through the point ( ; , 0).15. By using the results of Exercise 11, prove the n1ediansof a

triangle meet in a point.)

20. CIRCLESLet Po =

(xo, Yo) be the center of a circle of radius r

and P = (x, y) a generalpoint of the circle (see Figure)))

ANALYTIC GEOMETRY

Y)

101)

x)

FIGURE 59)

59). Then the vector equation of the circle is easily\037

found by imposing the condition that IPopl = r. Uti-lizing the position vectors of the points, we write the vec-tor equation of the circle as

Ip- Pol = Ir (52)

or V (P - Po) \302\267(P

- Po) = r. To get the analytic

form, we rewrite (52) as

I(x - xo)i+ (y-

yo)jl= r)

or)

Vex-

XO)2 + (y-

YO)2= r.)

Finally, by squaring both members, we get

(x -XO)2 + (y

-YO)2

= r 2)(53))

as the equation of the circle whosecenteris (xo, Yo) and

whose radius is r.)))


Concerning the equ\037tion of the circle, it should benoted that: (1) It isquadratic in x and y; and (2) it hasthe coefficients of x

2 and y2 both equal (to unity),when in the form (5\037).

Expanding (53), we get

x2 - 2xox+ X02 + y2-

2yoY + yo2 = r 2.

By groupingthe constantterms and designating coeffi-

cients of the variables by a single letter, we may write thisequation in the form

x2

+ y2 - Ax + By + C = 0, (54))

which is, of course, the equation of a circle but of such a

nature as to obscure its essentialcharacteristics,namely,

the radius and center of the circle.We pose two questions, which can be answered simul-

taneously. (1)Doesevery equation of the form (54)represent a circle? (2) If it does represent a circle, howcan we find the center and radius?

The answers to both questionsdependon whether we

can write (54) in the form of (53). We attempt to

accomplish this by completingthe squares in the paren-theses of the expression

(x2+ Ax) + (y2 + By) = -C.

In.ordertodo so we add A 2/4 and B 2/4 to both members:)

(

A2

) (

B2

)

A2

B2

x 2 + Ax+\"\"4 + y2 + By

+\"4 =\"\"4+ 4

-c)

or)

(

A

)

2

(

B

)

2A

2B

2

x+- + y+- =-+--C2 2 4 4')

which is in the form of (53), enablingus to say that the

center of the circle is (- A/2, - B/2) and the radius is)))

ANALYTIC GEOMETRY) .103)

V A 2/4 + B2/4 - C. The one difficulty that has beenoverlooked is the questionof whether this square root isreal or imaginary. If A 2/4 + B 2

/4 - C < 0, no real

circle exists. For example, x 2 + y2= -1 cannot repre-

sent a real locus. However, if A 2/4 + B 2/4 - C > 0,

we do have a real circle; and if this expression equals zero,the circle degenerates to a point.)

EXAMPLE 25. Examine the locus 3x2+ 3y2 - 6x + 9y + 2

_ 0- .Since the coefficients of x 2 and y2 are equal, we may rewrite

the equation with these coefficients equal to unity:)

2x 2

+ y2 - 2x + 3y +3

= o.)

Completing the squares,we get)

(

9

)

9 2(x2 - 2x + 1) + y2 + 3y + - = 1+ - - -,443

(3

)

231

(x- 1)2+ y + - =-.

2 12)or)

Hence, the locus is a circlewhose center is (1, -j) and whoseradiusis r =

\03703.)

If we consider a circle of radiusr centered at the origin

(as in Figure 60), the positionvector)

P = xi + yj)

can be written in terms of its angle of inclination 8:)

P = r cos 8i + r sin 8j.)

We therefore conclude that)

{

X = r cos 8

y=rsin8)(55))))


Y)

x)

FIGURE 60)

are the parametric equations of the givencirclewhere 8 is

the parameter.The single equation of the form (53) or (54) that repre-

sents a circle of radius r, centered at the origin, may berecovered from (55) by squaring alld adding as follows:)

x2 = r 2 cos 2 8

y2= r 2 sin 2

8,)

from which we deduce that

x 2+ y2 = r 2

(cos2

8 + sin 28).

But cos28 + sin 2 8 = 1. Thus)

x2

+ y2 = r2)

is the equation of the specified circle.)))


EXERCISES1. Write the equation of the circle with

(a) center (0, 1) and radius 2;

(b) center (0, -1) and radius2;(c) center (-2, 3) and radius 3.

2. Find the equation of the circle

(a) \\vith center at (2, -1) and passingthrough the origin;

(b) passing through (1, -1), (2,0), and (0, 3);

(c) having x-intercept 8, y-intercept -12 and passing

through the origin.

3. Find the centers and radii of circles having the followingequations:

(a) x2+ y2

- 16x+ 12y= 0;

(b) 2x2 - 2y2 = 9x;

(c) x 2 + y2- lOx + 7y + 9 = 0;

(d) 3x2+ 3y2 + 8x -

4y + 15 = O.

4. Find the points of intersection and the equation of the linethat contributes the common chord to the circleswhose equa-

tions are

(a) x 2+ y2 + 8y = 64

x2+ y2

- 6x = 16;(b) x

2+ y2 + 4x - 25 = 0

x2+ y2

-8y + 3 = o. (Hint. What is the algebraic

meaning of the interesectionof loci?)

5. Carry through the following vector approachto determinethe equation of the line tangent to a given circle at a given

point.Let the circleX be centeredat C =

(xo, Yo) and let Q =

(a, b) bea point of X. Call P = (x, y) the general point of theline gJ, which is tangent to X at (a, b). Using the fact that

\037 \037

the radius vector CQ is perpendicular to QP, prove that line

gJ is represented by the equation)

(xo - a)(x-a) + (Yo

- b)(y -b) =. o.)

6. Verify the equation for gJ given in Exercise 5 by using thepoint-slopeequation of the line and the fact that a radius isperpendicular to the tangent line at the point of contact.)))


7. Find the equation of the line tangent to(a) x2

+ y2 + lOx = 60 at (4,2);(b) x

2+ y2

- 12y= 36at (6,0);(c) x = 2 cos 8

}at CV2, V2);y=2sin8

(d) x = 2 cos 8

}at CV2, -V2).y=2sin8)

8. Prove that the parametric equations of a circlewith center

at (xo, Yo) and radius r is

{

X = Xo + r cos 8

y= Yo + r sin 8.)

9. Find a parametric representation of the equation of the

circle

(a) with center ( -1, 2) and radius 5;

(b) 3x 2 + 3y2 - 6x+ 9y- 2 = O.

10. Find the points of intersection of the loci whoseequationsare

(a) x2

+ y2- 2x -

2y + 1 = 0 and x + y= 1;

(b) x 2 + y2- 2x -

2y + 1 = 0 and x2+ y2 = 1)

21. SPHERES

The three-dimensional analogue of the circle is thesphere, which may be defined as the locus of points that

are equidistant from a fixed point. This definition is

easily recognized to be precisely that of the circle if the

discussion is restricted to a plane. Actually, the very

same definition carries over to four-dimensional,five-

dimensional, and n-dimensional spheres (sometimescalled hyperspheres).

Confiningour discussionto three dimensions, we call

Po = (xo, Yo, zo) the fixed point (center), P(x, y, z) thegeneralpoint of the sphere, and r the constant distance(radiusof the sphere) between Po and P (seeFigure 61).)))


2)

.y)

s)

FIGURE 61)

Then the vector equation of the sphere, in terms of the

position vectors of Po and P, is)

Ip-

pol = r) or) V(P-Po).(P-Po) =r,)

precisely the same as (52),\"\\vhich is the vector equation ofthe circle. In coordinateform we have

I(x- xo)i + (y

- yo)j + (z -zo)kl

=r,)

or)

Vex-

XO)2 + (y-

YO)2 + (z-

ZO)2= r.

Finally, by squaring both membersof this last equation,we get)

(x-

XO)2 + (y--.:.YO)2 + (z -

ZO)2= r 2

as tl1e equatio11of the sphere whose center is

Po = (xo,Yo, zo)

and ,vhose radius is r.)))


The problem of arriving at a general form for the equa-tion of a sphere (similar to that of (54)) and of converting

equations to that form would be quite repetitiousof the

discussion in Section 20 on circles. We therefore relegate

the details of such a discussionas an exercise for the

reader while we now turn attention to other matters

relating to the sphere.The unit sphereor sphere of unit radius, whose center is

at the origin,hasthe equation)

x2

+ y2 + Z2 = 1.) (56))

We shall analyze the components of the position vector)

P=xi+yj+zk)

of point P on the unit sphere, shown in Figure 62a.If a is the angle that P makes with the x-axis, (3 the

angle that P makes with the y-axis,and \"I the angle that

P makes with the z-axis,)

P \302\267i =Ipllil cos a = COS a)

and)

P \302\267i = (xi + yj + zk) \302\267i = x.)

Therefore x = COSa; similarly, y = cos(3 and z = cos \"I.

Hence P = cos ai + cos (3j + cos 'Y k .

Every vector V (see Figure 62b) in three dimensionsisparallel to the position vector of a point P on the unit

sphere, so that the direction of V is completely specified

by the allgles a,. (3, and \"I associated with the vector P.For this reasonwe call the angles ex, f3, and \"I the direction

angles of V (and also of P), and cas a, COS(3, and cos \"I

the direction cosines of V (alld of P).SinceIp!2 = p \302\267P = 1, the direction cosines of every

vector satisfy the relation

cos 2a + cos 2

{J + cos2'Y

= 1. (57))))


z)

y)

z)

x)

y)

(a))x)

(b))

FIGURE 62)

Suppose V = li + mj + nk. Then the unit vector to

which V is parallel is

V li + mj + nk

Ivl

=

VZ 2 + m 2 + n 2

\

Thus)

1cos a = ,

V l2 + m2

+ n 2)

mcos 13 ::::a: ,

V Z2

+\"m2

+ n 2

(58)n

and COB 'Y=

V2 2

'Z2 + m + n)))


from which we see that l:m:n = cosa:COS (3:cos 'Y;

that is, the numbers l, m, n areproportionaltothedirec-

tion cosines. It is for this reason that l, m, n, are

called direction numbers of V. We speak of the ordered

set {l, m, n} as a set of direction numbers. Since thedirection of tV = tli + tmj + tnk(t \037 0) is the same asthat of V (tV is parallel to V), we also call {tl,tm, tn}

(where t \037 0) a set of direction numbers for V. Ifsometripleof numbers is a set of direction numbersfor agiven vector, any nonzero multiple of the triple is also aset of direction numbers for the vector, for both triplesdesignatethe sa\037e direction (with the pOBsibility of

opposite sense) and both triplessatisfy the condition of

being proportional to the directioncosines.4

EXAMPLE 26. Let V = 2i - 3j + 6k.(a) We give three equivalent sets of directionnumbers for V.

From the given representation of V we see immediately that{2, - 3, 6} form a set of direction numbers. Two other sets

of direction numbers can be found by allowing t to take on twodistinct values in the orderedtriple {2t,- 3t, 6t} ; e.g., t = -1yields{-2, 3, - 6}and t = 2 yields {4, -6, 12}.

(b) Find the direction cosines of V.Making use of (58), we write

2 2 -3cosa = = -, cos{3 =-,V 22 + (-3)2 + 62 7 7

It should be noted that the unit vector)

6cos 'Y = -.

7)

!.V = \037i - \037J.

+ \037k7 7 7 7')

from which the direction cosinescan be read off directly, has)

4 In some developments, it may be advantageous to have aset of direction numbers for a vector V actually impart informationabout the senseof V. If this were the case, we would impose therestriction t > 0; for if t < 0, then {tl, tm, tn} .would imply a senseoppositeto the original triple {l, m, n}. However, our develop-ment of direction numbers is principally for applications to lines,which, in contrast to vectors, have no sense. Thus, we make the

simple restriction that t \037 o.)))

ANALYTIC GEOMETRY 111)

the same sense of direction as V. However,)

-1.V = -\037i + \037J. - \037k7777')

although oppositelysensed,is alsoa unit vector and thereforealso yields a set of direction cosines {- \037, \037,

-\037} for the vec-

tor V.)

EXERCISES

1. Find a set of direction cosinesfor the position vector ofthe points

(a) (4,3, 5),(b) (3, -4, 12),

(c) (0, 0, 1),(d) (0,2,0).

2. Using the method of Example 25, find the center and radiusof the sphereswhose equations are

(a) 4x 2 + 4y2 + 4z2 = 8y,

(b) x2 + y2 + z2 + 4x -6y

= 3.

3. Find the equation of the sphere with

(a) center at (1, -1, 0) and radius 2,

(b) center at (-1,2, -3) and radius V 2,(c) center at the origin and radius 12.

4. Find two sets of direction numbers and two sets of direction

cosines for the vector(a) V = 3i + j - k.(b) V = 2i - j + k.)

22. PLANES

The second locus, or surface, that we discuss is the

plane, which in some respects is analogousto thelinein

two dimensions. Although we have several equivalentchoicesfora definition, our needs of the momellt will bestbe servedby the following:

Definition. If Po is a fixed point andN a fixed vector, the--7

locus of points P so that PoP is perpendicular to N iscalled a plane. (In Figure 63, we picture N positionedso that its originat Po.))))


FIGURE 63)

An immediate consequence of the definition is that Po--7

itself is on the plane, for if P = Po, then PoP is the zerovector that is perpendicular to N (see Theorem5).

Writing the definition in vector language in terms of

position vectors, we have)

(P - Po) \302\267N =

0,) (59))

the vector equation of the plane through Po and perpendicu-

lar to N.In order to derivean al1alytic expression for the plane,

we follow the usual procedureof calling the variable

POillt of the plane P =(x, y, z). The fixed elements

will be Po =(xo, Yo, zo) al1d N = ai + bj + ck. Then

(59) becomes

[(x -xo)i + (y

- yo)j + (z- zo)k]

\302\267(ai + bj + ck) = o.

Expandingyields

a(x- xo) + bey -

Yo) + c(z-

zo)= 0, (60)

which is the analytic equation of the plane through (x.o,Yo, zo) and perpendicular to the vector whose directionnumbers are specified by {a, b, c}. The analogouscharacter of the equationsof the plane and the lil1e is)))


easily seen by comparing (60) with (48). Furthermore,

(60) may be written

ax + by + cz + d = 0, (61)whered = -

(axo + byo + cZo), which is the counterpartof (49),and which clearly shows that the equation of aplane is linear in all three variables. Finally, it shouldbe noted that the coefficients of the variables in both

equations 49 and 61 are related to the perpendicularvectors N.)

EXAMPLE 27. (a) What are the equationsof the coordinate

planes?The xy-plane can be describedas the planeperpendicular

to k and passing through the origin. Therefore, its equation

is found by simplifying [(x -O)i + (y

-O)j + (z -

O)k]\302\267k =

O. The result is z = O. Similarly, the reader can show that)

z)

z=zO)

(XO, YO,zo) \302\267)

y)

x)

FIGURE 64)))


the equation for the yz-plane is x = 0, and the equation for

xz-plane is y = o.(b) What are the equations of planes parallel to the coordi-

nate planes?We fix our attention on a plane parallelto the xy-plane and

passing through (xo, Yo, zo), as shown in Figure 64. Then,following the same reasoning as in part (a), we arrive at z = zo,which states that for all choices of x and y, the z-coordinateof a point on the plane is equalto Zo.

EXAMPLE 28. vVe shall discuss several questionsrelatingto the plane whose equation is

2x + y- 2z - 2 = O.

(a) To find a unit vector perpendicular to the given plane.A vector perpendicular to the plane can immediately be

written, by using the coefficients of the variables(see(60) or

(61)), as)

2i + j - 2k.

Two unit vectors perpendicular to the plane are then

2i+j-2k+ .-3)

(b) What are the intercepts of the given locus? That is,what are the pointsat which the plane intersects the coordinateaxes?

x-intercept. Since every point of the x-axis must have itsy- and z-coordinates equal to zero, we solve for the x-coordinate

of the Intercept by placing y= z = 0 in the given equation

of the plane. This gives 2x - 2 = 0 or x = 1. Thus the

x-intercept is (1, 0, 0).The reader can carry through the computation to verify

that (0,2, 0) and (0, 0, -1) are the other two intercepts.

(c) What are the traces of the plane whose equation is2x + y

- 2z - 2 = 0, in the coordinate planes? We definethe trace of a locus oCin a plane II to be the points of oC that liein II, or simplythe intersectionof oC and II.

