MTH10007 notes s1-2015

Department of MathematicsSchool of ScienceFaculty of Science, Engineering and Technology

MTH10007Engineering Mathematics 2

Study guide

ii

c Swinburne University of Technology, February 5, 2015.

Distributed by

Swinburne University of TechnologyHawthorn VIC 3122Australia

http://www.swin.edu.au

Copyrighted materials reproduced herein are used under the provisions of the Copyright Act 1968as amended, or as a result of application to the copyright owner.

No part of this publication may be reproduced, stored in a retrieval system or transmitted in anyform or by any means electronic, mechanical, photocopying, recording or otherwise without priorpermission.

Typeset by TechType, Canberra, using LATEX2e.

Table of Contents

1 Matrices 1

1.1 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Cramers rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Matrix inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Vectors 17

2.1 The vector (cross) product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Angular velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.3 Moment of a force (torque) . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4 Equation of a straight line in three dimensions . . . . . . . . . . . . . . . . 25

2.5 Equation of a plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.6 Vector spaces, linear independence, basis and rank . . . . . . . . . . . . . . 42

3 Complex Numbers 54

3.1 Arithmetic operations with complex numbers . . . . . . . . . . . . . . . . . 55

3.2 Geometrical representation of complex numbers . . . . . . . . . . . . . . . . 57

3.3 Powers and roots of complex numbers . . . . . . . . . . . . . . . . . . . . . 61

3.4 Exponential form of cis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.5 Engineering application: impedance . . . . . . . . . . . . . . . . . . . . . . 73

4 Differential Equations 77

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.2 First-order differential equations . . . . . . . . . . . . . . . . . . . . . . . . 80

4.3 Orthogonal families of curves . . . . . . . . . . . . . . . . . . . . . . . . . . 89

4.4 First-order linear differential equations: integrating factors . . . . . . . . . . 94

4.5 Second-order linear DEs with constant coefficients . . . . . . . . . . . . . . 98

4.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

iii

iv Table of Contents

5 Surfaces and Partial Differentiation 115

5.1 Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

5.2 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

5.3 Small increments formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

5.4 Errors of measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

5.5 Chain rules (rate-of-change problems) . . . . . . . . . . . . . . . . . . . . . 136

5.6 The directional derivative for a surface . . . . . . . . . . . . . . . . . . . . . 140

5.7 Stationary points (local extreme values) on a surface . . . . . . . . . . . . . 143

6 Curves 160

6.1 Polar curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

6.2 Two-dimensional (2D) parametric curves . . . . . . . . . . . . . . . . . . . . 167

6.3 Lissajous curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

6.4 Curves in space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

6.5 Velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

6.6 Projectiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

6.7 Curvature (bending) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Formulae 200

Module 1Matrices

1.1 Determinants

Associated with every square matrix is a number called its determinant. The determinantoccurs commonly in work on transformations and the solution of sets of equations. Thedeterminant of a matrix A can be denoted by det(A) or |A| or by writing the elements asan array as usual, but between two vertical lines, not brackets, for example

2 95 8. The

determinant can be positive, negative or zero.

A singular matrix has a zero determinant and a non-singular matrix has non-zero deter-minant.

The determinant of a 1 1 matrix is just the number (element) in the matrix.

Example 1.1.1. If A = (3) then det(A) = 3. In general, if A = (a) then det(A) = a.

The determinant of a 2 2 matrix ( a bc d ) is defined to be a bc d = ad bc,

i.e. the product of the elements on the main diagonal minus the product of the elements onthe other diagonal.

Example 1.1.2.

2 95 8 = 2 8 (9 5) = 61.

Example 1.1.3.

1 04 5 = 15 (0 4) = 5.

1

2 Module 1. Matrices

The determinant of a 3 3 matrix a1 b1 c1a2 b2 c2a3 b3 c3

is defined to be

a1 b1 c1a2 b2 c2a3 b3 c3

= a1 b2 c2b3 c3

b1 a2 c2a3 c3+ c1 a2 b2a3 b3

.Each element in the first row is multiplied by a 22 determinant, obtained by deleting thefirst row and one column in turn from the original matrix. These terms are then alternatelyadded and subtracted. We sometimes call this expanding the determinant by the first row.

Example 1.1.4.4 3 14 7 61 3 9

= 47 63 9

3 4 61 9+ (1) 4 71 3

= 4(63 18) 3(36 6) 1(12 + 7)= 287

Similarly, a 4 4 determinant is defined in terms of 3 3 determinants:a1 b1 c1 d1a2 b2 c2 d2a3 b3 c3 d3a4 b4 c4 d4

= a1b2 c2 d2b3 c3 d3b4 c4 d4

b1a2 c2 d2a3 c3 d3a4 c4 d4

+ c1

a2 b2 d2a3 b3 d3a4 b4 d4

d1a2 b2 c2a3 b3 c3a4 b4 c4

and then each 3 3 determinant is evaluated as above.This pattern continues for the larger matrices, but evaluating by hand becomes very cum-bersome. There are some properties of determinants, which you will study in some detaillater, which can simplify the calculations.

Example 1.1.5. If

A =

2 7 31 5 14 0 5

,find det(A) and det(AT)

det(A) =

2 7 31 5 14 0 5

= 2 5 10 5

(7) 1 14 5+ 3 1 54 0

= 2(25 0) + 7(5 + 4) + 3(0 20)= 53,

1.1. Determinants 3

AT =

2 1 47 5 03 1 5

,

det(AT) =

2 1 47 5 03 1 5

= 2

5 01 5 1 7 03 5

+ 4 7 53 1

= 2(25 + 0) 1(35 0) + 4(7 15)= 53.

This example illustrates one of the properties of determinants, which is that det(A) =det(AT) This means that a determinant can be evaluated using a row or a column.

Another property is that if two rows (or columns) are interchanged then the value of thedeterminant is multiplied by 1. In fact, any row or column can be used to evaluate adeterminant, provided the correct pattern of + and signs is followed.Example 1.1.6.

+1

4

+ 36 10 02 9 0

= 1 10 09 0

4 6 02 0+ (3) 6 102 9

= 1(0) 4(0) 3(54 20)= 222

or using column 3, because there are two zero elements, and following the correct patternof signs:

+1

4

+ 36 10

0

2 9 +0

= 3 6 102 9

0 + 0= 222

If two rows or columns of a determinant are identical, the determinant has value 0. Thisis proved by using the earlier result that swapping two rows or columns multiplies thedeterminant by 1.Sometimes the determinant to be evaluated contains variables, so there will not be a nu-merical answer, but the determinant is evaluated in the same way, giving an expression asthe result.

Example 1.1.7.k 4 36 k 02 9 k

= 6 4 39 k

+ k k 32 k 0 expanding by row 2

= 6(4k 27) + k(k2 + 6)= 24k + 162 + k3 + 6k= k3 18k + 162.


Exercise 1.1.1.

Evaluate the following determinants:

1.

3 12 4 2. 1 56 11

3. x 2y 3 4. 2 05 7

5.

5 0 12 1 13 2 0

6.

1 3 50 0 92 3 1

7.i j k1 1 23 2 1

8.1 1 13 5 34 7 4

9.

5 2 00 6 11 1 3

10.2 1 30 4 04 2 6

11. An element of a lamina is subjected to normal and tangential stresses on each of itsfour sides. The associated stress matrix is A =

[12 88 15

]. Calculate the principal stresses,

given that each principal stress, , satisfies det(A I) = 0

158

12

8

158

8

12

Answers to Exercise 1.1.1.

1. 10 2. 41 3. 3x 2y 4. 145. 11 6. 27 7. 5i+ 5j + 5k 8. 09. 87 10. 0 11. 21.64; 5.36

1.2 Cramers rule

1.2.1 Cramers rule for two unknowns

To solve

ax+ by = e (1.2.1)

cx+ dy = f (1.2.2)

multiply (1.2.1) by d and (1.2.2) by b

adx+ bdy = de (1.2.3)

bcx+ bdy = bf. (1.2.4)

1.2. Cramers rule 5

Equation (1.2.3) equation (1.2.4) adx bcx = de bf x(ad bc) = de bf x = de bf

ad bc (1.2.5)

In determinant form, (1.2.5) can be written as

x =

e bf d

a bc d

(1.2.6)or

x =1, (1.2.7)

where

1 =

e bf d and = a bc d

.The determinant contains the coefficients of x and y, while 1 is obtained from byreplacing its first column with the constants on the right-hand sides of (1.2.1) and (1.2.2).

Similarly, it can be shown that

y =2, (1.2.8)

where

2 =

a ec f .

The determinant 2 is obtained from by replacing its second column with the constantson the right-hand sides of (1.2.1) and (1.2.2).

Equations (1.2.7) and (1.2.8) taken together are known as Cramers rule for two unknowns.

Example 1.2.1. For the electric circuit shown, the equations for the electric currents (i1and i2 in amperes) are given:

8i1 3i2 = 2 3i1 + 10i2 = 6.

Solve the equations using Cramers rule for two unknowns.

2V2

5

3 6V5

i2i1


Solution

i1 =1

=

2 36 10

8 33 10 =

(2)(10) (3)(6)(8)(10) (3)(3)

=20 + 18

80 9=271

amps.

i2 =2

=

8 23 6 8 33 10 = 4271 amps1.2.2 Cramers rule for three unknowns

To solve

a1x+ b1y + c1z = k1, (1.2.9)

a2x+ b2y + c2z = k2, (1.2.10)

a3x+ b3y + c3z = k3, (1.2.11)

we can first eliminate z from the first two equations by multiplying the first by c2 and thesecond by c1 and then subtracting. We then obtain the equation

(a1c2 a2c1)x+ (b1c2 b2c1)y = k1c2 k2c1. (1.2.12)

Then we can eliminate z from equations (1.2.10) and (1.2.11) by multiplying (1.2.10) by c3and (1.2.11) by c2 and then subtracting. We then obtain the equation

(a2c3 a3c2)x+ (b2c3 b3c2)y = k2c3 k3c2 (1.2.13)

Equations (1.2.12) and (1.2.13) can be solved for x and y using Cramers rule for twounknowns, giving

x =(k1c2 k2c1)(b2c3 b3c2) (k2c3 k3c2)(b1c2 b2c1)(a1c2 a2c1)(b2c3 b3c2) (a2c3 a3c2)(b1c2 b2c1) (1.2.14)

y =(a1c2 a2c1)(k2c3 k3c2) (k1c2 k2c1)(a2c3 a3c2)(a1c2 a2c1)(b2c3 b3c2) (a2c3 a3c2)(b1c2 b2c1) . (1.2.15)

If we expand the terms in the numerator (top line) and denominator (bottom line) of(1.2.14), we obtain

x =c2(k1b2c3 k1b3c2 k2b1c3 + k2b3c1 + k3b1c2 k3b2c1)c2(a1b2c3 a1b3c2 a2b1c3 + a2b3c1 + a3b1c2 a3b2c1)

which simplifies to

x =k1b2c3 k1b3c2 k2b1c3 + k2b3c1 + k3b1c2 k3b2c1a1b2c3 a1b3c2 a2b1c3 + a2b3c1 + a3b1c2 a3b2c1 .

1.2. Cramers rule 7

In determinant form, this can be written as

x =1,

where

=

a1 b1 c1a2 b2 c2a3 b3 c3

and

1 =

k1 b1 c1k2 b2 c2k3 b3 c3

.The determinant 1 is obtained by replacing the first column of by the constant termson the right-hand side of the simultaneous equations.