If we ask specifically for the trace in the xy-plane, we areaskingfor the locus satisfying the simultaneous equations

{

2X + y- 2z - 2 = 0

z = 0 (xy-plane) .)))


z)

> y)

Trace in yz-plane)

x)

FIGURE 66)

Thus, we would sketch the graph of 2x + y- 2 = 0 in the

xy-plane (z= 0) to see the trace in that plane. Again, the

reader can verify that the other tracesareobtained by graphingx - z - 1 = 0 and y

- 2z - 2 = 0 in the xz-plane and yz-plane, respectively. Graphs of the traces in the coordinateplane are quite helpful in sketching a three-dimensional locus(see Figure65).)

23.DETERMINING A PLANE BY POINTS ON ITHow many points are actually necessary to determine

a plane? Geometric intuition indicatesthat three points

would be necessary and sufficient, but can this fact beshown algebraically?

If the four constants a, b, c, and d are known, the

equation of the plane ax + by + cz + d = 0 is com-pletely determined. This suggests that fOllr, rather thanthree, conditionsare necessary. However, our intuition

did not lead us astray,for one of the four constants canbe chosento be unity. Not all of a, b, and c are zero,so ,ve can divide through by one of them, say a \037 0,)))

116

and get)

ELEMENTARY VECTOR GEOMETRY)

b c dx + -

y +-

Z +- = o.

a a a)(62))

Making the substitution (3 = bfa, 'Y= cja, and 0 = dja,

(62)becomes)x + {3y + 'YZ + 0 = 0,)

from which we see that there are essel1tially three con-

stants to determine. In practice,it may turn out that

a = 0, in which casethe division just performed was

illegal. If this be thecase,then we could choose one of theother numbersasa divisor until we hit upon a legitimatedivision (seeExample30).EXAMPLE 29. Find the plane through (1, 2, 3), (1, -1, 0),and (2, -3, -4).

Assuming d = 1, we solve for a, b, and c in the equation)

ax + by + cz + 1 = o.) (63))

By imposing the condition that (1, 2, 3) renders (63) true, we

get)a + 2b + 3c + 1 = 0;

that (1, -1, 0)renders (63) true yields

a - b + 1 = 0;

and that (2, -3, 4) renders (63) true yields

2a \037 3b - 4c + 1 = O.)

We can reduce this system of three equations in three unknowns

to two equations in two unknowns most easily by using thesecond equation,which is already devoid of c, and then elimi-nating c from the first and third. To this end, we multiply

the first by 4 and the tl1irdby 3, getting

4a + 8b + 12c + 4 = 06a - 9b - 12c + 3 = O.)

....\\ddition yields)lOa - b + 7 =

0,)))

ANALYTIC GEOMETRY

which, together with a - b + 1 = 0, implies that)

117)

2a = --

3)and)

1b =-.

3)

Substituting these values into the original equations gives

c = -t. Therefore the desiredlineis -ix + ty-

tz + 1 =0,ormoresimply stated,

2x -y + z = 3.

EXAMPLE30. Find the plane through (0, 1, 2), (1, -1, -2),and (2, 2, 4).

Imposing the three pointson equation 63 yields the system

b+2c+l=0

a-b-2c+l =02a + 2b + 4c + 1 = o.

Adding the first two equations gives a + 2 = 0 or a = -2.However, adding twice the second equation to the third gives4a + 3 = 0 or a = -

\037, an inconsistency. So we must dis-card the assumption that d \037 o. Instead, let us assume thata r!= 0; particularly that a = 1. Then we seekb, c, d in)

x + by + cz + d = O.) (64))

Imposing the three given points on relation (64),we get the

system)b+2c+d=0

I-b-c+d=O2 + 2b + 4c + d = o.

Adding the first two equations of (65) gives 1 + 2d = 0 ord = -t; but adding twice the second to the third gives4 + 3d = 0 or d = -t, which is another inconsistency. So wediscard the assumptionthat a r!= 0, upon which the form (64)was based. We now try b \037 0, and use the form of the equa-tion with b = 1)

(65))

ax + y + cz + d = o.) (66))

Once again we impose the given points, this time on (66),getting the system)

1+2c+d=0a-1-2c+d=0

2a + 2 + 4c + d = o.)

(67))))


Adding the first two equations of (67) gives

a + 2d = 0,

and adding twice the second to the third yields)

(68))

4a + 3d = O.) (69))

Equations 68 and 69 now form a system of two equations in

two unknowns, the solution of which is a = d = o. Substi-tuting back in (67) permits us to find that c = -\037. Thereforewe put these values in (66) and find the desired plane to be)

y-

\037z= 0) or) 2y - z = O.)

(Actually, the computation could have been simplified some-

what if we had set d = 0 after the appearance of the first

inconsistency and then set a = 0 after the second inconsistency.

However, this might have been jumping too far and too fast for

an illustration.))

These cumbersome methods of finding the equation of

a plane through three given points will be superseded

by a far more simpleand elegantvector approach in the

next chapter.)

24. DISTANCEFROM A POINT TO A PLANEThe problem of determining the distance from a given

point Po =(xo, Yo, zo) to a given plane)

ax + by + cz + d = 0)

is handled in precisely the same manner as was the earlier

problem of finding the distance from a pointto a lineintheplane.

Call 8 the distance from Po to the givenplane,and let

P = (x,y,z) be an arbitrary point of)

ax + by + cz + d = 0)

--7

(see Figure 66). Then 8 =\\prNPopI,

where N is a

vector perpendicular to the plane. Letus-as before-)))


Po)

FIGURE 66)

use a unit vector for N. Then)

ai + bj + ckN= \302\267

V a 2 + b2 + c2)

To determine the desired projection we resort to inner

product)

\037 --7

IPop\302\267

Nt=

IprNPoP(= 8.)

Then)

8 = [(x -xo)i + (y

- yo)j

( )k]

ai + bj + ck+ z -

Zo\302\267

V a 2+ b2 + c2)

ax - axo + by- byo + cz -

CZo-

V a 2 + b2+ c 2)

But ax + by + cz = - d. Therefore,

I-axo- byo -

cZo-

dl taxo + byo + cZo + dl8

- - \302\267-

v' a 2 + br + c2

-

V a 2 + b2+ c

2)))


EXAMPLE 31. We find the distance from Po = (1, -2, -3)to the plane whose equation is)

x -2y

- 2z + 1 = O.)

Conforming to our policy of minimal memorization, we ignorethe formula derived above and instead reason from first

principles.\037

Since we seek the projection of PoP on the perpendicular,where P is any point of the given plane, we shall find someparticular P on x -

2y- 2z + 1 = 0, with which to work.

One choice for P that would result in simple computation canbe found by letting y = z = o. Then, x = -1 and P =

( -1, 0, 0).Therefore)

\037

8 = IprN PoPl

- [(-1 - l)i+ (0- (-2))j + (0 - (-3))k])

i -2j

- 2k.V 12

+ 2 2 + 22)

1-2- 4 -

61 12- ---- 4-3 -3-.)

EXERCISES

1. Sketch the following planes, determining their interceptsand traces in the coordinate planes:

(a) x + 2y- z = 3;

(b) x - z = 3;

(c) 2x -y + 2z + 9 = 0;

(d) y= -3.

2. Find the equation of the plane

(a) through (1, 1, 1), (3,-2, 1),and (2, -4, 3);

(b) parallel to -x + 2y- z = 5 and through (1, -1,0);

(c) perpendicular to 2i - j - k and th\037ough the origin;(d) perpendicular to 2x + 3y

- 6z = 12, through the origin

and (2, 1, -4).(e) through (-5, 0, 8) and perpendicular to a vector whose

directionnumbers are given by the set {4, -3, 12}.)))


z)

ce)

y)

\037)

FIGURE 67)

3. Find the distance from (3, 0, -2) to

(a) 3x + 4y- 12z = 52,

(b) 2x -y + 2z + 81 = 0,

(c) 4x - 3y = 100.)

25. THE STRAIGHT LINE IN THREE DIMENSIONS

Our discussion of the straight linein space shall follow

the same pattern as the discussionof the linewhen our

attention was confined to the plane. We therefore begin

by specifying a line \302\243by two of its points PI =(Xl, YI, Zl)

and seek an analytic representation of \302\243.

Let P = (x, y, z) represent the general point of \302\243.

Applying Theorem 4 to the position vectors of the threecollinear points (see Figure 67), we may write the vector

equation of \302\243as)

P = (1 - t)PI + tP2 .) (70))))


In terms of the basis vectors, (70)becomes)

xi + yj + zk = (1 - t)xli + (1-t)Ylj

+ (1- t)zlk + txli + tY2j + tZ2k

= [Xl + t(X2- xI)]i + [YI + t(Y2

- YI)]j

+ [Zl + t(Z2 - zl)]k.)Therefore)

x =Xl + t(X2

- Xl)Y

= YI + t(Y2- YI)

Z = Zl + t(Z2 -Zl).)

(71))

is the parametric representation of line \302\243in analytic form.Elimina tion of the parameter t can be accomplished

in a manner analogous to that in the discussionof the

line in the plane. We solve for t in the parametric

form, getting)

X - Xl Y-

YI Z - Zl= = = t.X2

- Xl Y2-

YI Z2- Zl

Thus we may write the equationof \302\243devoid of t as)

Y-

YI Z- Zl- - ,

X2- Xl Y2

-YI Z2

- Zl

which may be called the two-point form of the line inthree dimensions.

The readerwill observe that (72), which describes alocus in termsof coordinatesa11dno auxiliary variable,

is not a single equation. It actually consistsof threeequations,ofwhich two are sufficient to deduce the third:)

X-

Xl)

(72))

X -Xl Y

-YI X -

Xl Z - Zl- , - ,X2

- Xl Y2-

YI X2-

Xl Z2- Zl

and Y-

YI . Z - Zl(73)-

Y2-

YI Z2 -Zl)

It may appear strange that more than one equatio\037is

necessary to describe a locus, but a moment'sreflection)))


makes it appear quite reasonable. For when we con-

sidered two simultaneous linear equations in planeanalytic geometry, we stated that they defined a point,the pointof intersectionof two lines (if such existed).Each of the equations in (73) is linear and therefore

defin\037sa plane. Two of these considered simultaneously

consistof the locuscommon to the two, namely, a line.That is, the intersectionof two planes is a line. Each

plane in (73) is parallelto oneof the coordinateaxes,

but such special planes need not be the only ones used

to define a line. In fact, any two simultaneous linear

equations)

{

a1x + b 1y + CIZ + d 1 = 0

a2X + b 2y + C2Z + d 2 = 0)

(74))

define a line unless the equations represent parallelplanes. We shall have more to say later on such repre-sentationsof lines.

\037

The vector P IP 2 that is parallel to (or along) line \302\243is\037

P 1P 2 =(X2

- x1)i + (Y2- Yl)j + (Z2

- zl)k and there-fore has a set of directionnumbers {X2

- Xl, Y2- Yl,

Z2- Zl}. If the concept of di,.ectionnumbers is applied

to lines (see Figure 68), we see that the denominators in

the two-point form (72) are precisely the directionnum-

bers of the line. Given a set of directionnumbersfor \302\243,

direction cosines for the line can be found by equations

58.

Let \302\2431be a line with direction l1l1mbers {ll, m1,nl}and \302\2432a line with direction 11l1mbers {l2, m2, n2}. \302\2431is

parallel to \302\2432if and only if vectors VI = lli + m1j+ n1kand V 2 = l2i + m2j + n2\037 are parallel. .This latter

condition is satisfied if and only if V 1 is a (nonzero)multipleof V 2, say)

V 1 = tV 2,)))


z)

Pl(Xl,Y1,Zl))

,\302\243)

Y)

x)

FIGURE 68)

in which case II =tl2, ml = tm2, and n1 =

tn2 or)

II m1 n1_-_ -_e- -l2 m2 n2)

(75))

Hence a necessary and sufficient condition for lines \302\2431

and \302\2432to be parallel is that their direction numbers be

proportional.)

26. ANGLE BETWEEN TWO LINESThe angle betweentwo lines is defi11ed as the angle

between two vectors, one parallel to each of the givenlines. As a consequenceof this definition, we may speakof the angle betweentwo lines even if the lines do notintersect.

Consideringthe lines \302\2431and \302\2432as .specified earlier,we shall derive a formula for the cosineof the angle (J

between \302\2431and \302\2432in terms of the direction numbers of

the lines.)))


The angle (J is the angle between V I and V 2, where

V I = lli + mIj + n1kand V 2 = l2i + m2j + n2k. Usingscalar products, we have)

V I .V 2COB0 =

Iv11lv2 1)

VI\302\267

V2= ,

VV I\302\267

VI V' V 2 \302\267V 2)

or)

lIl2 + mlm2 + nIn2COB (J = \302\267

(76)Vl I

2+ mI

2 + nI 2V 12

2+ m2

2+ n2

2

If the given direction numbersare direction cosines

as well, (76) simplifies to become

cos(J = cos al cos a2 + COS {3I COS {32

+ COS'Y I COS 'Y2. (77)

As a corollary, we may concludethat the two lines are)

z)

\302\2431)

, .)

VI)

y)

x)

FIGURE 69)))


perpendicular if and only if

lll2 + mlm2 + nln2 = o.) (78))

EXAMPLE 32. Find the equation on the line \302\243joining

PI = (2, -1,3) tOP 2 = (1,0, -5).(a) Parametric Form. From (71) '\037le have)

{

X = 2 + (1 -2)t

\302\243: y=-l+(O-(-l))t

z = 3 + (-5 -3)t.)

Simplifying yields)

{

X=2-t

\302\243: y= -1 + t

z = 3 - 8t.)

We see that {-I, 1, 8} is a set of direction numbers for \302\243.)

A t f d.. ..

{

1 1 8

}se 0 lrectlonCOSInes IS -

_ /_' _ I-=' _ /-\302\267

V 66 V 66 V 66)

(b) Two-point form. By means of (72)'we have)

x-2 y+1.--1 1.)

z-3--8)

EXAMPLE 33 a . 1. 01 X - 1. y + 3fi d ( )\302\267 1ven Ine d...J:

2= _ 1

=z, n a)

a parametric representation of \302\243';(b) the cosine of the angle (J

between \302\243'and line oC of Example 32.)

x-I y+3(a) We set = = z = t

2 -1')

which can be restated as a parametricrepresentationof \302\243')

x=I+2t

y = -3 - t

z = t.)))


(b) A set of direction numbers for \302\243'is { 2, -1, I} . We

therefore apply (76), which gives)

O -1(2)+1(-1)+8.1cos =Y(-1)2 + 12

+ 8 2y2

2+ (-1)2 + 12)

5.

6Y ll)

EXAMPLE 34. Find the line\302\243\"parallel to \302\243(of Example 32)and passing through (1,'0, -2).

Since\302\243\"is parallel to \302\243,\302\243\"is of the formx

= tl = Y

\037

YI

=Z

\037

zt, where (Xl, YI, Zl) is a point of \302\243\". But (1, 0, -2)

is given as a point of \302\243\". Therefore the equations of the linex-I z+2

are =y=-1 8

Thus far we have ignoredthe possibilityof one of the

direction numbers being zero, in which case the two-

point form (72) could not apply becauseof the illegality

of division by zero. Before proceeding with the algebraicproblem involved, let us determine the geometric mean-ing of, say, the first direction number being zero. Thiswould mean that cosa = 0, which implies a = 1r/2.The line in questionwo1ildthereforebe perpendicular to

the x-axis; or equivalently stated, the line would be

parallel to (or actually in) the yz-plane. In summary, a

line with a set of direction numbers {O, m, n} is parallelto the yz-plane. Similarly,the reader can show that a

line with a set of directionnumbers {I, 0, n} is parallelto the xz-pla11e;and a linewith a set of direction numbers

{I, m, O} is parallelto the xy-plane. If two of the direc-tion numbers are zero,then the line is parallel to one ofthe axes.)

EXAMPLE 35. Find equations of the line through (1, 2, -1)and (3, 2, -2).

A set of direction numbers is easily found to be {2, 0, -1}, sothat the two-point form (72) cannot be used. However, a)))


parametric form of the line can be written

x=I+2t

y=2

z = -1 - t.Elimination of the parameter yields

{

X-l = Z+l2 -1

y= 2,

which is a nonparametric form of the line and is as closeaswe may come to writing the two-point form. It is easy to seefrom both these representations that y

= 2 no matter what thevalues of x and z are. That is, the line must be in the planey

= 2 and is, therefore, parallel to the xz-plane.EXAMPLE 36. Find the equations for the line through

(1, -1,2), with direction numbers given by the set to,4,OJ.An equivalent set of direction numbers is to, 1,O};indeed,

to, m, O} (when m \037 0) is an equivalent set of directionnum-bers. We may thus write a parametric form as:

x = 1)

y= -1 + t

z = 2.

Actually, the equation y= -1 + t is quite superfluous, for

the line is completely determined as the intersection of theplanesx = 1 and z = 2. The y-coordinate may take on any

value, which is what may happen in the foregoing parametric

form of the line.We now return to the representationof a line as the inter-

section of two planes (see equation 74). This idea has, of

course, been employed throughout; for instance, in Example35, (79)is the equation of a line in terms of two planes throughit. Let us examine a more generalcase.EXAMPLE 37. Given a line \302\243whose algebraic representation).IS)

2x + 3y - z = 1

x -y + z =

2,)))


we shall exhibit a method finding other forms for the equationsof aC.

Eliminating z by adding the two equations gives

3x + 2y= 3.

Eliminating x from the given equations yields

5y- 3z = -3.

Solving for y in each of these allows us to write

3x-3 3z-3=y=-2 5)

which may now be put in the form (72) by dividing numerator

and denominator of the first and third membersby 3:

x-I z-1

-(2/3)=

y=

5/3.)(80))

Therefore a set of directionnumbers is { -i, 1, i}.By setting)

x-I z-1

-(2/3)=

y=

5/3

= t,

we arrive at a parametricrepresentationof the line

x = 1 - itI

Y= t

z = 1 + it.)EXERCISES1. Find the equations, in two-point form, of the lines deter-mined as follows:

(a) through (-4, 1, -2) and parallel to z-axis;

(b) through (-4,1, -2) and (5,3, -1);

(c) through (-4, 1, -2) and (3, 1, 2);

(d) through (3, 2, -5) and perpendicular to 3x -

2y- 6z

= 1;(e) through (3, 2, -5) and having cos a =

\037and cos {3 =

\037(two possibilities);

(f) through (-4, 1, -2) and having direction numbers

3, 4, 12;(g) parallelto 5i -

4j + 6k and through the origin.)))


2. Find parametric representationsof the lines in Exercise 1.

3. Let AB, AC, and AD be three edges of a cube. Prove that

plane through D and the line joining the midpointsof AB and

AC is tangent to the sphereinscribedin the cube.4. (a) Find a set of direction numbers for the line)

{

X - 3y + 7 = 0

x - 2z - 4 = o.)

(b) Find a parametricrepresentationfor the line of part (a).

5. (a) Find equations of the form (72) that represent

3x -2y

- 6z = -9

x+4y-8z=-16.)

(b) Find a parametric representation of the line of part (a).

6. Prove that the following are vertices of a right triangle:(a) (0,7, 6), (-2, 1, 3), (6, 4, 8);(b) (0, -2,4), (-3, -4, -2), (2,4,1).

7. Three vertices of a parallelogram ill order are (4, 3, 5),(0,6,0), and (-8, 1, 4). Find the fourth vertex.

8. Find the direction cosines of a line perpendicular, simul-

taneously, to)

x-I y-l z-l x Y z- - and -----.- - - -2 3 6 3 4 12)

9. Find the cosine of the angles made by the two lines of

Exercise 8.)

27. INTERSECTION OF A LINE WITH A PLANE

Since a line may be representedby the equations of

two planes through it, the pointof intersection (if there

is one) of the line with a plane may be found by solvingthe threelinear equations simultaneously. However, we

illustrate another method for solving this problemby

utilizing the parametric representation of the line.)))

AN AL YTIC GEOMETRY) 131)

EXAMPLE 38. Find the point of intersection of the line

{

2x + 3y- z = 1

\302\243:

x-y+z=2)and the plane)

II: x + 2y- 2z = -3.

Solving these three linear equations simultaneously yields(j-,1,I) as the point of intersection. However, for the alter-native approach, we use the parametric form of \302\243:

x=l-jt

y=t

.z= 1 + -it,

which was found in Example 37.Substitutingtheseequations in the equation of plane II gives

(1 - it) + 2t - 2(1 + it) = -3, the solution of which will

give the value of the parameter t, which yields the point of \302\243

that is also on II (seeFigure70). This value is t = 1. Thus,the point we seek is given by x = j-, y

= 1, z = -i.)

y)

%)

FIGURE 70)))


28. ANGLE BETWEEN A LINE AND A PLANE

How can we extend our understandingof the angle of

intersection of a pair of lines to define the concept of theanglebetween a line and a plane? If \302\243is the line and II

the plane, there are an infinity of lines \302\24311\"in the plane II

that might conceivably be usedto find an angle between

\302\243and II (see Figure 71). Since such an infinity of choices

for the angle between a line and a plane is undesirable,

the one usually selected as the angle of intersection

between \302\243and II is the minimum angle {3 of the infinity

formed by \302\243and the lines \302\24311'. However, such a defini-

tion renders the problemof finding {3 extremely difficult-

indeed, impossible at this point. We therefore turn to

the more convenient definition: The angle {3 between

line \302\243and plane II is defined to be the complementof the

angle between \302\243and a line m perpendicular to II.)

EXAMPLE 39. Find the angle of intersection betweentheline \302\243and the plane II of Example 38.

A vector parallel to \302\243is L = -fi + j + j-k. A vector per-

pendicular to plane II is N = i + 2j - 2k. Thus the comple-)

ffi,)

FIGURE 71)))

ANALYTIC GEOMETRY

ment of (j may be'found by dot products)

133)

cos

(

r _(3

)

= \\-2/3 + 2 -\037O/3\\ = 2_ ,

2 09/9 y9 y39)

which is equivalent to sin (j = 2/09.)

EXERCISES1. In the computation of Example 39, we were careful to take

the absolute value of the numerator in the determination of

COB

(;

-(3).

Why?

This indicates that our definition of {3 has a slight short-

coming, which the reader should eliminate by a small modifica-tion of the definition.

2. Find a general formula for the sin {3, where {3 is the angleformed by a line meeting a plane.3. Using two different methods, find the coordinates of the

point of intersection of line, with a representation)

x-3

2)

y+2

-1)

z-3

3)

and the plane whose equation is 3x + 2y- z = 5.

4. Find the anglebetween the line and plane given in Exercise3 above.5. Find the point of intersection and the angle of intersection

of the line and plane whose equationsare)

x+3

2)

y+3

3)

z+3

6)

and) lOx + 2y - l1z = 3.)

6. (a) Give a suitable definition for the angle between two

planes.)))

134) ELEMENTARY\" VECTOR GEOMETRY)

(b) Find a generalformula for the angle between two planes,whoseequationsare)

and)

alx + b1y + CIZ + d 1 = 0

a2X + b 2y + C2Z + d 2 = o.)

(Hint. Use two vectors, each perpendicular to one of the given

planes. )

(c) Find the angle made by the intersection of the planeswhoseequationsare2x+ 2y + z = 1 and 2x + 10y

- llz = 3.(d) Find the anglebetween the two planes whose equations

are x -4y + 8z = 2 and 2x + y

- 2z = 18.

7. Show that the line given by

{

X - 3y + 7 = 0x - 2z = 4,)

(a) is parallel to the plane whose equation is 3x -6y

- 2z= 0,

(b) lies in the plane whose equation is 12x + 4y + 6z = 1.8. Apply fundamental vector techniques to the followingproblems.

(a) Find the anglebetween the diagonal of a cube and oneof its edges.

(b) Find the angle between the diagonal of a cube and a

diagonal of one of its faces.

(c) Find the angle betweenthe diagonal of a cube and oneof its faces.)))

cross

products)

29. CROSS PRODUCTS

As we have seen, the inner product of two vectors yieldsa scalar quantity, hence the synonym scalar product.

In the present section we introduce another kind of

multiplication of vectors, which leads to (1) a vector

quantity and (2) a relation involving the si11eof the angle

between two vectors (as opposed to the appearanceofthe cosinein dot products). Since, by now, the readerhas gainedsomesecurity in dealing with vectors, we takethe liberty of a longerexcursion through the algebra of

cross products before illustratingthe theory.Definition of cross product. Let A and B be any two

vectors, and call 8 the smaller angle betweenthem.

Then the cross product of A and B, denoted A X B, is avector)

IAIIBI sin 8 U,) (81))

where U is a unit vector perpendicular to the plane of135)))


AxB)

U B\037

\037\037A \037)

B

\037\037U A \037)

BxA)

FIGURE 72)

A and B, and pointed in such direction as to make

{A, B, U} a right-handed triple. That is, if the observer

who walks from the terminus of A to the terminus of

B through the angle 0 (when the two vectors both

emanate from a singlepoint0) always has 0 to his left,then he is always standing upright parallel to U, with his\"\037pward direction\" being precisely the direction of U.Thus, {A, B, A X B} form a right-handed triple if none of

the three vectors are the zerovector,for A X B (if non-

null) points in the same direction as U. This assertionfollows from the factthat the coefficient of U in (81) can-not be negativebecause0 < sin0 < 1 when 0 < 0 < 1r,and lengthsof vectors are non-negative scalars.

Because the cross product results in a vector quantity,it is often termed the vector product. We shall continue

our earlier policy of utilizing both names so that the

reader will become familiar with all the standard-andequally popular-terminologies.

An immediate corollary of the definition is that thevector product defines a noncommutative multiplication.)))

CROSSPRODUCTS) 137)

In fact, we have theTheorem 8. A X B = - B X A.

This resultfollows quite simply from the observationthat B X A has the same magnitude as A X B but pointsin the opposite direction (seeFigure 72).

Query: Suppose the word \"smaller\" were left out of

the definition of vector product. Would the resultingdefinition be equivalent to the original one?

Attempting to gain further physical or geometric

insight into the notion of cross product, we observethat)

A is parallel to B implies A X B = 0; (82)

for if A and B are parallel, then 0 = 0 and sin (J = o.

(Is the converse of (82) true?) A geometric interpreta-tion of the magnitudeof the crossproductisgiven by

Theorem 9. The area of the parallelogram generated by

vectors A and B is equalto IA X B \\.

Proof. (See Figure 73.) The area of the paral-lelogramgenerated by A and B is equal to)

IAlh=

IAIIBI sinO =IA X BI.)

30. TRIPLE SCALAR PRODUCTThe product A. B X C is called the triple scalar

product of A, B, and C.

N.B. (1) Although parenthesesmight appearneces-sary-at leasttoassistthe beginner-to indicate which

operatioll (the \"cross\"or the \"dot\") is performed first,

it should be noted that such parelltheseswould be com-)

\037h) )A)

r\\hl) ))A)

FIGURE 73)))


II

I

\037-----------------//B/

\"\" C)

FIGURE 74)

pletely superfluous, for there is only one possiblelogicalorder for the two operations. \"Cross\" must precede

\"dot\"; otherwise we would have a scalar crossed with a

vector, which has no meaningfor us.

(2) The product A X B \302\267C also yields a scalar quan-tity and thus is alsocalledthe triplescalar product of

A, B, and C (in that order). Callingbothquantities,A

\302\267B X C and A X B \302\267C, the triple scalar product would

be reasonableand warrantedonly if they were always

equal-and so they are, as we shall soon see.

Theorem 10. The magnitude of A \302\267B X C represents the

volume V of the parallelepiped generated by the vectors

A, B, and C.

Proof. (See Figure 74.))

IB X ci = area of base (Theorem 9)and

IAIcos a = + altitude of parallelepiped.

Therefore + V =IAIIB X CI cos a, which states that

V =IA. B xci.

The reader should investigatethe questionof sign by

answering the following question: What kind of orienta-

tion of A, B, and C leads to a positivetriplescalar prod-

uct? What kind to a l1egative triple scalar product?)))

CROSS PRODUCTS 139

In which case does the triple scalar product equal V

and in which case doesit equal- V?

Corollary. A. B X C = A X B.C.

Proof. Since both triple scalar products represent

the same volume, and since the sign dependsonly on the

orien\037ation of the triple {A, B, C}, the two products

must be equal.-Our final result on the algebra of vector products is

embodied in the following theorem.Theorem 11. The vector product is distributive with

respect to addition. That is,)

(i) A X (B + C) = A X B + A X C, and(ii) (A + B) X C = A X C + B X C.

Proof of (i). Let

D = A X (B + C) - A X B - A X C.)

Our proof will be completed if we can show that D isnecessarily the zerovector.

We take the scalar product of D with an arbitrary

vector V.)

v \302\267D = V \302\267[A X (B + C) - A X B - A X C]

= V \302\267A X (B + C) -V

\302\267A X B -V \302\267A X C

(Theorem 7)

= V X A \302\267(B + C) -

V X A \302\267B - V X A \302\267C

(Corollary to Theorem 10).

Now, using the distributivity of dot product! on thefirst term of the right member,we get)

V.D =VXA.B+VXA.C -VXA.B-

V X A \302\267C.)

Hence)

v \302\267D = O.) (83))))


Recalling that V is arbitrary, we see then that (83)statesthat the vector D is perpendicular to any vector.Consequently, D = 0 and we have)

A X (B + C) = A X B + A X C.)

Part (ii) is left as an exercise, for it follows readilyfrom (i) and Theorem8.

Wearenow in possession of sufficient tools to developa formulafor the crossproductof two vectors that are

represented in terms of our {i, j, k}-basis. First, wenote that

i X i = j X j = k X k = O. (84)

Applying the definition (81), we further see that

i X j= k, j X k = i, and k X i = j, (85)

and imposing Theorem 8 on the relationsof (85) yields

j X i = -k, k X j = -i, and i X k = -j. (86)

Now let)

A = a1i + a2j + ask

Then

A X B = (a1i + a2j + ask) X (b1i + b 2j + bsk)

= (a1i+ a2j + ask) X b1i+ (a1i + a2j + ask) X b 2j

+ (a1i + a2j + ask) X b 3k)

and) B = b1i + b 2j + bsk.)

= a1i X b1i+ a2j X b1i + ask X b1i+ a1i X b 2j + a2j X b2 j + ask X b2j

+ a1i X bsk + a2j X bsk + ask X bsk.

By using (84), (85), and (86), we simplify this expansion

to obtain)

A X B =(a2bS

- a Sb2)i + (aSbl - a1bs)j+ (a 1b 2

- a 2 b 1)k. (87))))

CROSS PRODUCTS) 141)

The reader who is familiar with determinants will be

pleased to see how convenientit istoexpress the formula

for A X B in the language of determinants.1

Equation

(87) states)

A. X B = i a2 a3 -j al a3+k

al a2b2 b 3 b1 b3 b 1 b2

or, more simply . . k1 J

AXB= al a2 a3 . (88)b i b 2 b 3)

EXAMPLE 40. Our first illustration of the use of cross

products will be to apply this conceptto the problem of Exam-

ple 28, namely, that of finding a vector perpendicular to theplanewhose equation is 2x + y

- 2z - 2 = O.

Since the cross product of two vectors is a vector perpen-dicular to the plane of the given two, we need only determine

two vectors in the plane 2x + y- 2z - 2 = 0 and take their

cross product. We therefore select three points, arbitrarily

in the given plane: A = (0, 0, -1), B = (0,2,0), C = (1, 0, 0).\037 \037

Then AB = 2j + k and AC = i + k.Hence)

\037 \037 i j k

AB X AC = 0 2 1 = 2i+ j - 2k

101)

is a vector perpendicularto the plane of A, B, and C. (It isreassuringto note that this answer agrees with the one foundin our first solution to this problem).)

Although the area of a triangle is a simple theoreticalproblem, it is quiteoften a cumbersome task to computean area by the usual formulas when given the coordinatesof the vertices. We exhibit the power of vector productsin attacking such a problem as)

1 For the reader who is unfamiliar with determinants, \\ve have a

brief word of assistance in the appendix.)))


A)

FIGURE 75)

EXAMPLE 41. Let the vertices of a triangle be given by)

A = (1, -5, 2), B = (-2, 3,-1), and C = (5, 0, 2).)

---7 ---7Then AB = 3i+ 8j - 3k and AC \037 4i + 5j.)

---7 ---7 i j kAB X AC = -3 8 -3 = 15i- 12j + 17k.

4 5 0)

---7 ---7Since lAB X ACI equals the area of the parallelogram (see

Figure 75), three of whose vertices are A, B, and C,the areaK---7 ---7

of the shaded triangle we desire is precisely \037IAB X ACI.Therefore)

K = i-115i- 12j+ 17kl= i V 658 .)

In our earlier work we solved, by analytic methods,

the problem of determining the equation of a plane

through three given points. The vector approachtothisproblemisdemonstrated in the next example.)

EXAMPLE 42. Find the plane through A = (1, 2,3),B = (1, -1, 0), and C = (2, -3, -4). (The analytic solu-

tion to this problem was given in Example 29.) The vector---7 ---7

AB X AC is perpendicularto the desiredplaneand is there-fore perpendicular to every vector in the desired plane (see)))


Figure 76). Hence P =(x, y, z) is a general point in the plane

\037 \037 \037

if and only if BP. AB X AC = o.Writing this condition explicitly gives

[(x - l)i + (y + l)j + zk] \302\267(-3j

- 3k) X (i -5j

- 7k) = o.In terms of determinants, this relation becomes)

x-I y+1o -31 -5)

z

-3 = O.,-7)

or, expandingby means of the formulas for cross product anddot product, we get -2x + y

- z + 3 = 0 as the equation

of the plane determined by A, B, and C.)