Similarly, y, as obtained in (1.2.15), can be written in determinant form, and if the ex-pressions in (1.2.14) and (1.2.15) are substituted into (1.2.9), we obtain an expression forz which can be written in determinant form.

The determinant expressions for y and z are

y =2

and z =3

with

2 =

a1 k1 c1a2 k2 c2a3 k3 c3

and

3 =

a1 b1 k1a2 b2 k2a3 b3 k3

.The determinant 2 is obtained by replacing the second column of by the constant termson the right-hand side of the simultaneous equations, while 3 is obtained by replacingthe third column of by the constant terms on the right-hand side of the simultaneousequations.


Example 1.2.2. For the electric circuit shown below, the equations for the mesh currents(i1, i2, and i3, in amperes) are given:

10i1 3i2 i3 = 33i1 + 14i2 2i3 = 6i1 2i2 + 6i3 = 4.

Solve the equations using Cramers rule.

2

3

35

4

42

1

3V

2V 4V

i2

i1

i3

Solution

i1 =1, i2 =

2, i3 =

3,

where

=

10 3 13 14 21 2 6

= 7201 =

3 3 16 14 24 2 6

= 2002 =

10 3 13 6 21 4 6

= 202and

3 =

10 3 33 14 61 2 4

= 446Hence,

i1 =1

=200

720 0.278 amps,

i2 =2

=202720

0.281 amps,

i3 =3

=446

720 0.619 amps.

1.3. Matrix inversion 9

Exercises

1. Using Cramers rule for two unknowns solve

(a) 2x+ 3y = 8; 4x+ 5y = 14.

(b) 2x y = 5; 3x+ 2y = 4.2. For the electric circuit shown, the equations for the mesh currents (i1, i2, and i3, in

amperes) are given by

14i1 4i2 5i3 = 6,4i1 + 19i2 3i3 = 15,5i1 3i2 + 20i3 = 3.

Solve the equations using Cramers rule, giving your answers correct to 3 significantfigures.

3

4

126

6

32

5

6V

12V 3V

i2

i1

i3

Answers.1.(a) x = 1, y = 2; (b) x = 2, y = 1.2. 0.262 amps, 0.717 amps, 0.108 amps.

1.3 Matrix inversion

1.3.1 Matrix inverse

If A is a square matrix, and there exists a second matrix of the same order as A whichsatisfies the matrix equations

AB = I and BA = I,

where I is a unit matrix of the same order as A, then B is said to be the inverse of A andis written as A1.


1.3.2 Inverse of a 22 matrix

Let A be a 2 2 matrix, with

A =

[a bc d

]and A1 =

[e fg h

].

As AA1 = I, hence, [a bc d

] [e fg h

]=

[1 00 1

]giving [

ae+ bg af + bhce+ dg cf + dh

]=

[1 00 1

].

Hence,ae+ bg = 1, af + bh = 0, ce+ dg = 0, cf + dh = 1.

Solving the first and third equations for e and g by Cramers rule for two unknowns andthe second and fourth equations for f and h by Cramers rule for two unknowns gives

e =d

ad bc , f =b

ad bc , g =c

ad bc , h =a

ad bc .

As detA = ad bc, we findA1 =

1

detA

[d bc a

].

If detA = 0, e, f , g and h are not defined and hence A1 does not exist.

Using matrix multiplication, it can be checked that

1

detA

[d bc a

]A = I.

Example 1.3.1. If A =[8 24 3

]find A1.

Solution.

A1 =1

(8)(3) (2)(4)[

3 24 8

]=

1

16

[3 24 8

].

1.3.3 Inverse of a 33 matrix

The inverse of a 3 3 matrix can be found using the adjoint-determinant method. To findthe adjoint of a 33 matrix one must first find the minors of its elements and the cofactorsof its elements. The method is set out below.

Let the 3 3 matrix A be given by

A =

a11 a12 a13a21 a22 a23a31 a32 a33

.


Then

detA =

a11 a12 a13a21 a22 a23a31 a32 a33

,e.g. if

A =

1 2 37 5 64 8 9

then

detA =

1 2 37 5 64 8 9

= 27.The minor of element aij in matrix A is written as Mij . Mij is defined as the 2 2determinant obtained by deleting row i and column j of det A.

For example, for the matrix A given above

M11 =

5 68 9 = 3; M12 = 7 64 9

= 39; M32 = 1 37 6 = 15, etc.

The cofactor of element aij in matrix A is written as Cij . Cij is defined by Cij = (1)i+jMij .The (1)i+j term has the following pattern:

+1 1 +11 +1 1+1 1 +1

.This pattern is also written as

+ + + + +

.For example, for the matrix A above

C11 = +M11 = +

5 68 9 = 3; C12 = M12 = 7 64 9

= 39;C32 = M32 =

1 37 6 = 15; etc.

The matrix C, which is called the matrix of cofactors of A, has the cofactors as its elements,i.e.

C =

C11 C12 C13C21 C22 C23C31 C32 C33

= M11 M12 M13M21 M22 M23M31 M32 M33

The adjoint matrix of A, written adjA, is defined by adjA = CT (i.e. adjA is the transposeof C, i.e. the matrix obtained by changing Cs rows into columns), i.e

adjA =

C11 C21 C31C12 C22 C32C13 C23 C33

.


It can be shown that A1 = adjA/ detA, which we will do in the following.

Proof.

AadjA

detA=

1

detA

a11 a12 a13a21 a22 a23a31 a32 a33

C11 C21 C31C12 C22 C32C13 C23 C33

.The matrix product on the right-hand side contains elements of two types:

(i) e.g. a11C11 + a12C12 + a13C13 = detA,

(ii) e.g. a21C11 + a22C12 + a23C13 = 0.

(This can be checked by multiplying out the individual terms.)