31. DISTANCE FROM A POINT TO A PLANE

The distance from a point to a planewas found by pro-

jecting on a unit perpendicularto the plane. But now

this procedure may be facilitated, for by means of cross

products, the unit perpendicular to a planemay be found

directly from three points of the plane without any need

for determining the equation of the plane.EXAMPLE 43. Let a plane be determined by the points

A = (-3,0,1), B = (-1,1, -1), and C = (2,1, i); and letP = (1,- 2,

- 3) . We shall determinethe distance d from

Po to the plane of A, B, and C. (The reader is referred toExample31,which is the present problem in disguise).)

\037 \037AB x AC)

A)

P(x,y,z))

B)

FIGURE 76)))


\037\037AB X AC)

i

---- D

d---:-\\

1;:\\\\\037)

B)

FIGURE 77)

A vector perpendicular to the given plane may be found by\037 \037

taking the cross product AB X AC (see Figure77, which

symbolizes the geometry)

\037 \037 i j kAB X AC = 2 1 0 = ji -

3j- 3k = 3(ti - j - k).

5 1 \037

Then a unit perpendicular is)

\037 \037

N=ABXAC = .!i-j-k =!(i-2.-2k)

lAB X ACI v i + 1+ 1 3 ]

(Compare with Example 31.))

\037 \037

d = IprN PoBI = IPoB. NI

=1(-2i + 3j + 2k) --l(i - 2j- 2k)1

=t(2 + 6 + 4)

= 4.)

32. DISTANCE BETWEEN TWO LINESWe now directattentiontooneofthenastiestproblems

ofelementary analytic geometry, namely, the problem of)))


finding the distance between two nonintersectinglinesof

space. We refer to this as a nasty problembecause,ifpure analytic geometry is the means of analysis, the

beginning student is usually plagued by the problem of

visualization that is required to \"see\"the derivation of

the fQrmula, and then he is plagued by another compli-

cated formula to memorize. Once again, the vector

approach assists with both difficulties.If two lines \302\2431and \302\2432are given, the minimum distance

d between them is the distancebetween \302\2431and \302\2432along

the mutual perpendicular (QR in Figure 78). Thus, if

we call A and B points of \302\2431and \302\2432,respectively,)

\037

d =IprQRABI.)

(Readers who have difficulty understanding this point

are encouraged to check back on the definition of pro-

jection.) The procedure for computing d is given in the

following example.

EXAMPLE 44. Let \302\2431= AC, where A = (2, -1, 3) and

C = (1, 0, -5); and let \302\2432= BD, where B = (1, 3, 0) andD = (3,-4, 1). We shall determine the distance between\302\2431and \302\2432.)

FIGURE 78)))


A vector that is simultaneously perpendicular to \302\2431and \302\2432is)

\037 \037 i j

N = AC X BD = -1 12 -7)

k

- 8 = 55i - 15j+ 5k.1)

Then)

---7d =

\\prlf ABI=)

NABo

1N !

_ I(-i + 4j - 3k). 5(11i- 3j +'k)1-5 v'121 + 9 + 1

1-11-.12 -3\\

26

v' 131=

v' 131.)

EXERCISES

1. Evaluate the following(a) i X (j + k),(b) (i + j) X k,(c) (i + j + k) X (i + j + k),(d) (2i - 3j+ k) X (2j

- k),(e) (-i + 2j+ 4k) X (i + 2j + 4k).

2. Usingthe fact that a parallelepiped can be sliced into sixtetrahedraof equal volume, determine a formula for the volume

of a tetrahedron, the four vertices of which are A = (aI, a2 aa),B =

(b l , b 2 , b a), C = (CI,C2, ca), and D = (dl, d2 , d a ).

3. Find the volume of a regular pyramid whosebaseis a squareof side a and whose height is h. (A pyramid is called regularwhen its base is a regularpolygon and when the altitude fromthe apexmeetsthe base in its center.)

4. Prove that the answerto Exercise3is unchanged if the sec-

ond condition of \"regular\" is omitted. That is, a change inthe positionof the apex does not alter the volume of the pyra-mid as long as the apexremains at a fixed height h above the

base.

5. Determine the equation of the line .\302\243through Po = (xo, Yo, zo)\037

and parallel to V = li + mj + nk by observing that PoP X V= 0, where P = (x,y, z) is any point of cC.)))


6. (a) If P = (x,y, z) is an arbitrary point of plane ABC,\037 ---? \037

justify the equation AP. AB X AC = 0 as being a vectorequation of the plane. (See Example 42.)

(b) Usingthe method of part (a), q.erive the equation of the

plane determined by A = (1, 0, 0), B = (0, 1,0), and C =

(0, 0, \037).

7. Using the methods of this chapter, determine a vectorformula for the distance from the origin to planeABC.8. Let U and V be parallel to plane III, while Wand Yare

parallel to plane II2 . Under the further assumption thatIII 1. 112 , prove (U X V) \302\267

(W X Y) = o.

9. Find the distancebetween)

(a) x = 1 -2t, Y

= t, z = -tand x =

t, Y= 1 -

2t, z = - t.x-I Y

(b)= - = z + 1

2 2x y+2

and - = = z.2 -1)

(c) line AB and line CD, where A =(0, 1, 2), B = (1, -1,

4), C = (0, 2, 0), and D ;= (-1, 2, 2).

10. If C is perpendicularto the plane of A and B, prove that

(A X B) \302\267(C X D) = O.

11. Find a set of direction numbers for the line of intersectionof the planes whose equations are alX + bly + CIZ + dl = 0and a2X + b 2y + C2Z + d 2 = 0, by a direct application of

cross products.)

33. TRIPLE CROSS PRODUCTSFor the applicationofvector algebra to more complex

problems of geometry and, particularly,to thedevelop-ment of trigonometry-as the reader will see in the nextchapter-we shall find \"it useful to expand cross productsof three and morevectors.)))


We begin by obtaining an expansionfor the triple

vector product A X (B X C). Unfortunately, there is

no simple and well-motivated development for su.chaformula. A summary of the efforts of several mathe-maticianswho have contributed to various simplifica-tions of the problem appears in a short paper, \"On theVectorTripleProduct,\" by Murrey S. Klamkin (Ameri-can Mathematical Monthly, December 1954). We adoptan approach, the virtue of which is simplicity.

If we are given the threevectors A, B, and C, we con-sider the three emanating from a point 0, which is takento bethe originofa rectangular coordinate system. We

impose the coordinate system in the following way (see

Figure 79) :)

(i) the x-axis is taken along B;(ii) the y-axis is taken in the plane of Band C;(iii) the z-axis is taken so that the xyz-coordinate

system,with {i, j, k} as a basis, is right-handed.

It follows from (i) that B = b1i, and from (ii) that

C = c1i + C2j. Since A is free from any stipulation, it)

C = cli + C\037)

FIGURE 79)))

CROSS PRODUCTS 149

must be written A = ali + a2j + a3k. Then). .1 J

B X C = b l 0)

ko =

blC2k,

CI C2 0)

and)

i j k)

A X (B X C) = al Q.2

o 0)

Q.:\037= a2blC2i - alb1c2j. (89)

b 1c2)

We observe that A X (B X C) must be perpendicular \037o

B X C, that is, A X (B X C) is perpendicular to thecoordinatevector k. Thus the triple cross product,A X (B X C), must be in the plane.of Band C. Hence,it must be a linearcombination of Band C. We there-fore simplify (89) by factoring the right member so thatit is exhibitedas a linear combination of Band C. Thus)

A X (B X C) = (alcl + a2c2)b l i - a lb1(cli + C2j).)

Finally, \\ve have)

A X (B X C) = (A.C)B- (A \302\267

B)C, (90))

an expansion of the triple cross product in elementary

tern1S.

By using Theorem 8 and (90), the reader ca11 easily

sho,v that the triple cross product (A X B) X C has theexpanSIon)

(A X B) X C = (A. C)B - (B\302\267C)A.

EXAMPLE 45. Given A = 2i - 3j + k, B = i - j, and

C = -4i + k, ,ve computeA X (B X C) and (A X B) X C.A X (B X C) = (A. C)B - (A.B)C

= (-8- 3)(i -

j)- (2 + 3)(-4i + k)

= - IIi+ IIj + 20i - 5k= 9i+ lIj - 5k

(A X B) X C = (A. C)B -(B

\302\267C)A

= -IIi + IIj - (-4)(2i - 3j+ k)

= - 3i - j + 4k.)))


EXERCISES1. Verify that (A X B) X C = (A. C)B

- (B \302\267C)A.

2. If A, B, and C are given as in Example 45, compute(B X C) X A and B X (C X A).

3. Provethe identities:)

(A X B) X (C X D) =(A X B \302\267

D)C- (A X B \302\267

C)D (91))

and)

(A X B) \302\267(C X D) = (A. C)(B \302\267

D)- (A. D)(B \302\267

C) (92)

4. By making use of (91), show that (A X B) X (A X C) =

(A \302\267B X C)A.

5. Prove Lagrange's identity:

(a2 b s- aSb2)2 + (aSb l - a1bs)2 + (a1b 2

- a2b1)2= (a12

+ a22 + as 2

)(b 12 + b2

2+ bS

2) - (a1b1+ a2 b 2 + a sb s )2.)

(Hint. Use (92), which is, indeed,sometimesreferred to as

the Generalized Identity of Lagrange.)6. If A, B, C, and D are coplanar\" prove that)

(A X B) X (C X D) = o.)))

trigonometry)

This short chapter is devoted to illustrating the

application of vector notions to the development ofstandardformulas of plane and spherical trigonometry.We shall see that spherical trigonometry, particularly,

admits to a simple analysis in termsof vectors.)

34. PLANE TRIGONOMETRY

Law of cosines. Consider the triangleof Figure 80,

where IAI= a, IBI = b, and lei = c. It isclear that

C = B -A,)

.

gIves)

e \302\267e = (B -A)

\302\267(B

- A)

c2 = a 2+ b2 - 2A \302\267B.)

and that)

When we expand the last dotproduct,we get the familiar

law of cosines)c2 = a 2

+ b2 - 2abcosI') (93))

Law of sines. Here we seek a relation involving the

151)))


FIGURE 80)

sides of the triangle of Figure 80, and the sinesof itsangles. We therefore employ the cross product)

c X C = C X (B- A),)

which implies)o = C X B - C X A,)

or)

C X A = C X B.) (94))

Equating the magnitudes of left and right membersof(94) yields)

ca sin {3 = cb sin a,

whichis equivalent to the law of sines

a b)

sin a sin {3

By repeated application, or simply by the symmetry of

the variables, we have the completerelationship)

c) b) c)

(95))-. 'SIn 'Y)

.sIn a) sin \037)))

TRIGONOMETRY) 153)

which states that the sides of a triangleare proportional

to the sines of their respective oppositeangles.Sum and difference formulas. The usual treatment of

sum and difference formulas in high school textbooksrequires somewhatmessy geometric arguments that

are fU:fther complicated by the need to considersev-

eral cases, depending on the quadrants in which theangles lie. Once again matters are greatly simplifiedby vectors, by means of which all cases are treatedsimultaneously.

Let Q and R be points on the unit circle (i.e., on a circleof radius one centeredat the origin) as shown in Figure

81, so that)

Then,)

Q= cos {3i + sin {3j

R = cos ai + sin aj.

Q \302\267R =IQIIRI cos (a -

(3).)

:y)

R)

x)

FIGURE 81)))


Hence)

cos (a -(3)

= COS a COS{3 + sin a sin {3. (96)

Once again, referringto Figure 81, we may write)

Q X R = IQ11RIsin (a .- (3)k= sin (a -

(3)k.)

In terms of coordinates this cross product becomesi j k

cos {3 sin {3 0 = (sin a cos {3- sin {3 cos a)k.

cos a sin a 0)

Thus. sin (a- (3)k=

(sin a cos {3- sin {3 cos a)k,

which implies.)

sin (a- (3) = sin a cos{3

- sin {3 cos a. (97)

EXERCISES1. Using (96) and (97), deduce formulas for cos (a + (3),

sin (a + (3), cos 2a, sin2a.2. By using vector methods, derive direc.tly the sameformulas

as requested in Exercise 1.

3. Observewhat happens if {3- a is chosenfor the expansions

(96) and (97). Is the result consistentwith the facts cos 8 =cos (- 0) and sin 8 = - sin (

- 8) for all angles8?

Area. The familiar formula for the area K of thetriangle of Figure 80 in terms of two sides and theincluded angle is an immediateconsequence of cross

products, for

K =\037\\A X B\\

=\037\\A\\ \\Bl sin \"I.)

Hence) K = lab sin 'Y.)

Since a triangle is completely determined by its threesides,itsarea must therefore be completely expressible interms of the sides. We shall now apply vector productsto the determinationof such a formula.)))

TRIGONOMETRY) 155)

Consider once more the triangle of Figure 80. Itsarea K may be written K = ilA X BI, which implies

2K =1A X BI.

Using the scalar product to expressthe length of A X B,we have)

4K 2 =IA X B\\2

= (A X B) \302\267(A X B)

= (A. A)(B \302\267B)

- (A. B)2 (by (92))= a2

b2

-. (A . B)2=

(ab- A \302\267

B)(ab + A \302\267B). (98))

But, in the derivation\037f

the law of cosines, we found that

a2+ b

2 - c2

A.B = ,2)

and substituting ill (98) gives)

(

a2 + b2 - c2

) (

a2 + b2 - c 2

)4K 2 = ab - ab + \302\267

2 2)

Simplifying, we get

4K 2 =\037(2ab

- a 2 - b2

+ c 2)(2ab + a 2

+ b2 - c2

)

=\037( -a + b + c) (a - b + c) (a + b - c)(a + b + c).)

Calling t = a + b + c, we make the substitutionst - 2a = - a + b + c, t - 2b = a - b + c, and

t- 2c = a + b

- c.)

Thus) 4K 2 =\037(t

- 2a)(t - 2b)(t -2c)t,) (99))

and we observe that a greater simplificationwould have

resulted if we were clever enough to introducea varia-

ble half that of t. Because, if 28 = t, that is,

8 = i(a + b + c),)

then)

4K 2 =\037(28

- 2a)(28 - 2b)(28-2c)28.)))


A)

FIGURE 82)

Hence) K 2 = S(S - a)(s -b)(s

- c),)

and we arrive at Hero'sFormul a for the area of a triangle

K = v s(s-a)(s

- b)(s - c),wheres = j-(a+ b + c).

A theorem in Euclidean geometry states that the

area of a rhombus is equal to one-half the product of its

diagonals. Before giving a simpleproofofthis theorem,

we first prove a preliminary result: The diagonalsof arhombus are perpendicular.

Let the sides of the rhombusbe designatedby A and

B, as shown in Figure 82. Sinceall sidesofa rhombus

are equal, we haveIAI

=IBI. To determine the per-

pendicularityof the diagonals,we take the inner product

(A + B) \302\267(A

- B) =IAI2

-IBI2

= 0,

which showsthe diagonal A + B to be perpendicular tothe diagonalA-B.

Now, the area K of the rhombus is.given by

K = 21i-(A + B) X leA - B)I,the quantity inside the absolute value signs being twice)))

TRIGONOMETRY 157

the area of the shaded regionin Figure82. Then

K = i-ICA + B) X (A-

B)( = i-IA + BIIA -B( sin

1r.

2

Hence K =\037IA + B(IA

- B(,which is the desired result.)

35. SPHERICAL TRIGONOMETRYWe wish to derive the basic relations involving the

sides.and angles of a spherical triangle. We thereforeconsiderA, B, and C as points on a unit spherewhose

center is O. Call a, {3,and 'Y the great circle arcs 2 that

form the sides of the spherical triangleABC;a being

opposite A, {3 being opposite B, and \"I opposite C. Since

the sphere is of unit radius, the arcsa, {3, and \"I are also

the radian measures of the central anglesformed by

Band C, A an C, and A and B, respectively. We further

stipulate that a, {3, and 'Y be less than 1r.

Amassing all the given information in terms of thevectors emanating from the center 0 of the sphere, wehave A \302\267B = cos \"I, B \302\267C = cos a, and A \302\267C = cos {3.

As we saw earlier, the angle between two planes is

determined most easily by finding the angle between

perpendiculars to the planes. Thus the interiorangle A

between plalle OAB and plane OAC is the same as the

allgle between A X B and A X C. Since by equation 92

(A X B)\302\267

(A X C) = B \302\267C - (A. B) (A\302\267

C) (100)

sin \"I sin {3 cos A = cos a -cOS'Y cos {3, (101))

2 A great circle on a sphere is the intersection of the sphere witha plane through the center of the sphere. Meridian circleson the

globe are all great circles, but parallelsof latitude (other than the

equator) are not great circles. Great circles are the so-calledshortest distance paths on the surface of a sphere. Thus, if A

and B are two points of a sphere, the shortest path, lying on the

sphere that joins A to B is along the arc of the unique (unless Aand B are antipodal, or diametrically opposite, points) greatcircle that passes through A and B. Finally, we see that thesides of a sphericaltriangle are great circle arcs.)))


FIGURE 83)

which is often termed the law of cosines for spherical

triangles.

EXERCISEThe perpendicularvectors to two planes may intersect in anangle equal or supplementary to the angle between the givenplanes. Verify, by considering the direction of the perpendicu-lars, that the angle determined by the scalar product (100)of

A X B and A X C does, in fact, yield angle A and not thesupplementof A.

We shall now derive the law of sines for spherical

triangles by first considering the expansion)

(A X B) X (A X C) =(A X B \302\267

C)A. (102)

(See equation 91 or Exercise 4 of Section33.)Taking the magnitudes of both members of (102), we

get)

IA X BIIA X ci sin A =IA X BI.lcos 01,)))

TRIGONOMETRY 159

where 0 is the angle betweenA X Band C. Thus, wehave)

IA X cl sin A = Icos OJ,)

which is equivalent to

sin ,8 sin A =I cos 01.) (103))

Now, evaluating the magnitudes of both membersofthe expression)

(B X A) X (B X C) = -(A X B \302\267

C)B) (see (91)))

yields

IB X A((B X ci sinB =IA X BllCllcos of,)

which is equivalent to)

IB X cl sinB = Icos81.)

Simplifying, and rewriting in terms of the anglesof thetriangle, we have

sin a sin B = Icos01 (104)

From (103) and (104) we deduce)

sin a sinB = sin,8sinA,)

which is the law of sines for sphericaltriangles. Again,

by an interchange of letters, we may write the completelaw of sines in the more usual form)

.SIn a

sin A)

sin ,8-sin B)

.SIn 'Y-sin C)))

more geometry)

We devote this chapter to a potpourri of geometricconsiderations.)

36.LOCI DEFINED BY INEQUALITIESIn our earlier work with lines,planes,and spheres, we

discussed loci that were defined by algebraic (and vector)

equations. It has no doubt occurredto the reader that

relations other than equality may be utilizedto specify

the conditions defining a locus. For example, if P is the

position vector of a point in three-dimensionalspace,then Ipl

< r defines a locus that consistsof a sphere of

radius r and all points interior to the sphere,whereas

I pi> r defines the locus that consistsofthe sphere and all

points exterior to it. Analytically, these two relations

would be expressed

x 2 + y2 + Z2 < r and x2+ y2 + Z2 > r,

respectively. Further examples of loci sodefinedfollow.EXAMPLE 46. yv e sketch the locus defined by

-2x - 2 < y < x + 1.160)))

MORE GEOMETRY 161

y.\\\\\\ \037\\

I

\302\2431\\

.'

Iv

1

I

\037

i\037\\II

r-1 0 1

,- x)

(a)

y\302\2432

y

2 2

1

1

x1 1

x0 0

(b) (c)

FIGURE84)

We consider, first, that y < x + 1. The equation y= x + 1

represents the familiar straight line, which is sketched inFigure 84a). Sincey is larger than or equal to x + 1, the

desired locus consists of points above the line as well as on the

line itself (see Figure84a).The locus - 2x + 2 < y, by similar reasoning, consists of

the points below (but not including) the line y = -2x + 2.)))


(See Figure 84b. We have used a broken line to indicatethat\302\2432is not included in the locus.)

The originalproblem imposes the two inequalities simul-

taneously, so the final solution may be found by combiningthe two sketches to determine the region of the plane thatsatisfies both inequalities simultaneously. This region isshadedin Figure84c. The desired locus includes the darkenedlower left portion of line \302\2431but does not include any point of\302\2432. The latter is indicated by the fact that the \302\2432-portion of

the boundary is drawn as a brokenline.)

EXERCISES

1. Sketch the regions defined by the relations(a) x > 0 and y < o.

(b) x < y < 1.(c) -2 < x < y.

(d) 4x - 1 < 2y < x + 2.

2. What is the shape of the region 1 <Ipl

< 2 if

(a) the vector P is a planepositionvector?(b) the vector P is a position vector in three dimensions?

3. Let0 be the origin of a rectangular coordinate system and

\037

P, Q any points in three dimensions such that lopi = 2 and

-4

IPQI = 1. What is the locus of Q?

4. Expressanalytically the fact that the points (x, y)

(a) lie inside an annular region whoseinner radius is 2 and

outer radius is 3 (take the centerat (0,0));(b) lie inside triangle ABC, where A = (1, 0), B = (1, 2),

and C = (0, 1).

5. Confining attention to the x-axis, give a geometric inter-

pretation of the points (x,0) such that

(a) x - a = b;(b) x - a < b. (Assume b > 0 in both parts.)

6. What is the locus of points (x, y) suchthat(a) x < I?(b) x = y?

(c) x < y?(d) 2xl < I y

-II?)))

MORE GEOMETRY) 163)

(For assistance with Exercises 5 and 6, the reader is referredto Section37 c.))

37. A FEW BOOBY TRAPS(a) How shallwe sketch the plane locus defined by

the equation)x + y

= x + y?)

At first glance it appearstobea straight line because the

appearance of the equation is linear. But lookonce

again! By grouping like terms, that is, by subtracting

x + y from both members, we get)

0=0.) (105))

If you suggest that there isnolocus,you are wrong! Let

us consider the problem in the lightof primary question:

What ordered pairs (x, y) satisfy the relation? Clearly,any ordered pair may be substituted in (105), and thestatementremains true. It is an identity, true for everychoice of values for the variables. Consequently, thegraph consistsof the entire xy-plane.

(b) What is the locus of the parametric eqllations)

x = cos2

t

Y= sin 2

t,)

in which t ranges over all the real numbers?Adding the two equations yields x + y = 1, which is

the straight line in Figure 85a. But this cannotbe thelocus,for there is no value of t that yields x = 2 orx = - 1! The slip-up occurred when we blithelyignored the fact that cos2 t alld sin 2

t are non-negative.

The locus (see Figure 85b) is, therefore,the segment of

x + y = 1, with the further restrictionsx > 0 and

y > 0, for the given parametricequationsplace such

restriction on x and y.)))


y) y)

1)

x)o) 1)

%)

(a)) (b))

FIGURE 85)

(c) What is the graph of the relation

Ix-

It < 2?)

In order to gain geometric insight into this relation,we begin with an algebraicanalysis. For which x is

it true that Ix - 11= 2? The reader may approach this

question in several ways:(i) Usingthe definition of absolute value, he reasons

that x-I = 2 orx-I = -2, which results in x = 3or x = -1.

(ii) Or, he may be aware of the fact thatlal

=Ibl if

and only if a 2 = b2

, in which case he reduces the problemto thesolutionof the polynomial equation

(x - 1)2=4,)

which again yields x = 3 or x = -1. In either case we

become aware of the fact thatIx

-11

...:..-2 states that thepoint x (on the axis) is at a distance of two units frompoint1. In general, the quantity Ix

-al symbolizes

the distance between x and a. Thus it is reasoned that)))

MORE GEOMETRY) 165)

2A.)

2A.) X

\\

L-3)

xr

-' -1)I

o)

,rI

1)

I

2)

FIGURE 86)

the locusI x-II < 2 is the segment betweenx = - 1

and x = 3, including the endpoints. The bobby trapariseswhen the reader overlooks the question, \"What isthe dimensionin which the problem has been stated?\"

Hence if our discussion is of the plane, we seek thesetofall (x, y) such that Ix

- 11< 2. The answer now

is that -1 < x < 3,1and y may be any real number,which impliesa locusconsistingnotonly of the segment

on the x axis but alsoof the entirestrip between and

including the lines x = -1 and x = 3 (seeFigure 87).)

)')

I ,

\"':I Y\037\037/,

\037\037\037/

I ....

-2 -1 3\".

\037I'//, i%:'l)

x)

FIGURE 87

1 In terms of set theory, we seek {(x, y) Ilx - 11< 2},which is

the sanle set as {(x,y)ll < x < 3}.)))


(d) What is the graph of)

(x + y) (x -y)

= O?) (106))

Again we resort to the fundamental question:What

ordered pairs (x, y) render the given relation true? We

note that (106) is true if and only if)

x+y=O) or) x-y=O) (107))

that is, if and only if y = -x or y= x. Thus, the locus

satisfying the given relation is a pair of lines through theorigin. If point(x, y) is on either one of these lines,therelation is satisfied. (Observe that this is quite differentfrom b\037th relations being satisfied simultaneously; thenwe would have had an \"and\" between the equations in

(107) and the locuswould be the point of intersection ofthe two lines in the figure.))

y)

x)

FIGURE 88)))

MORE GEOMETRY) 167)

The language of set theory helpsto clarify matters of

this sort. Since the locus is definedwith an \"or,\" we

must have the union of two sets. Therefore the locus

may be written as the set)

s= {(x,y)ly=x}U{(x,y)ly= -x}.)

This description of the locus illustrates the simplicityand clarity that often resultfrom an elementary applica-

tion of the ideas of set theory.)

38. SEGMENTS AND CONVEXITYTheorem 4 providedus with a necessary and sufficient

condition for a vector to have its endpoint on a line. Wenow turn to confining the vector endpoint to a specificsegment.

Let A and B be two points and C somepointof thesegment AB. This situation can be described by stating-4 -4thatBC=

tBA, where 0 < t < 1.In the languageof positionvectors relative to some

fixed point 0, we write)

C = B + teA - B),C = tA + (1

- t)B.)o < t < 1.)

(108))

Thus the segment AB consistsof all points that can be rep-resented by position vectors of the form tA + (1

- t)B,where 0 < t < 1. The endpoints occur when t = 0and t = 1. That is,

segment AB = {pIp = tA + (1- t)B}.)

Definition. A set is called convex if and only if it con-

tains the entire line segment that joinsany. two points

of the set. That is, a setS issaidto be convex if, given

any two points A and B of S, the segment AB is entirelycontained in \302\243.)))


EXAMPLE 47. We prove that the points of a sphereand its

interior form a convex set.Let the sphereof radius r be centered at the origin, and let

A and B be any two points of the set. Then,)

IAI< r) and) IBI < r.)

We must show that any point of the segment AB is also adistanceof less than or equal to r from the origin. The seriesof inequalities)

ItA + (1 -t)BI

< ItAr + 1(1 -t)BI

=tlAI + (1 -

t)IBI< tr + (1 - t)r = r)

establishes the desired result.)

EXAMPLE 48. We continue the consequences of our abilityto write segmentsin vector languageby finding an algebraiccondition that a point lie interior to, or on the boundary of, a

triangle.Let A, B, and C be three distinct points not on one line.

We assume, for the purposes of illustration, that a triangle (seeFigure89) consists of the points of all the segmentsCD,where D may be any point of segment AB. Again, using)

c)

A)

B)

FIGURE 89)))

MORE GEOMETRY) 169)

position vectors to represent points under consideration, V\\Te

represent any point D of segment AB by the equation)

D = tA + (1 -t)B,) where 0 < t < 1.)

Thus, if X is any point of the triangle, there is someD so thatX is on the segmentCD, and we may write)

x = mC + (1 -m)D,) where 0 < m < 1.)

Then)x = mC + (1

-m)[tA + (1 - t)B]

= (1 -m)tA + (1

- m)(1 -t)B + mC.

Noting that (1 -m)t + (1

- m)(1 -t) + m = 1, we draw

theConclusion. If X is ariy point of the interior or boundary

of triangle ABC, there are scalars r, s, and t so that X = rA +

{

r + s + t = 1,. . 0 < r < 1,sB + tC, In whIch

0 < s < 1,o < t < 1.)

EXERCISES

1. Justify each step of the series of inequalities in Example 47.

2. Graph the loci

(a) x 2 ---y2

= 0

(b) (x2 + y2

- 1)(x2 + y2

- 2) = 0(c) (x + 2y + 1)( -x + y + 2) = 0

3. Prove that all spheres-not only those centered at the origin-are convex.)

4. Representthe points of a tetrahedron and of a squarein amanner similar to the one exhibited in Example 48.5. What condition on the coefficients of A, B, and C would

guarantee that a point would be on the boundary of triangleABC?6. Stateand prove the converse of the conclusion of Example

48.

7. One might expect that the relation betweenX, A, B, and Cof Example 48 would depend somehow on the location of theauxiliary point (the origin of the position vectors). However,)))


the result of the example indicates otherwise. Can you explainwhy the relation between the four vectors is independentof

the auxiliary point? (Hint. Consider the meaning of r + 8 +t = 1.)

8. What modification, if any, shouldbe made to the result of

Example 48 when the position vectors all emanate from A?

That is, when A = o.

9. Let triangleABC be defined by its vertices as follows:A = (1, 0), B = (1, 1), and C = (0, 1)

(a) Find the median point M (point of intersection of the

medians), and write its position vector in the formM = rA + 8B + tC.

(b) Let r = j-, \"8 =!, and t = t. Check the result of

Example 46 by locating the point associatedwith these values

of the coefficients.(c) Find the point of i\037tersection of the altitudes of ABC,

and determinewhether it is inside, outside, or on the boundaryof the triangle.

10. Let tetrahedron ABCD be defined by its vertices as follows:A = (0, 0, 0), B = (1,0,0), C = (0,1,0) and

D = (0, 0, 1).(a) Write an expression, stipulating necessary conditions,

for point X to be on the faceABC of the tetrahedron.

(b) Do likewise for X to be on the face BCD.(c) How would you guarantee that X be inside(not on the

boundary) of the given tetrahedron?)

39. LINEAR PROGRAMMINGConsiderthe following problem.

2

A hospital is concerned with minimizing the costofitsmeat (beef and pork) diet. The average hospital diet

2 This problem was suggested by a Jack Spratt problem in

\"Linear Programming Problems for First-Year Algebra\" by

Donovan Lichtenberg and Marilyn Zweng, published in The

Mathematics Teacher, March 1960. This fine article is recom-mended to the readerwho is interested in the details of a successfulexperiencein the teaching of linear programming to high school

students.)))

MORE GEOMET:ay) 171)

requires 2 pounds of lean meat and 1.5 pounds of fatmeat per personper week. The beef, which costs $1.00

pe\037 pound is 0.2 fat and 0.8 lean. The pork,which costs

$.75 per pound is 0.6 fat and 0.4 lean. If the hospitalhas200 patients on this diet, and if it cannot purchasemore than 900 pounds of meat per week because ofrefrigerator space,find out how many pounds of beefand how many pounds of pork should be purchased sothat the cost is at a minimum.

This problem is typical of the elementary problemsoflinear programming. The word linear is used becausethe relations involving the variables will be linear rela-tions as we shall soon see. The word programmingstems from the fact that we are trying to determine aprogramfor optimum operations. Here the hospitalseeks a program(for the purchasing of meats) which is

optimum in the sense that it would satisfy the needs of

the hospital while minimizingthe cost.Let'sextract the pertinent data from the hospital

problem. LetB = poundsof beef to purchase each week, andP = poundsof porkto purchase each week.

I t is clear that)

B > O) and) P > o.) (a))

The restriction of refrigerator space tells us that)

B + P < 900.) (b))

The rest of the given data can be summarized in the

table)

Fat Lean CostjIb

Beef 0.2 0.8 $1.00

,.,'\"'. Pork 0.6 0.4 0.75)))


Since a total.of 200 \302\2672 = 400 pounds of lean meat isrequired,we see that)

0.8B + JP > 400;) (c))

and since a total of 200 \302\2671.5 = 300 pounds of fat meatis required,we have)

0.2B + 0.6P > 300.) (d))

Finally, we write an expression for the cost C(indollars)of the purchase)

.C(B, P) = B + :P .) (e))

In accordance with the earlier prediction, all the rela-tions involving Band P are linear relations. Theinequalities (a)

- (d) place restrictions on Band P.Our problem, then, is to select the ordered pair (B,P)that yields a minimum value for C(B, P), provided(a)-(d) hold. Thus we turn our attention to finding,graphically, the set R of ordered pairs (B, P) that satisfythe given restrictions. This set is)

R = {(B, p)IB > 0,P > 0, B + P < 900,2B + P > 1000,B + 3P > 1500}.)

The set R is the darkest region shown in Figure 90. Wehave thus narrowedthe choiceof orderedpairsto thoseof the triangle R, but we still have an infinite number of

choices-certainly too many for a trial and error pro-cedure unless we were willing to settle for an approximateanswer. Fortunately, however, there is a theorem toassist us in narrowing the search considerably.