Hence,

AadjA

detA=

1

detA

detA 0 00 detA 00 0 detA

=1 0 00 1 0

0 0 1

= Iand so adjA/ detA = A1.

Example 1.3.2. Given that

A =

1 2 37 5 64 8 9

find C, adjA, detA and A1.

Solution.

C =

+

5 68 9 7 64 9

+ 7 54 8

2 38 9

+ 1 34 9 1 24 8

+

2 35 6 1 37 6

+ 1 27 5

C =

3 39 366 3 03 15 9

adjA = CT =

3 6 339 3 1536 0 9

detA = 1

5 68 9 2 7 64 9

+ 3 7 54 8

= (5)(9) (6)(8) 2{(7)(9) (4)(6)}+ 3{(7)(8) (5)(4)}= 45 48 2{63 24}+ 3{56 20}= 30 2(39) + 3(36)


= 3 78 + 108= 27

A1 =adjA

detA=

1

27

3 6 339 3 1536 0 9

=

3

27

6

27 3

27

3927

327

15

2736

27

0

27 9

27

=1

9

2

91

9

1391

9

5

94

30 1

3

.

1.3.4 Solving matrix equations by matrix inversion

Example 1.3.3. For the circuit shown below, the currents i1, i2, and i3 (in amperes)satisfy these equations:

10.5i1 + 4i3 = 12.5,i1 i2 + i3 = 0,4i2 + 4i3 = 0.

Solve these equations using matrix inversion.

8 0.5

2

41.5

2.5

12.5V

i2

i3

i1

Solution. These equations can be written in matrix form thus:10.5i1 + 0i2 + 4i31i1 1i2 + 1i30i1 + 4i2 + 4i3

=12.50

0

.


The left-hand matrix can be written as a product of two matrices, giving10.5 0 41 1 10 4 4

Matrix A

i1i2i3

Matrix X

=

12.500

Matrix B

i.e. AX = B.

To solve for X, premultiply both sides by A1

A1AX = A1B IX = A1B X = A1B

i1i2i3

=0.08 0.16 0.040.04 0.42 0.145

0.04 0.42 0.105

A1, as obtained bythe adjdet method

12.500

= 10.5

0.5

i1 = 1 amp, i2 = 0.5 amp and i3 = 0.5 amp.

Exercises

1. Find the inverses of the following matrices and in each case show that AA1 = I andA1A = I:

(a) A =

[8 35 2

]

(b) A =

[u 35 v

]

(c) A =

1 2 32 1 42 5 1

(d) A =

1 6 97 5 23 4 8

2. Using matrix inversion, solve

(a) 2x+ 3y = 8; 4x+ 5y = 14.

(b) 2x y = 5; 3x+ 2y = 4.


3. For the electric circuit shown, the equations for the mesh currents (i1, i2, and i3, inamperes) are given by

14i1 4i2 5i3 = 6,4i1 + 19i2 3i3 = 15,5i1 3i2 + 20i3 = 3.

3

4

126

6

32

5

6V

12V 3V

i2

i1

i3

Solve the equations using matrix inversion, and give your answers correct to 3 signif-icant figures.

4. For the circuit shown below, the currents i1, i2, and i3 (in amperes) satisfy theseequations:

i1 + i2 + i3 = 0,

6i1 8i3 = 24,7i2 2i3 = 9.

6 12

12V

8 16

24V

i2

i3

i1

Solve the equations using matrix inversion, giving all answers correct to 3 decimalplaces.


Answers.

1. (a)

[2 35 8

]

(b)1

uv 15[v 35 u

]

(c)1

7

19 17 116 5 28 9 5

(d)1

151

32 12 3350 19 6113 14 37

2. (a) x = 1, y = 2

(b) x = 2, y = 13. 0.262 amps, 0.717 amps, 0.108 amps.4. 1.309 amps, 0.709 amps, 2.018 amps.

Module 2Vectors

2.1 The vector (cross) product

If a and b are two vectors, with the angle between them (see Figure 2.1.1), then the vector(cross) product of a and b is written as a b and is defined bya b= (ab sin )n, (2.1.1)

ab

Figure 2.1.1

where n is a unit vector perpendicular to both a and b, and a and b are the magnitudesof a and b respectively. The direction of n is given by the right-hand rule; i.e. if we takea to be pointing in the direction of the thumb on the right hand, and b to be pointing inthe direction of the first finger on the right hand, then n is pointing in the direction of thesecond finger on the right hand (see Figure 2.1.2).

Thumb

First finger

Second finger

n

ab

ab

ab

Figure 2.1.2

17

18 Module 2. Vectors

(Picture the vector b as pointing into the page, while in Figure 2.1.3 i points out of thepage.)For example (see Figure 2.1.3),

i j = (1)(1)(

sinpi

2

)k = (1)(1)(1)k

= k

i j

k

Figure 2.1.3

Similarly it can be shown that

j k = i, k i= j, j i= k, k j = i, i k = j,and

i i= 0 = j j = k k.Here the negative signs follow from the right-hand rule.

Ifa = a1i+ a2j+ a3k

andb= b1i+ b2j+ b3k,

then

a b= (a1i+ a2j+ a3k) (b1i+ b2j+ b3k)= a1b1i i+ a1b2i j+ a1b3i k

+ a2b1j i+ a2b2j j+ a2b3j k+ a3b1k i+ a3b2k j+ a3b3k k

Using the previous results for the cross products of the unit vectors, this simplifies to

a b= a1b10+ a1b2k+ a1b3(j) + a2b1(k) + a2b20+ a2b3(i)+ a3b1j+ a3b2(i) + a3b30

= (a2b3 a3b2)i+ (a3b1 a1b3)j+ (a1b2 a2b1)kThis can be rewritten in determinant form as

a b=i j ka1 a2 a3b1 b2 b3

(2.1.2)Note that here we used (a1b3 a3b1) = (a3b1 b3a1). It is usually more useful to use thecross product in the form given in (2.1.2) than in the form given in (2.1.1).