Theorem 12. Let f be a linear function whose domain

of definition is a convex polygon. The minimum value

of f(x, y) is attained at a vertex of the polygon. (The)))

MORE GEOMETRY

P)

173)

1000)

500)

1000) 1500)B)

(0,0))

FIGURE 90)

word \"maximum\" may be substitutedfor \"minimum,\"

without disturbing the validity of the theorem.)Returning to the original problem, we determine the

vertices of R as: S = (100,800), T = (600,300), andU = (300,400).Applying Theorem 12, which has

narrowed our search to the pointsS, T, and U, we

evaluate)

C(100,800) = 1 \302\267100 +

3\302\267

800 = $7004

C(600,300) = 1 \302\267600 +

\037

\302\267300 = $825

3C(300,400) = 1 \302\267300 +

4\302\267

400 = $600,)

giving the conclusion that 300 pounds of beef and 400

pounds of pork make up the optimum purchasingpro-'.gram for the hospital.)))


Proof of Theorem 12. Let the value of the linear 3

function at P = (x, y) be given by)

f(x, y) = ax + by + c.)

We first consider the values of f(x, y) when P is con-

.fined to the segment P1P2 , where PI = (Xl, YI) andP 2 (X2, Y2). We show that, under this restriction, themaximum and minimum values of f(x, y) are attained atthe end-pointsof the segment.

In Section 38 we leari1edthat if P = (x, y) belongs tothe segmentPIP2 , then)

X =lXI + (1

-l)X2

= X2 + l(XI -X2),

y= lYl + (1 -

l)Y2= Y2 + l(Yl -

Y2),

in which 0 < l < 1. Then the value of f at P is

j(x, y) = a[x2+ l(XI- X2)] + b[Y2 + l(Yl

-Y2)] + c

= aX2 + bY2 + l(axi.-

aX2 + bYl-

bY2) + c= (ax2 + bY2 + c) + l[(axi + bYI + c)

- (ax2 + bY2 + c)]= f(X2, Y2) + l[f(XI YI) -

f(X2, Y2)].

Hence f(x, y) ranges between f(Xl, YI) and f(X2, Y2) as

o < l < 1. The extreme (maximum and minimum)

values of the function are attainedat the endpointsof

the segment.

Query. What happens if f(Xl, YI) =f(X2, Y2)? Ex-

amine the proof to see if your answer is substantianted.

The remainder of the proof for Theorem 12 is left asan exercisefor the reader, with the hint that the defini-tionsof polygon and convexity are essential.

3Strictly speaking, this function is linear.if and only if c = o.

Since the term \"linear\" is often used in a loosemanner to include

expressions of the form ax + by + c, when c F- 0, we have chosento exhibit a proof of Theorem 12 for such a loose interpretation

of \"linear.\)

MORE GEOMETRY) 175)

ADDITIONAL EXERCISEMake the following replacements in the statement of Theorem

12:)

\"polyhedron\" for \"polygon,\" and

\"f(x, y, z)\" for\" f(x, y).\

Provethe resultingstatement.)

EXAMPLE 49. A moving van company charges25 centsperpound for moving furniture from N ew York to San Franciscoand 15 cents per pound for moving crates on the samecross-country trip. If at least one quarter of each load is furnitureand at least one quarter of each load consists of crates, find

the minimum and maximum cost per pound for a load.

Let F = that fraction of a load that is furniture,C = that fraction of a load that consists of crates.)

Then)1 3-<F<-,4- -4

1 3-<C<-.4- -4)and)

Furthermore, since a load consists entirely of furniture and

crates, we have the additionalrelation)

F + C = 1.)

These three relationsdefine a segment, as shown in Figure 91.Our problemreducesto a consideration of cost on a one-dimen-sional set, a segmentwhose endpoints are (-1-, : ) and (i, \037).

The cost K(F, C) is given by the relation K(F, C) = .25F+.15C.

(1 3

) (

1

) (3)

0.70K4

'4

= 0.254 + 0.15

4

=4

= 0.175

(3 1

) (3

) (

1

)0.90

K4

'4

= 0.254 + 0.15

4

=4

\037 0.225.)

Hence, the minimum cost for a load would be 17 \037cents per

pound, while the maximum would be 22! cents per pound.)))


c)

1)

(i,t))34\

1'2)

14\"

(f,t)

(0,0) 1 1 3 1F

4 '2 4)

FIGURE 91)

EXAMPLE 50. A sporting goods company makes baseballbats and softball bats, which are processed first by a lathethat cuts the wood to size and then by a finishing machine that

paints, dries, polishes, and labels the bats. The lathe oper-

ates 6 minutes to produce a baseball bat and 3 minutes to

produce a softball bat. The finishing machine operates for3 minutes on a bat, no matter which kind. However, becauseof loading and maintenance problems,the lathe can operateonly 4 hours per day and the finishing machine can operateonly 3 hours per day. If the profit on a baseball bat is $1.00and the profit on a softball bat 75 cents, and if the companycan sell all the bats it makes, find how many of each kind itshould producedaily in order to realize maximum profit.

Although there seems to be a mass of detail, the problem canbe summarized easilyin the following manner. Let

b = the number of baseballbats produceddaily,

8 = the number of softball bats produceddaily.

The restriction (4 \302\26760 minutes per day) on the lathe operationglves)

6b + 38 < 240 or 2b + 8 < 80.) (a))))

MORE GEOMETRY) 177)

The restriction (3. 60 minutes per day) on the finishingmachine gives)

3b + 38 < 180

We know also that)

or) b + 8 < 60.) (b))

b > o.

8 > o.

The profit PCb, 8) per day would be

PCb, 8)= b + :8 .)

(c)

(d))

(e))

Thus all the data are summarized in terms of inequalities

(a)-(d) and the relation (e). We seek to maximize PCb, 8) onthe convex polygon det\037rmined by the restrictions (a)-(d).The polygon is exhibitedin Figure92. The vertices of the

polygon are

o = (0,0), A = (40, 0), B = (20,40), and C = (0, 60).

Evaluating the profit at these four points shows that amaximum would be achieved by the production of softball

bats only.)

s)

80)

40 . B)

0(0,0) 20) 60) 80)b)

20)

FIGURE 92)))


EXAMPLE 51. A chocolate manufacturing concern makesmilk chocolate, semisweetchocolate,and bitter chocolate.

It can produce 500 poundsper day, but demand for the varioustypes is such that the maximum amounts that can be sold areas follows)

milk-400 pounds per daysemisweet-300 pounds per day

bitter-200 pounds per day.

If the profit is 80 cents per pound for bitter, 75centsper pound

for semi-sweet, and 60 cents per pound for milk chocolate,determine the program for productionthat would maximize

the profit.

Although this problem couldbe handled by only two vari-

ables, we shall treat it as a three-dimensionalproblem for pur-

poses of illustration. Let

m = pounds of milk chocolate produced per day

8 = pounds of semisweet chocolate producedper day

b = pounds of bitter chocolate producedper day.

Then)

m+8+b < 500

0 < m < 400

o < 8 < 300

o < b < 200.)

(a)

(b)

(c)

(d))

The profit function P is defined by the relationP(m,8, b)=

\037m + : 8 + \037b (in dollars).The graph of restrictions (a)-(d) is shown in Figure 93.

An algebraic solution for the vertices of the polyhedron shows

A = (400,0,100), B = (400,0,0), C = (400,100,0),D = (200,300,0), E =

(0, 300, 200), F = (0, 0, 200),and G = (300, 0, 200).

We may concludethat the concernwould achieve a maximum

profit by producing only two types or one type of chocolate.The latter alternative is excluded because producing less than500 poundswould always result in a loss. We may thereforeconcludethat two kinds of chocolate should be manufactured.So the searchis confined to an investigation of the profit at)))

MORE GEOMETRY) 179)

b)

E)

t I

400 500):> 8)

m)

FIGURE 93)

points A, C, D, E, andG-the points of a plane represented bym + 8 + b = 500.

The computation of profit at these vertices is left as an exer-cisefor the reader.)

EXERCISES

1. (a) Carry out Example51asa two variable problem.

(b) Carry out Example 51 with the restriction that themaximum amount of bitter chocolate that can be sold is 100pounds per day.

2. Find the minimum and maximum values of I(x, y) = 2x +4y

- 1 on the set defined by the inequalities

- 2x + 3y < 6

x+y < 2

x + 2y > 3.)))


3. The delivery trucks owned by an oil company have acapacityof 1000 gallons. On each trip a truck must carry atleast 400 gallons of regular gasoline, at least 200 gallons of

high test gasoline, and, at most, 300 gallonsof white gasoline.

If the profit per gallon is 3 cents for regular, 4 cents for hightest, and 5 cents for white gas, find the program for loadingthe trucks that would yield a maximum profit. (This can bedonewith only two variables.)

4. Assume the trucks in Exercise3 areall loaded to capacity.What kind of loadingwould yield the least profit?

5. A mountain climbing party wishes to purchase A-rationsand B-rations for an expedition. The food values and costsper unit are classified as follows:)

A B

Units of carbohydrate 1 3Units of protein 3 4

Units of fat 3 1Cost $2 $1)

The minimum requirements are 10 carbohydrate units, 18protein units, and 6 fat units. Find the minimum cost diet

satisfying the requirements of the mountain climbers.

(Observe that the convex region is not a closed polygon

in this case. How do you know that you have the minimum?))

40. THEOREMS ARISING IN MOREGENERAL GEOMETRIES

The study of projective geometry deals with someofthe same entities discussed in Euclidean geometry,namely points, li11es,and planes. However, projective

geometry discards the notions of distance and anglemeasurements, and focuses attention on incidence prop-erties: points lying on \037ines, lines passing through points,intersections of various sorts, and so forth. For ele-

mentary discussions of projectivegeometry, the reader is

referred to Chapter IV of What is Mathematics? by)))

MORE GEOMETRY) 181)

Courant and Robbins, and to the Real Proiective Plane

by H. S. M. Coxeter. We shall look into two of theprincipal results of this field of study: the theorems of

Desargues and Pappus.Desargues' theorem. If two triangles have corresponding

vertic\037s joined by concurrent lines, then the intersections of

correspond\037ng sides are collinear (see Figure 94).The readershouldobserve that the theorem is devoid

of metric concepts. It is concerned with neither lengths

of sides nor with the sizeof the angles of the triangle.

One of the axioms of projectivegeometrystatesthatevery two lines intersect, that is, parallelism is ruledout.Consequently, the intersections referred to in Desargues'theorem, namely points P, Q, and R, must exist in pro-jective geometry. But, if we consider the hypothesis of

Desargues' theorem in Euclideangeometry, parallelism

may interfere with the existence of P, Q, and R. Wetherefore precede the generalized projective version of)

s)

FIGURE 94)))

182) ELEMENTARY. VECTOR GEOMETRY)

o)

//////./

/)A')

B')

FIGURE 96)

Desargues' theorem with a specialcase that arIses111Euclidean geometry.

Theorem D. Let ABC and A'B'C' be two triangles with

AB parallel to A' B' and BC parallel to B'G' while lines

AA', BB', and CC' meetin O. Then AC is parallel to

A' C' .In terms of our standard use of position vectors

emanating from 0, we have the following collinearities:

0, C, and G'; 0, A, and A'; 0, B, and B'. These imply

respectively: A = rA', B = sB' and C.= tC'. And AB

parallel to A' B' implies B - A = l(B'- A'). But

B - A = sB' - rA'. Hence,r = s = l. Similarly,

C - B = m(C' - B') = tC'- sB',)))

MORE GEOMETRY) 183)

which implies 8 = t = m.have l = m = r = s = t.)

Combinillg these results, we

Now,)

C - A = tC' - rA' = r(C' - A'),which states that AG is parallel to A'G', the resultwe

set out to prove. (Note that the giventrianglesmayormay not be in the same plane. The vectorproofisvalid

in both cases.)

Under the assumption that every two lines meet,we

present a vector proof of general form of Desargues'

theorem.

As in Figure 94, let, the two triangles be ABC and

A'B'G' with the correspondingverticesjoinedby con-

current lines meeting at S. Call P, Q, and R the inter-

sections of corresponding sides, as follows:

P the intersection of AB and A' B'

Q the intersectionof BC and B'G', and

R the intersection of CA and C'A'.)

Again we use position vectors emanating from somepoint0 (which does not appear in the figure).

The fact that S is collinear with every pair of cor-

responding vertices allowsus to write)

S = rA + (1 -r)A'

= sB + (1 - s)B'= tC + (1

- t)C'.)

This tripleequality implies the following three relations:)

rA - sB = (1 - s)B' - (1-r)A'

sB - tC = (1 - t)C'- (1- s)B'

tC - rA = (1- r)A' - (1- t)C'.

Noting that the sum of the coefficients of left and right

members in each of these three equationsis equal,we)))


have the opportunity to useTheorem 4, by dividing and

having the sum of the coefficients equal to 1.

rA _

8B =

1 - 8B' _ 1 -

r

A'. (109)r-s r-s r-s r-s)

8 B _ tC =

1 - tC' _ 1 - 8

B'. (110)s-t s-t s-t s-t)

t rC-

t-r t-r)A =

1 - rA' _ 1 -

t

C'.t-r t-r) (111))

Each memberof (109)representsa vector emanating

from 0, with its .endpoint on A'B' and on AB, simul-

taneously. This vector is P. That is,)

p=)r) s

A- Br - s)r - s)

or (r - s)P = rA - sB. (112)

Similar reasoning with (110)and (111)yields

(s- t)Q = sB - tC (113))

and)

(t- r)R = tC

- rA.

Finally, (112), (113), and (114)give us

(r- s)P = - (s -

t)Q- (t - r)R,

p = t- 8

Q +r -

t

R,r-s r-s)

(114))

or)

which establishes the collinearity of P, Q, and R, forthe coefficients t - sir - sand r - tlr - s sum to

unity.

EXERCISE

What justification is there in performing tIle divisions byr -

8, 8 -t, and t - r to obtain (109), (110),and (Ill)?

What would be implied by any of these denominators beingzero?)))

MORE GEOMET.RY) 185)

Pappus' theorem. Pappus of Alexandria (3rd century

A.D.) was the last of the remarkableGreek mathema-

ticians of antiquity. Among his original contributionsisa theorem that can be stated as a pure incidencetheorem,

devoid of metric concepts, and therefore falling into thecateg9ry of projective theorems.

If alternate vertices of a plane hexagon lie on two lines,the three pairs ofoppositesidesmeet in three collinear points.

We may restate the theorem in specific terms as

follows:

If A, B, G are distinct points of line \302\2431and A', B', C'

are distinct points of another line \302\2432,the three points ofintersection of the pairs of lines AB' and A' B, and BC'and B'G,GA',and G' A are collinear.

Before proceeding with the proof,we point out once

again that every two lines are assumed to intersect in

projective geometry. If one considersPappus'theorem

in the strict realm of Euclidean geometry, one must con-siderspecialcaseswhere parallelism may arise and inter-fere with the existenceof certain intersections (see

Exercise 4, page 191).Proof of Pappus' Theorem. Referring to Figure 96,

we call U a unit vector along \302\2431and V a unit vector)

FIGURE 96)))

186 ELEMENTARY'VECTOR GEOMETRY

along \302\2432. Thus we may write)

A = aU, B =bU,) C = eU ,)

and) A' = a'V B' = b'V C' = e'V., ,)

Since L is collinear with Band 0', we may invoke

Theorem 4 to write)

L = lB + (1- l)C'= lbU + (1 - l)e'V.)

And since L is also collinear with B' and C,we may also

write)

L = XC + (1 -X)B'

= XeU + (1 -X)b'V.)

These two representations imply)

lb=Xe

(1 - l)e' = (1-X)b',)

which solving for l and X, yields)

b'e - ee'l=

bb' - ee')and)

bb' - be'X=

\302\267

bb' - ee')Thus)

be(b' - e') b'e'(b - e)L =

bb' , U + bb' , V.- ce - ee)

Similarly, it can be shown that

ae(e' - a') a'e'(e- a)M = U + V,

ee' - aa' ee' - aa')

and)ab(a'

- b') a'b'(a -b)

N =, bb'

U + , bb'V.

aa - aa -)

To determine whether L, M, and N are collinear,we

seek r, 8, and t, such that)

rL + 8M + tN = O.)))

MORE GEOMETRY) 187)

(What other algebraic relation must r, 8, and t satisfy

before we may conclude collinearity?)Perseverance with elementary algebra yields the

solution)

r = aa'(bb' - ee'), s =bb'(ec'

- aa'),t = ee'(aa'

- bb'),

from which we do conclude that L, M, and N are col-linear. (Now, do we know that r, 8, and t are not allzero ?)