2.1. The vector (cross) product 19

Example

Given that a = 2i+ 3j+ 4k and b= i+ 5j+ 6k, find1. a b and2. a unit vector perpendicular to both both a and b.

Solution.

1.

a b=i j k2 3 41 5 6

= i3 45 6

j2 41 6

+ k2 31 5

= i(3 6 4 5) j(2 6 1 4) + k(2 5 3 1)= 2i 8j+ 7k

2. As a b= |a b|n and n is a unit vector perpendicular to both both a and b, n is ananswer (n is also an answer, as it is a unit vector perpendicular to both a and b).n = a b|a b| =

2i 8j+ 7k(2)2 + (8)2 + (7)2 =

2i 8j+ 7k117

.

Exercises

1. Find the vector product a b where(a) a = 2i j b= 3i+ 4j(b) a = 3i j k b= i+ j k

2. Noting that the area of a triangle is 0.5(ab sin C), which is 0.5|CACB|, find the areaof a triangle whose vertices have position vectors:

(a) 0, i+ j+ k, 2i j+ 3k(b) i+ j+ k, 2i j 3k, 3i+ 2j+ 4k

3. Simplify:

(a) a (2a+ b)(b) (a+ b) (2a b)(c) a (a b)

4. If a = 2i+ 4j+ 3k, b= 3i 6j 4.5k, c= 1.5i+ 9j+ 3k, and d = i+ 6j+ 2k, showthat (a c) + (b d) = 05. Find a unit vector perpendicular to the plane containing the vectors 2i+ 4j 3k andi j k6. If a = 3i+ 6j+ 2k and b= 6i+ 2j+ 3k, find

(a) a unit vector parallel to 2a+ b, and(b) a unit vector perpendicular to both a and b

7. P, Q and R are points in space with coordinates (6, 3,4), (7, 5,5) and (8, 4,3),respectively. Find the area of the triangle PQR.


8. (a) Given that a = 2i+ 4j+ 4k, b= 5i+ 2j+ 6k, and c= 7i+ 2j, find(i) a b

(ii) a

(iii) b

(iv) (the angle between a and b)(v) a b(vi) (a b) c

(b) Given that v = 2i+ 2j+ k and w = 3i j+ k, find a unit vector perpendicularto both v and w9. A charge q = 151020 coulombs moves in a magnetic fieldB = 300i+100j500k teslawith a velocity v = i+ j 2k m/s. Find the resulting force F on that charge giventhat F = qvB .

Answers

1. (a) 11k (b) 2i+ 2j+ 4k2. (a) 0.5

26 (b) 2.5

6

3. (a) a b (b) 3(a b) (c) 05. 1

86(7i+ j+ 6k)

6. (a)15

(2j+ k) (b) 17(2i 3j+ 6k)7. 1.5

3

8. (a) (i) 42 (ii) 6 (iii)

65 (iv) 2945 (v) 16i+ 8j 16k (vi) 128(b)

174

(3i j+ 8k)9. 15 1018(3i+ j+ 2k) newtons

2.2. Angular velocity 21

2.2 Angular velocity

Suppose that a point P travels along a circular path with point O the centre of the circleof radius r. Suppose that at time t = 0 seconds, P is at position P0 and that at time t, Pis at position Pt. Suppose that angle P0OPt = radians and arc length P0Pt = s. Thens = r.

O

r

Pt

s

P0

Figure 2.2.1

Suppose that P travels with a constant speed v and with angular speed = /t in radiansper second. Then

v =s

t= r

t= r

= r.

Example

Find the speeds and angular speeds for the second and minute hands of a clock, theirlengths being 5 and 8 cm respectively.

Solution. The second hand travels 2pi radians in 1 minute, i.e.2pi

60radians in 1 second,

i.e. =pi

30radians/s.

Using v = r,

v =

(pi

30

)5 =

pi

6cm/s.

The minute hand travels 2pi radians in 1 hour, i.e.2pi

60radians in 1 minute, i.e.

2pi

60 60radians in 1 second, i.e. =

2pi

60 60 =pi

1800radians/s. Using v = r,

v =

(pi

1800

)8 =

pi

225cm/s.

Angular velocity as a vector

Consider a rigid body (e.g. a wheel) rotating with an angular speed of radians/s about afixed axis LM, which passes through a fixed point A.


The point P on the rigid body, with position vector r relative to A, will move in a circle(perpendicular to ML) whose centre N is on LM, with speed v = (NP) = (AP) sin , i.e.v = r sin (2.2.1)

The velocity is a vector v with magnitude v and direction that of the tangent to the circleat P, with the tangent pointing in the direction of travel.The angular velocity is a vector with magnitude and direction along the axis of rotationsuch that

v = r, (2.2.2)i.e. is in the direction of the vector LM.Note that the magnitude of v given by (2.1.2) is consistent with the definition (2.1.1).

L

A

N

M

P

rv

Figure 2.2.2

Example

A rigid body rotates with an angular velocity of 4 radians/s about an axis in the directionof i+ 2j+ 4k and passing through the point A(3, 5, 8). Find the (linear) velocity v of thepoint P(4, 6, 10) of the body.Solution.

r = AP = OPOA= 4i+ 6j+ 10k (3i+ 5j+ 8k)= (4 3)i+ (6 5)j+ (10 8)k= i+ j+ 2k = = 4

i+ 2j+ 4k12 + 22 + 42

=421

(i+ 2j+ 4k).

2.3. Moment of a force (torque) 23

Hence,

v = r = 421i j k1 2 41 1 2

= 421 (2j k).The reader should be aware that it is always wise to distinguish between scalar and vectorquantities, hence the use of the words speed/angular speed and velocity/angular velocity.In practice, it is often the case that people get a bit sloppy and talk about velocity/angularvelocity when they actually mean the scalar quantity. In the case of acceleration, un-forunately, the same word is used for both the scalar and the vector versions.

Exercises

1. Find the angular speed and speed of the hour hand of a clock, its length being 6 cm

2. The blade of a lawnmower has a rotation rate of 3600 revolutions per minute. Giventhat the blade is 0.3 m. in radius, what is the angular speed and speed at the tip ofthe blade?

3. A rigid body is rotating with an angular velocity 5 radians/s about an axis in thedirection of vector i+ 4j+ 2k and passing through the point A(3, 5, 9). Dimensionsare in metres. Find the linear velocity v of the point P(4, 8, 11) on the body (usev = r).

Answers

1. 1.454 104 radians/s, pi3600

cm/s

2. 120pi radians/s, 113 m/s

3.1021i 521 k m/s

2.3 Moment of a force (torque)

If a force F acts through a point P, which has position vector r, then M = r F , whereM is the moment of the force F about O (see Figure 2.3.1).F

P

O rFigure 2.3.1

The moment of force about a point is a measure of the tendency of the force to rotatean object about the point. If a body in equilibrium is acted upon by several forces, then:(i) the sum of the forces is 0; and (ii) the sum of the moments of the forces about any pointis 0. These relationships can be used to calculate unknown forces (e.g. the forces on thesupports of the body) from known forces (e.g. gravity).


Example

A force of 4 newtons acts through P(2, 3, 5) in the direction of a = 4i+ 7j+ 9k. Find themoment of the force about A(6, 10, 3). Dimensions are in metres.F

Pra

A

Figure 2.3.2

Solution.

r = AP = OPOA= 2i+ 3j+ 5k (6i+ 10j+ 3k)= 4i 7j+ 2k

F = 4a = 4 4i+ 7j+ 9k42 + 72 + 92 = 4146 (4i+ 7j+ 9k).Hence,

M = r F = 4146i j k4 7 24 7 9

= 4146 (77i+ 44j).Exercises

1. A force F with magnitude 4 newtons acts through P(1, 2, 4) in the direction of a =2i+3j+5k. Find the moment of the force about A(0, 7, 4). Dimensions are in metres.2. A cable AB helps support a tower. The point A is at the top of the tower, while B

is the point on the ground to which the cable is attached. The origin is the centreof the base of the tower. Given that A has the coordinates (0, 0, 80) and B has thecoordinates (0, 40, 0), with both coordinates in metres, and the tension in the cable is4 kN, find

(a) the tension in vector form, and

(b) the moment about O due to the force exerted by the cable at A.

Answers

1. 438

(25i+ 5j 13k) Nm2. (a)

45

(j 2k) kN(b) 320

5i kNm

2.4. Equation of a straight line in three dimensions 25

2.4 Equation of a straight line in three dimensions

Let a straight line in the direction of a, where a = a1i+ a2j+ a3k pass through a fixedpoint C(c1, c2, c3), and let P(x, y, z) be any point on the line (see Figure 2.4.1).z

y

x

P(x,y,z)C(c1,c2,c3)

ac r

Straightline O

Figure 2.4.1

By the triangle rule for vector addition,

OP =

OC +

CP.

That is,

r = c+ (a multiple of a)= c+ ta. (2.4.1)

Equation (2.4.1) is the vector equation of a straight line in three dimensions. The term t

is called a parameter. AsCP = ta, if, for example, CP = 3a, then t = 3.

If the vectors c and a in (2.4.1) are each expanded out as a sum of vector components,(2.4.1) becomesr = c1i+ c2j+ c3k+ t(a1i+ a2j+ a3k). (2.4.2)

Equation (2.4.2) is also the vector equation of a straight line in three dimensions. Similarly,if r is expanded out as a sum of vector components, then (2.4.2) becomes

xi+ yj+ zk = c1i+ c2j+ c3k+ t(a1i+ a2j+ a3k). (2.4.3)If the left-hand and right-hand coefficients of i, j, and k in (2.4.2) are respectively equated,we obtain

x = c1 + ta1

y = c2 + ta2

z = c3 + ta3

(2.4.4)

Equations (2.4.4) are together called the scalar parametric equations of the straight line.They can be rewritten as

t =x c1a1

t =y c2a2

t =z c3a3

,

(2.4.5)


as long as the components of a are non-zero. Equations (2.4.5) imply thatx c1a1

=y c2a2

=z c3a3

(2.4.6)

If any component of a is zero, then the corresponding coordinate x, y or z is constant (givenby the corresponding component of c). The equations in (2.4.6) are known as the equa-tions without parameter of the straight line (or the Cartesian equations or the symmetricequations).

Example

1. Find the equation of the straight line that passes through the point C(4, 2, 3) andwhich is in the direction of the vector a = 4i 2j+ 3k in(a) vector form

(b) scalar parametric form, and

(c) in Cartesian form (without parameter).

2. Draw a graph of the straight line.

Solution.

1. As C has the coordinates (4, 2, 3), c= 4i+ 2j+ 3k.(a) Using (2.4.1), the vector equation is

r = 4i+ 2j+ 3k+ t(4i 2j+ 3k).(b) Using (2.4.4), the scalar parametric equations are

x = 4 + 4t

y = 2 2tz = 3 + 3t.

(2.4.7)

(c) Using (2.4.6), the Cartesian form is

x 44

=y 22 =

z 33

.