The theorem we have just proved possessesmathe-matical significance far' beyond any dreams that Pappuscould have had back in the third century. Approxi-mately one hundred years ago a method of building upnumbersystemsfrom geometry was first discovered by aGerman mathematician,Karl G.C.von Staudt. Since

that time much creative work has been done on thefoundations of geometry and its interrelationship withthe algebraicstructureof number systems that can be

built up from geometry. One of the great achievements

of this study is a remarkabletheorem, first proved by

David Hilbert (1862-1943) around the turn of the

century, which states:)

A number system related to a geometry satisfies the

law a. b = b \302\267a if and only if Pappus' theorem isvalid in the geometry. (The \"dot\" here stands for

multiplication.))

The geometry we havebeenworking with is Euclidean,and the number system for our analytic geometryisthereal number system, in which multiplication is commuta-tive, that is,a \302\267

b = b \302\267a. Thus we were able to provePappus' theorem;and, conversely, if we had developed anumber system with the aid of Pappus' theorem, its

multiplication would necessarily be commutative.)))


Since it is hardly possibleto enter into a detailed dis-

cussion of the foundationsof geometry in this short work,we shall have to contentourselves by closing with the

remark that it is a rich subject, which hfts resulted in thediscovery of many strange numbersystems as well as

geometries. There are geometries for which the coor-dinatenumber system has 1 + 1 \037 2; in fact, where1 + 1 = O. There are geometries in which neither

Pappus' theorem nor Desargues' theoremholdstrue.And stranger yet is the fact that such geometriesandtheir accompanying number systems have fou11d a widevariety of applicat\037ons

in fields as diverse as the designof

agricultural and other experiments, military logistics,psychology, and the study of mathematical machines.

The theorem of Menelaus. Menelaus of Alexandria,

who wrote a treatise on spheres and actuallymade some

discoveries in spherical trigonometry in the 1st centuryA.D., is also noted for having discovered an interestingtheoremconcerning transversals.)

If a transversal cuts the sides AB, BC, and CA of

triangle ABC in the points L, M, and N, respectively,)

AL BM CN

LB.

MC.

NA

= -1, (115))

where all segmentsreferredto are directed segments.

Conversely, if points L, M, and N are on the respec-tive sides AB, BC, and CA or triangle ABC, and)

AL BM CN_e_e_ -LB MC NA)

-1 ,)

then L, ]f, and N are collinear.)

We shall prove only the first part of the theorem, leav-

ing the converse as an exercisefor the reader.)))

MORE GEOMETRY) 189)

A)

B)

M)

FIGURE 97)

Consider all position vectors of points emanating frompoint A, and

L divides AB in the ratio l: 1 - lMdividesBCin the ratio m: 1 - m

N dividesAC in the ratio n: 1 -n.)

These conditio11s state)

AL BM CN_e_e_ -LB ]lifC N A)

l m

l-ll-m)

1 - n)

n)

Then)

\037 \037

L = lB, N = nC, and BM = mBC=m(C

- B).

(116))

(Figure 97 illlistrates the casewhere l < 1, n < 1, andm > 1.) Tllerefore M - B = mC - 11\037B a11d)

M = (1 - m)B + mC.) ( 117))

Now, since L, It.1,alld N arecollinear,we know that there

exists a real number r such that)

M = rL + (1 - r)N.) (118))))


From (116), (117), and (118),we deduce)

Hence)

(1- m)B + mC = rlB+ (1-

r)nC.

m = (1- r)n and rl = 1 -

m,)

h.

h\302\267

I l1 - m

w IC Imp y= -

r)

1 - (1 - r)n .)

r)

Weare now in a position to examinethe productofthe three ratios in terms of rand n alone:)

1 - (1-r)n

l m I-n r. -1 - l 1- m n

I-I - (1 -

r)n

r

(1- r)n I-n

1 - (1-r)n n

1 - (1 -r)n-

r - 1 + (1-r)n

(1- r)n I-n

1 - (1-r)n n

(1- r)(l -

n)-1- -

(r - 1)(1- n)

,)

which is the result we wished to prove.)

EXERCISES1. Prove the converse of the Menelaus theorem.

Hint: M = aL + bN for some scalarsa and b. Prove that

l m I-na + b = 1, by using the relation

1 _ l. 1 \037 m

.

n= -1.

(You might try to eliminate one of the scalars,say l, by first

solving for it in terms of m and n.))))

MORE GEOMETRY

A)

191)

\"\"

\"\"

\"\"

,\"C)

A'\",, , ,

\"\"

C')

B) B')

FIGURE 98)

2. Prove the specialEuclideancaseof Desargues' theorem,

which reads: Let ABC and A'B'C' be two triangles with AB

parallel to A'B', and BC parallel to B'C' while lines AA',

BB', and CC' are parallel. Then AC must be parallel toA'C' (see Figure98).3. Prove the following special case of Pappus' theorem. Iftwo sets of three points A, B, C and A', B', C'on two coplanar

lines \302\243and \302\243',respectively, are so related that the linesAA',

BB', CC' meet in a point, the points of intersection of the pairsof lines AB' and BA', BC' and CB', CA', and AC' are collinear

with the point of intersection of \302\243and \302\243'.

4. Prove by vector methods the following special Euclidean

case of Pappus' theorem: If A, B, C are distinct points of line\302\243and A', B', C' are distinct points of \302\243'such that AB' is

parallel to BC' and A'B is parallel to B'C, then AA' must beparallelto CC'.

Why is this a special case of Pappus' theorem?)

41. APPLICATIONS OF PARAMETRIC

EQUATIONS TO LOCUSPROBLEMSThe cycloid. Considera wheel set on a line and per-mitted to rollwithout slippage. An interesti11g problem,)))


FIGURE 99)

and one whose results are applicable to certainengineer-ing problems, is the following: What is the locusofa fixed

point P on the rim as the wheelrollsalong the line? (See

Figure 99.)To simplify matters we allow the line to be the x-axis

and let P beginat the origin of the coordinate system.We call C the centerand r the radius of the wheel. Theparameter8isusedtodenotetheangle through which the

radius CP has rotated. Referring to Figure 100, the)

y)

o M(x,O)) F)x)

FIGURE 100)))

MORE GEOMET.RY) 193)......-...)

horizontal distance OF is equal to the arc FP, by the

assumption that the wheel does not slip; and from......-...

trigonometry we know that FP = r8, where8ismeasured

in radians. Thus OF = r8. We shall use this importantfact in determining the x- alld y-coordinates of P.)

x = OF - MF = r8 -PQ.)

But from triangle PCQ we see that)

. PQ PQSIn 8 = - = -.

PC r)

Thus, PQ = r sin 8, and the x-coordinateofP isx = r8 - r sin 8 = r(8 - sin8).

Again, referring to Figure 100, we note that)

Hence)

y = MP = FQ = CF -QC

= r - QCj

CQ= CP cos 8 = r cos8.

y= r - r cos8 = r(l -

cos 8),)

but)

and we have)

{

X = r(8- Sill8)

Y= r(l

- cos 8))

as the parametric representation of the desired locus,which is called a cycloid.Involutes. We shall confineour attentionto involutes

of circles, for these curves are easily treated without the

use of calculus. The more general theory is part of thefield of differential geometry alld requires-as the namesuggests-differential calculllSas its instrument of

analysis.

Consider a string (of a mathematicaloridealsort,with

zero thickness) to be tightly wound about a circularpost,)))


Y)

x)

(a)) (b))

FIGURE 101)

and call P the endpointof the string. The involute of a

circle is the curve described by point P as the string isunwound whilebeingheldtaut (seeFigure 101).

We shall now derive parametric equations for theinvolute, using the fact that the portion of the stringalreadyunwound is tangential to the circle (because it isbeingheldtaut).

Letthecircleofradius r be centered at the origin, andsupposethe initialpositionofP to be at A = (r, 0) onthe x-axis. We arbitrarily choose the unwinding to takeplace in a counter-clockwisedirection,calling the

unwound portion of the string PT, whereOTisa radius

making an angle 8 with the horizontal radius OA.

Referring to triangle TGP in Figure 101b,we observethat angle GTP is equal to 8, which allowsus to write)

GP = TP sin 8) and) GT = .TP cos 8. (119))

Since the unwound string TP is equal in length to the......-... ......-...

arc AT, we have TP = AT = r8, which permits us to)))

MORE GEOMETRY) 195)

rewrite (119) in the form)

GP = r8 sin 8 and GT = r8cos8.)

We- are now ready to determine expressions for thecoordinatesofP in terms of the parameter 8.)

x = OF + GP = r cos8+ r8 sin 8 = r(cos 8 + 8 sin 8),and y

= FT - GT = r sin8 - r.8 cos 8

= resin 8 - 8 cos 8).)

Thus a parametric representation of the involute of the

circle is.)

{

X = r(cos 8 + 8 sin 8)

y= resin 8 - 8 cos 8).)

(120))

Vector methods are often of assistance in determininglociof the type just considered. We therefore attempta derivation of (120)fromthe vector point of view in the

hope that it may be instructive to the reader.\037

The position vector OP = xi + yj\037 \037

= OT + TP)\037

= r cos 8i + r sin 8j + TP.)

\037

We call TP = ai + bj and determinea and b.)

\037

ITPI= Va 2 + b2 =

r8,) (121))

\037 \037

and TP \302\267OT = ar cos 8 + br sin 8 ::cO.)

Since r r!= 0, we may divide by r, getting)

or)

a cos 8 + b sin 8 = 0

a = -b tan (J.)))


Substituting this va lue for a in (1 21) yields

V b2 tan 2 8 + b2 = r8)

or b sec 8 = r8,

which implies b = r8 cos 8

and a = -r8 sin8.)

\037

Thus TP = -r8 sin 8i + r8 cos8j.We are now equipped to return to the problemof getting

\037

an explicit formulation for OPe)\037

OP = r cos 8i + r sin 8j - r8sin8i+ r cos8j)

= r(cos 8 - 8 sin 8)i + resin 8 + 8 cos 8)j.Therefore)

{

X = r(cos 8 - 8 sin 8)

y= resin 8 + 8 cos 8),

which differs from the parametric form (120)! Why?See Exercise1 below.)

(122))

EXERCISES

1. The parametric representation (122)is in error, for some

careless algebra was put forth in the argument. Find theerrorand give a valid vector-type derivation of the involute

of a circle.

2. (a) A wheel of radius r rolls along a line without slipping.

Find the locus described by a point P on a spokeof the wheel,

where P is at a distancea from the center of the wheel. Thiscurve is known as a trochoid or prolate cycloid (see Figure l02a).

(b) What is the locus if a > r, as is the \037ase when P is onthe rim or flange of a locomotive wheel (see figure l02b).

This curve is called a curate cycloid. (Observe that a descrip-tion of this locus shows that some part of a locomotive is mov-

ing backward, no matter how fast the train is moving forward!))))

MORE GEOMETRY) 197)

(a)) (b))

FIGURE 102)

42. RIGID MOTIONSEuclid's theory of congruenceis predicatedon the

ability to move a figure from one position to anotherwithout disturbing its metric properties (e.g. length of

edges, size of angles). Such motions are generally

termed rigid motions and consist of two types or com-

binations of these two types. The first, called trans-

lation, refers to displacing a figure B= by allo,ving a fixed

vector T to act on each pointof the given figure B=.

That is, if P is a point of B=, the point P is translatedto Q, where)

\037 \037

OQ= OP + T. (See Figure 103.))

1'hesecondtype of rigid motion is the rotation, in which afigure is rotated about a fixed point in the plane. Figure104shows an ellipse and a triangle rotated simultaneouslyabout the centerofthe ellipse.

We shall obtain analytic expressions for these rigidmotions.

Supposeit isdesiredthat the point (h, k) be nl0ved bytranslation to the origin of the (x, y)-coordillate system.To keep matters straightwe designate the. new coordi-

Ilates of a point with primes. Thus)

translation

(h, k) ) (h', k') = (0,0),)))


Y)

x)

Q)

FIGURE 103)

which, in terms of the definition,states that the trans-lationvector T has the effect)

Hence)

hi + kj + T = O.T = -hi - kj.)

In general, the point(x, y) moves to point (x', y'), and)

y) y)

x) x)

FIGURE 104)))

MORE GEOMETRY) 199)

their relationship follows from the equation)

Thus)

xi + yj + T = x'i + y'j,

(x- h)i + (y - k)j = x'i+ yj.

{

X'=X-h

y'=y-k)

or)

are the equations of translation.In order to discovertheequationsdescribing a rotation

about the origin of the coordinatesystem,letP =(x, y)

rotate clockwise through the angle 8 to its new position

p' = (x', y'). For purposes of our analysis,let a be theangle that the position vector P makes with the positivex-axis (see Figure 105). Then)

x = p \302\267i =Ipl cos a = Ip/l cosa

(since Ipl=

'p/l) (123)

y = P \302\267j

=Ipl sin a = Ip/l sin a,)

y)

P(x,y))

P'(x',y'))

x)

FIGURE 105)))

200) ELEMENTARY.VECTOR GEOMETRY)

and)

X' = P' \302\267i =Ip/t cos(a -

(J)=

\\p/\\ COS a COS(J (124)

+ \\p/lsin a sin (J

y' = P' \302\267j

=Ip/l sill(a -

(J)=

\\p/l sin a cos (J

-Ip/l cos a sin 8.)

Imposing equations (123) in the right members of (124),

we arrive at)

{

X' = x cos (J + y sin (J

y' = - x sin (J + y cos (J,)

(125))

which analytically describes the rotation stipulatedabove.)

EXERCISE

Rotations are usually described by a pair of equations that

follow from (125) by solving for x and y. (a) Carry out thisalgebraicprocedure. (b) Derive these equations directly byvector methodsanalogousto thoseusedto determine(125).)

EXAMPLE 52. Take the circle (x - 1)2+ (y + 2)2 = 4.We discuss the translation of the circle so that its new centeris at the origin.

The given circle is centered at (1, -2). Thepoint (1, -2)becomes the origin under the translation)

{

X' = x-I

y' =y + 2.)

Thus, the translated circle is X'2 + y'2 = 4. Observe thatthe algebraictransformation did not affect the radius, which is,after all, the circle's metric property that was to remain

unal tered.)

EXAMPLE 53. If the points of the plane are rotated 7r/6 (or30\302\260)clock\\vise, what will be the new location of the point

(1, -2)'?)))

MORE GEOMETRY

The equations of the rotation through ?r/6 are:)

201)

-v-3 1x' = x - + y

-2 2

1 \03713

y' = -x- + y\037

2 2)

Therefore)

X' = V3 - 1 =approximately -0.13

2)

1\037/-

y' = -2

- v.3

= approximately -2.24,)

which states that)

rotation

(1, -2) ) (-0.13, -2.24).thru ?r 16)

EXERCISES

1. Find the coordinatesof the point (2, -3) if the translationmoves the origin to

(a) (4, 5); (b) (-3, 3); (c) (-5, 4).2. Find the coordinates of

(a) (2, 4); (b) (-3, 6); (c) (-2, 0)when the axes are rotated counterclockwise through the angle

\302\2674arcsm '5.

3. Do the sameasExercise2when the axes are rotated counter-clockwisethrough the angle ?r14.

4. Show that the equation x2+ y2 = r 2 is unchanged by any

rotation of the axes.

5. A cylindrical surface is generatedby a line moving in sucha way as to be always parallel to its original position. If, as

the generating line moves, it remains perpendic\037lar to a fixed

plane and traces out a circlein the fixed plane, the cylinder istermed right circular.

(a) Provide a reasonable definition for the radius of a right

circular cylinder.)))


(b) By analogy-perhaps loose-with the cone, define axis

for a right circular cylinder. (Seethe first of the Miscellaneous

Exercises that follow.)(c) Find, by vector methods, the equation of the right

circular cylinderthat cutsthe xz-plane in the eircle (x - 1)2+(z + 1)2 = 1 and whoseaxisis parallelto the y-axis.

6. Let Po = (xo, Yo, zo) be the center of a sphere that hasPI =

(Xl, YI, Zl) as one of its points. Using vector techniques,

prove that

(x -XO)(XI

-XO) + (y

-YO)(YI

- Yo) + (z - ZO)(ZI.-ZO) =0

is the equation of the planetangentto the sphereat Pl.)

MISCELLANEOUS' GEOMETRIC EXERCISES

1. If a line X varies in such a way that it always intersectsafixed line cC in a fixed point V at an angle j-a( < 1r/2), then X

is said to generate a right circular cone with axis cC,vertex V,

and vertex angle a (see Figure 106).Prove by vector methods that the right circular cone, with

vertex at the origin, axis the z-axis,and vertex angle 21r/3,is represented by the equation X2 + y2

- 3z2 = o.)

.\302\243)

FIGURE 106)))

MORE GEOMETRY) 203)

(Hint. Let P bethe position vector of P = (x, y, z) on the cone.Then p. k =

(pI cos ?r/3 is the vector equation of the cone.)

2. Find the equation that representsthe cone whose axis is

the x-axis, vertex is at the origin, and vertex angle is ?r 12.

3. Find the equation that representsthe cone,whose axis is

the z-axis, vertex is at (0,0, 2), and vertex angle is 2?r/3.

4. Given points A, B, and R in space, find a formula for thedistance d from R to line AB.

\037 \037 \037

(Hint. If AR and AB make an angle 8, then d =IARl sin 8;

\037 \037

but sin 8 may be determined from AR X AB.)Observe that (i) the equation of line AB is never needed;

and (ii) interchanging the'roles of A and B yields a check.5. Prove, by vector methods, that the area of a triangle formed

by joining the midpoint of one of the nonparallel sides of atrapezoidto the endpoints of the opposite side is half that of

the trapezoid.

6. Prove by vector methods: If a straight line is equally

inclined to three coplanar lines, it is perpendicularto theirplane.7. Prove: The sum of the squares of the edges of any tetra-

hedron is equal to four times the sum of the squares of joins

of the midpoints of opposite edges.8. Find the angle between two nonintersecting edges of a

regular tetrahedron.)))

appendix-

expansion

of

determinants)

For the reader \\vho is unfamiliar with the theory of

determinants, we provide a brief description of the

method of determinant expansion. This discussion doesnot attempt to communicate mathematical insight into

theory but merely provides a mnemollicdevice for deal-

ing with cross products by means of a compact notation.

The two by two determinant, written:\037 ::

repre-

sents a quantity according to the following expansion:)

al a2.

b b= a 1b2

- a2b 1.1 2)

The three by three determinant is expanded as fo110v,\"s:

204)))

APPENDIX-EXPANSION OF DETERMINANTS) 205)

al a2 ag b2 b g b 1 b gb l b 2 b g = al - a2Cl C2 Cg

C2 Cg Cl Cg

+ agb1 b2 = a 1b2C3 - a1b 3 C 2

- a 2 b 1c3Cl C2

+ a2bgCl + aab 1C2 -aab 2 Cl.)

(Observe that interchanging the second and third rowsleadstoal1expression which is the negative of the one wehave computed. This is the algebraic equivalent of

A X B = -B X A.)There are various \037pproaches to expanding deter-

minants, and there are severaldevices by which one

recalls the expansion. However, the readerwho is not

familiar with determinants would be wiseto stickto the

particular expansion of cross products in this volume.In accordance with the given expansion, we view thecrossproduct)

(ali + a2j + aak) X (bli + b 2j + bak))

as equal to)

i j k

\\

= i\\

a2 a3 _j

al a3+ k al a2al a2 aa

b 2 b a b l b g b i b2b1 b2 ba)

- (a2 b 3 - a3b 2)i + (agb l - a1ba)j+ (a 1b 2

- a2bl)k.)))

answers)

SECTION 3)

4. .y 89 . 6. 1000.)SECTION 4

2 '3 4 3 5-36. (a) l =\"5'

m =5.

(b) l =\"7'

m =7.

. (c) l =2'

m =2.)

SECTION 6)

6. They are equal.)

SECTION 81. (b) A = 6i + 4j + 10k,B = -6i + 4j - 10k,C = 4i - 6j - 10k,D = 10j + 4k. (c) 4i + 12j- 6k.

\037

(d) A - B = 12i+ 20k, BD = 6i + 6j + 14k.(f) (-18, 4, -30). (g) (-18, 4, -30).2. (a) 4i - 8j. (b) 3i + 6j. (c) -7i + 2j. (e) Yes. (f) Yes.

158(g) (2, -3, 3). 3. No.4. (a) No. (b) V =

3A + 3

B + 3C.

SECTION 14

34

(. \302\267

- 4(

. .) 4 .y2i 0 ( )\302\267

\037/- 31 + 2J) or\037/-

31 + 2J. \302\2677

\302\2671. a-II.\037 13 \037 13

206)))

ANSWERS) 207)

(b)-11 V6

and-11 V29 . (c)

;;\037(2i + j - k).6 29 29

(d) 4 units. 11. Because the dot product is distributive.)

SECTION 16

1. (a) ? = (1-t)(5i +4j) +t(3i +j). (b) x = 5 -

2t,

\037 -5 33 33y

= 4 + 5t. (c) 5x + 2y= 33. 2. (a) _

2.

(b) -, -.5 2

3. (a) 9x+ 5y + 25 = o. (b) y= 2x + 6. (c) 2x -

y= 1.

4. (6, 0). 6. (a) (2, -1) (b) x + 3y + 1 = o.(c) x = -2 + (7 V 34 + 5v'58)t, y

= 1 - 3(V 34 + V 58)t.

7. (a) 3y = V3 x - (6 + V3). (b) x = 1. (c) y= -2.

9. (c) x = o.)

SECTION 17

1. N = ai + bj. 2. (a) x + 2y = o. 3. x + 4y= 6.

4VS

(. .). +

521-

J .)

SECTION 18

8 111.

0. 2.O. 3. O. 4. o. 6.

5\"\

SECTION 19

m2 -ml 25 9 1r

6. tan 8 = \302\2678. (a) arc tan -, arc tan -,-.1 + mlm2 9 252

\037\037 5V3 +9V29. -7 + V 50. 10.y = .

9x. 12. x -

8y= O.

V313. (a) m(x

- 1) + n(y - 2) = o. (b) y=

3x + b.

14. (a) 5x + 4y - 2 = O. (b) 7x - 32y - 3 = O.

SECTION 20

1. (a) x 2+ (y - 1)2= 4. (b) x 2

+ (y + 1)2 = 4.(c) (x+ 2)2 + (y - 3)2 = 9. 2. (a) (x

- 2)2 + (y + 1)2 = 5.18

(b) 5x 2+ 5y2 -

\037x- : y

-5

= o.

(c) x2+ y2 - 8x + 12y

= o. 3. (a) (8, -6).)))


(

4 2

)V6510.(d)

-3' 3' 3

\302\2674. (a) 3x + 4y = 24.

7. (a) 9z + 2y= 40. (c) y

= -x + 2 V2.9. (a) x = -1 + 5 COB 8, Y = 2 + 5 COB 8.

_/- 31.../-(b) x = 1 + \037V 141 9, Y

= -2

+ 6 V 1418.

10.(a) (1, 0) and (0, 1). (b) (1, 0) and (0, 1).)

SECTION 214311.(a)

_ /_' _ /_ \"'\"j_.(c) 0, 0, 1. (d) 0, 1,O.

5V2 5V2 5V22. (a) (0, 1,0), 1. 3. (a) (x

- 1)2 + (y + 1)2 + Z2 = 4.

(c) x2+ y2 + z2 = 144.

4. (a) 3, 1, -1 and 3m, m, -m (m ;:C0).)

SECTION 242. (a) 6x+ 4y + 7z = 17. (c) 2x -

y- z = O.

19 88(e) 4z -

3y + 12z = 76. 3. (a)-. (c)-.. 13 5)

SECTION 26

1. (a) Two-point form is non-existent in this case. The line is

given by x = -4, y= 1. (c) Again, two-point form is non-

. ...b

x+4 z+2deXIStent. LIne IS gIven y - an y

= 1.3+4 2+2

x-3 y-2 2+5 x-3 y-2 z+5(e)

= = and = =3 - 2 2 - 10 -5 + 9 3 - 2 2 - 10 -5 + 1x-O y-O z-O

(g)5 - 0

=-4 _ 0

=6 _ O

. 2. (b) x = -4 + 9t,

y= 1 + 2t, Z = -2 + t. (c) x = 3 + 3t, 1i = 2 -

2t,

z = -5 - 6t. (f) x = -4 + 3t, y= 1 + 4t, Z == -2 + 12t.

34 . 394. (a) 6, 2, :1. 6. (b) x = -

7\"+ 20t, Y

= -14

+ 18t,

-12 6 1z = t. 7. (-12,4, -1). 8.

_ /_ _ /_ _ /-V 181' V 181'V 181.)))

ANSWERS) 209)

SECTION 28)

1r1. In order to have

2-

(3 be first quadrant angle.

(

33 69 39

). 83. (9, -5, 12). 6. -

10' -

20' -20

' SIn (3 =21

.

136. (a) Use vectors normal to the planes. (c) cos 9 = _ .

V45)

1 . 18. (a) cos9 =

V3

' (c) 8m fJ =

V3)

SECTION 32

1. (a) k - j. (c) O. (e) -8j- 4k. 2. Usetriple scalar product.

\037 \037

AB X AC6. (b) x + y + z = 1. 7. d = A -

\037 \037. 9. (a) o.

lAB X Acl

2(c)

3.11. blC2

-b2 C l, a2Cl - alC2,alb2 + a2bl-)

SECTION 33

2. -13i - 7j + 5k. 6. A X Band C X D are parallel. Hencetheir cross product is the zero vector.)

SECTION362. (a) Annular region centered at origin (including the boundingcircles. (b) Spherical shell including the bounding spheres.3. A torus (doughnut) whose circular cross-section has a radius of

one. 4. (a) 4 < x2+ y2 < 9.

(b) 0 < x < 1and 1 - x < y < x + 1.)

SECTION 384. If ABCD is a tetrahedron or a square then X = TlA + T2 B +

TaC + T4D, with 0 < Ti < 1 and Tl + T2 + Ta + T4 = 1.

6. At least one coefficient equals zero. 9. (a)(\037, \037).

M = j-A + iB + -lC. (b) (1, 1).10. (a) X = TlA + T2B + TaC,with 0 < Ti < 1and Tl + T2 + Ta = 1. (c) Insist that 0 < Ti < 1.)))


SECTION 391. (a) Note that m + 8 + b = 500 for maximum profit.(b) 100milk, 300 semi-sweet, 100 bitter.2. max = 7, min = 5. 3. 400 gal regular, 300 gal high test,300 gal white. 4. 800gal regular, 200 gal high test.6. i A-ration, 4 B-ration.)

SECTION 41

2. (a) x = r8 - a sin 8, y = r - a cos8.(b) x = r8 - a sin 8, y

= r - a cos 8.)

SECTION 42

1. (a) (6, 2). (c) (:-3, 1). 2. (a)

e51 ,\037}(c)

( 56.

58}6. (c) (x

- 1)2 + (z + 1)2= 1.)

MISCELLANEOUS EXERCISES

2. x2 =y2 + z2. 3. 3(z -

2)2= x 2

+ y2.\037 \037

4. d = \\AR X ARI . 8 \037.

IABI

\302\2672)))

Absolute value, 14, 57-9Addition of vectors, 9, 11-13Altitudes, 75Auxiliary point technique 30

Axioms, 1')

Basis, 46-9

Bisectors, perpendicular, 66-7Bound vector, 9

Bypass lemma, 19)

Circle, 100 fi.Commutativity, 9, 63, 187

Complex plane, 54-9Components, 68-70, 73

Cone, 202

Convexity, 167-9Coordinate-free methods, 76

Coplanar vectors, 27, 34Courant, R., 181

Coxeter, H. S. M., 181Cross product, 135 ff.

triple, 147-50)

index)

Curate cycloid,196Cycloid, 191-3

Curate, 196

Prolate, 196Cylinder,201)

Deductive science, 1

Dependence, 21 fi.Desargues'theorem,181fI.,188,

191Determinants, 141, appendix

Dimension, 49Direction angles, 108Direction cosines, 108-10, 125Direction numbers, 110-1,123-5Distance,60-1Distributivity, 65, 139

Division of segments, 27-9, 34Dodecahedron,54Dot product, 62 fi.')

Equilibrium, 10, 20)

211)))

212)

Force, 7-8, 10, 20, 73Freevector, 6-7, 9)

Galileo, 11

Generates (= spans), 48Great circle, 157)

Hall, D. W., 19Hero's formula, 156Hilbert, David, 187)

\"If and only if\", 6

Incidence, 180, 185

Inclination, angle of, 81

Inner product, 60 fi.Intercept form, 87

Inverse, additive, 16

Involute, 193-6)

Klamkin, Murray S., 148)

Lagrange, identity of, 150Law of cosines, 150, 158Law of sines, 150-1,158-9Left-handed triple, 43 fi.

Lichtenberg, Donovan, 170Line of action, 7-8

Linear programming, 170 fi.Length, 57-9Linear combination, 20-21, 35Linear dependence, 21 fi.)

Mechanics, 7-8, 10, 68-9, 73Menelaustheorem, 188-90

Model, 3

Multiplication by scalar, 12 fI.)

Negative triple, 43 ff.

Newman, J. R., 53)

Ordered set, 43

Orientation, 40 ff.

Pappus' theorem, 185, 188, 191Parallelism,86)

INDEX)

Parallel postulate, 2

Parameter, 78Parametric representation, 78-

80, 122, 146, 191Pencil of lines, 100

Perpendicularity, 63, 89Planes, 111ff.

Point-slope form\037 87

Position vector, 46, 49Positivetriple, 43 fi.

Postulate, parallel, 2

Postulates, 1Programming, linear, 170 fi.

Projection, 64-6Projective geometry,180fi.

Prolate cycloid, 196

Pyramid, 146)

Radial vectors, 52-4, 56

Radians, 43Resultant, 10, 20

Rhombus, 156-7

Right-handed triple, 43 fi.Rigid motions,197-201Robbins, H., 181

Rotation, 197, 199-201)

Scalar,4-5Scalar,product, 63 fI.

triple, 137-43Segment division, 27-9, 34Set theory, 21, 35, 85, 165, 167Shortcut lemma, 19

Slope, 81

Slope-intercept form, 83Space,49

Span, 48

Sphere, 106 fi.

unit, 108-10

Statics, 10

Staudt, K. G. C. von, 187

Stevin, Simon, 10, 11Straight line, 77 fi., 121 fi.)))

INDEX)

Subtraction,\" 16-17

Symmetry, 53)

Terminus (= endpoint), 5, 9Traces,114-15Translation, 197-9

Triple cross product, 147-50Triple scalarproduct, 137-43Trochoid, 196

Two-point form, 80, 126)

Union, 167Uniqueness of representation,

34 ff.

Unit sphere, 108Unit vector, 5)

Vector(s), addition of, 9, 11-13

bound, 9definition of, 4

direction of, 4, 6,..13)

213)

Vec.tor(s), equality of, 6, 10

free, 6-7, 9line of action of, 7-8linear combinations of, 20 ff.,

35

linear dependence of, 21 \302\243f.

magnitude of, 4, 6multiplication by scalar,12ff.

origin of, 6

radial, 52-3subtraction of, 7, 16

sum of, 9unit, 5

zero, 5, 12, 14, 22Vector product, 136 fi.

Vector product, triple, 147-50)

Weyl, Hermann, 53

Work, 73)

Zero vector, 5, 12,14,22Zweng, Marilyn, 170)))

seymour schLlster)

Dealing primarily with the development of vector algebra as a mathe-

matical tool in geometry, this elementary text features applications to

trigonometry, both plane and spherical,and algebra. Appropriate for highschool students and college undergraduates, it offers greater insights into

theorems by employing vector and analytic proofs, rather than syntheticproofs.)

Starting with an overview of elementary operations,the text defines vec-

tors, explores their fundamental properties and linear combinations, anddiscusses auxiliary point technique and uniqueness of representations.Subsequentchaptersexamine vectors In coordinate systems, propertiesand formulas of inner products, and elements of analytic geometry-the

straight line, analytic methods of proof,and circles, spheres, and planes.The text concludeswith surveys of cross products, plane and sphericaltrigonometry, and additional geometric concepts, including segments

and convexity and linear programming. ExercIsesappearthroughout the

book, with solutions at the end.)

Dover (2008) republication of the edition published-byJohn Wiley & Sons,Inc., New York, 1962. 224pp. 53/8 x 81f2. Paperbound)

also availableA VECTOR SPACE APPROACH TO GEOMETRY, Melvin Hausner. 416pp. 53/8 x 81f2.

0-486-40452-8

VECTOR METHODS ApPLIED TO DIFFERENTIAL GEOMETRY, MECHANICS, AND

POTENTIAL THEORY, D. E. Rutherford. 144pp. 53/8 x 8 112.0-486-43903-8)

For current price information write to Dover Publications, or log on to

www.doverpublications.com andsee every Doverbookin print. $12.95 lJSA)

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ISBN-13:978-0-486-46672-9ISBN-10:0-486-46672-8

I I

5 1 2 9.\037.,

9 780486 466729)PRINTED IN THE U.S.A.)))

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Elementary Vector Geometry