2. To graph the straight line, first set up a table of x, y, and z values for points C and D,corresponding respectively to t = 0 and t = 1. From (2.4.7) we obtain Table 2.4.1):

Table 2.4.1

Point C B

t 0 1x 4 8y 2 0z 3 6


The graph is shown in Figure 2.4.2.

z

y

x

DC

4

8

2

3

6O

Figure 2.4.2

The angle between two lines

Two lines have vector equations r = c+ ta and r = d+ tb (see Figure 2.4.3). The angle between the two lines is found using the dot product, i.e.cos =

a bab

C D

O

ac

bd

Figure 2.4.3

Example

Given two straight lines with equations

r = i+ 2j k+ t(i+ 3j+ 2k)and

r = 3i+ 4j+ 7k+ t(i+ 2j+ 3k),find , the angle between the two lines.


Solution.

cos =(i+ 3j+ 2k) (i+ 2j+ 3k)

12 + 32 + 22

12 + 22 + 32

=(1)(1) + (3)(2) + (2)(3)

14

14

=13

14

= = cos1 1314

2147.

Intersection of two lines

Example

An underground pipeline, P1, passes through the points A(0, 8,10) and B(2, 10,12).A second pipeline, P2, yet to be constructed, is designed to pass through the pointsC(1, 8,11) and D(7, 16,17).

1. Find the equation of the line P1 (use t as the parameter) in vector and parametricform.

2. Find the equation of the second pipeline (use s as the parameter) in vector and para-metric form.

3. The designers are concerned that the two pipelines might intersect. Is their concernjustified, and if so, where would they intersect?

Solution.

1. Using (2.4.1), P1 has vector equation

r = OA + t(AB)=OA + t(

OBOA)

= 0i+ 8j 10k+ t(2i+ 10j 12k (0i+ 8j 10k))= 0i+ 8j 10k+ t(2i+ 2j 2k). (2.4.8)

Equation (2.4.8) implies that

xi+ yj+ zk = 2ti+ (8 + 2t)j+ (10 2t)k. (2.4.9)Equating respectively the i, j, and k components on the left- and right-hand sidesof (2.4.9) gives

x = 2t (2.4.10)

y = 8 + 2t (2.4.11)

z = 10 2t, (2.4.12)which are the parametric equations of P1.


2. From (2.4.1), P2 has vector equation

r = OC + s(CD)=OC + s(

ODOC)

= 1i+ 8j 11k+ s(7i+ 16j 17k (1i+ 8j 11k))= 1i+ 8j 11k+ s(6i+ 8j 6k). (2.4.13)

Equation (2.4.13) implies that

xi+ yj+ zk = (1 + 6s)i+ (8 + 8s)j+ (11 6s)k (2.4.14)Equating respectively the i, j, and k components on the left- and right-hand sides of(2.4.14) gives

x = 1 + 6s (2.4.15)

y = 8 + 8s (2.4.16)

z = 11 6s, (2.4.17)which are the parametric equations of P2.

3. The two pipelines will intersect if there is a point where their x, y, and z coordinatesare the same. Equating (2.4.10) and (2.4.15), (2.4.11) and (2.4.16), and (2.4.12) and(2.4.17), we obtain

2t = 1 + 6s

8 + 2t = 8 + 8s

10 2t = 11 6s.These equations can be rewritten as

6s+ 2t = 1 (2.4.18)8s+ 2t = 0 (2.4.19)6s 2t = 1. (2.4.20)

These are three equations with two unknowns, which, if they are consistent, can besolved using Gaussian elimination. We row reduce the matrix of coefficients from theequations 6 2 | 18 2 | 0

6 2 | 1

to get 1 0 | 120 1 | 2

0 0 | 0

which tells us that there is a consistent solution s = 12 , t = 2 and so the pipelines doindeed intersect.Substituting this value for s into (2.4.15), (2.4.16), and (2.4.17) (or the value of t into(2.4.10), (2.4.11), and (2.4.12)) gives x = 4, y = 12, z = 14, the coordinates of thepoint of intersection of the two lines. If the pipelines had not intersected, then wewould have found no consistent solution for s and t.


Exercises

1. Find the equation of the straight line through (1, 2, 3) with direction 4i j+ k in(a) vector form,

(b) scalar parametric form, and

(c) without parameter (i.e. Cartesian form).

2. Find the vector (parametric) equation of the pipeline passing through (0, 3, 4) and(1, 5, 0).

3. Convert the straight line equation

x 31

=y + 2

2=z 3

4

to vector form.

4. Find the angle between the lines

r = i 2j k+ t(i+ 3j+ 2k)and

r = i j 3k+ t(2i+ j+ 3k)5. An underground pipeline passes through the points (0, 16,30) and (20, 20,32) with

the x- and y-axes in the horizontal surface of the ground. Find the equation of thepipeline.

A second pipeline, yet to be constructed, has been designed to pass through the points(10, 10,32) and (25, 17,33). Find the equation of the second pipeline (use s as theparameter). The designers are concerned that the two pipelines might intersect. Istheir concern justified, and if so, where would they intersect?

Answers

1. (a) r = (1 + 4t)i+ (2 t)j+ (3 + t)k(b) x = 1 + 4t, y = 2 t, z = 3 + t(c)

x 14

=y 21 =

z 31

2. r = ti+ (2t+ 3)j+ (4t+ 4)k3. r = (t+ 3)i+ (2t 2)j+ (4t+ 3)k4. 381248

5. r = 16j 30k+ t(20i+ 4j 2k), r = 10i+ 10j 32k+ s(15i+ 7j k),Yes, at (40, 24,34)2.5 Equation of a plane

Suppose that vector n is normal (i.e. perpendicular) to a plane, with n = ai+ bj + ck.Suppose also that E(e1, e2, e3) is a fixed point on the plane, with

OE = e, and let P(x, y, z)

be any point on the plane, withOP = r. The plane is shown in Figure 2.5.1.

2.5. Equation of a plane 31

z

y

x

ner

Part of(infinite)plane

O

E

P

Figure 2.5.1

The vectorEP = r e.

AsEP is perpendicular to n, then EPn = 0, i.e.

(r e) n = 0= r n e n = 0= r n = e n= r n = d,

where d = e n.Expanding r n in terms of the vector components gives

(xi+ yj+ zk) (ai+ bj+ ck) = d,i.e.

ax+ by + cz = d. (2.5.1)

Equation (2.5.1) is the equation of the plane.

Suppose that p is the perpendicular distance of the plane from the origin, with Q the footof the perpendicular from O to the plane, i.e. p = OQ.

There are two posssibilities: (i) n is in the same direction as OQ, or (ii) n is in the oppositedirection to

OQ.


The first possibility is illustrated in Figure 2.5.2, with the plane shown side-on.

z

yx

ne

Plane(side on)

OE

Q

p

Figure 2.5.2

p = OQ

= e cos

= ee ne|n|

=e n|n|

= e n= (e1i+ e2j+ e3k)

( ai+ bj+ cka2 + b2 + c2

)=ae1 + be2 + ce3a2 + b2 + c2

=d

a2 + b2 + c2(using (2.5.1)).

If n is in the opposite direction to OQ, then d turns out to be negative and it can be shownthatp = d

a2 + b2 + c2.

As p is always positive or 0, the last two formul can be summarised in one formula, i.e.

p =|d|

a2 + b2 + c2. (2.5.2)

Note. Equation (2.5.2) implies that `, the distance between two parallel planes ax+by+cz =d1 and ax+ by + cz = d2, is given by

` =|d1 d2|a2 + b2 + c2

.

Example

Given that 2i+ 3j+ k is normal to a plane which passes through the point (4, 7, 8), findthe equation of the plane and its distance from the origin.


Solution. Using (2.5.1), the equation of the plane is

2x+ 3y + z = d (2.5.3)

As (4, 7, 8) is on the plane, substituting x = 4, y = 7, and z = 8 into (2.5.3) gives

d = (2)(4) + (3)(7) + (1)(8) = 8 + 21 + 8 = 37.

Hence, the equation of the plane is 2x+ 3y + 1z = 37.

Using (2.5.2), the distance of the plane from the origin is

p =|37|

22 + 32 + 12=

3714.

Example

1. Find

(a) the equation of a plane normal to 4i+ 12j+ 3k and a distance 4 from the origin,and(b) the intercepts of the plane with the origin.

2. Sketch the plane.

Solution.

1. (a) Using (2.5.1), the equation of the plane is 4x+ 12y + 3z = d. Using (2.5.2),

4 =|d|

42 + 122 + 32=|d|13

= |d| = 52= d = 52.

Hence there are two planes with a normal 4i+ 12j+ 3k and a distance 4 fromthe origin. The equation of one plane is4x+ 12y + 3z = 52 (2.5.4)

and the equation of the other plane is

4x+ 12y + 3z = 52.(b) For the plane defined by (2.5.4), the plane intercepts (cuts) the x-axis where

y = 0 and z = 0, and hence, by (2.5.4),

4x = 52 = x = 13.The plane cuts the y-axis where x = 0 and z = 0, and hence, by (2.5.4),

12y = 52 = y = 133.

The plane cuts the z-axis where x = 0 and y = 0, and hence, by (2.5.4),

3z = 52 = z = 523.


2. The intercepts can be sketched on a 3D diagram as shown in Figure 2.5.3. These threepoints can be joined together (two at a time) by straight lines. The three straightlines form a triangle, which is part of the plane.

z

y

x

O 52/3

13/3

13

Figure 2.5.3

The angle between two planes

Example

Find the angle between the planes 4x+ 5y + 12z = 3 and x+ 3y + 2z = 8.

Solution. Figure 2.5.4 shows a diagram of the planes side on, with the first plane called P1,with normal n1, and the second plane called P2, with normal n2.

pi2 P1

P2

n1 n 2

Figure 2.5.4

The angle is defined as the angle between the two planes. From the diagram, it can beseen that is also the angle between the normals. The angle between the normals can befound using the dot product, i.e.

cos =n1 n2|n1||n2| =

(4i+ 5j+ 12k) (i+ 3j+ 2k)42 + 52 + 122

12 + 32 + 22

=(4)(1) + (5)(3) + (12)(2)

185

14

=4 + 15 + 24

185

14

=43

185

14

= = cos1(

43185

14

) 3220.


Finding the equation of a plane given three points on the plane

Example

Find the equation of the plane which passes through the points A(1, 1, 1), B(0, 1, 2), andC(1, 4,1).

Solution. A diagram of the plane is shown in Figure 2.5.5, with n being defined byn = ABAC.

nPlane

C

AB

Figure 2.5.5

Thus, n is a vector that is normal to the plane, andAB =

OBOA = 0i+ 1j+ 2k (1i+ 1j+ 1k)

= 1i+ 0j+ 1kAC =

OCOA = 1i+ 4j 1k (1i+ 1j+ 1k)

= 2i+ 3j 2k.Hence,

n = ABAC =i j k1 0 12 3 2

= i0 13 2

j1 12 2

+ k1 02 3

= 3i 4j 3k.

Hence, the equation of the plane is 3x 4y 3z = d. As (1, 1, 1) is on the plane,

d = 3(1) 4(1) 3(1)= 10.

Hence, the equation of the plane is 3x 4y 3z = 10.


Intersection of a line and plane

Example

Given a planex+ y + 2z = 37 (2.5.5)

and the straight line

x = 2 + 3t, y = 1 + t, z = 1 + 2t, (2.5.6)

find

(a) the coordinates of the point of intersection, and

(b) the angle between the plane and the line.

Solution. The plane and straight line are shown in Figure 2.5.6.

Plane

Point ofintersection

Straight line

Figure 2.5.6

(a) Substituting (2.5.6) into (2.5.5) gives

(2 + 3t) + (1 + t) + 2(1 + 2t) = 37

= 5 + 8t = 37= 8t = 32= t = 4 (2.5.7)

Substituting (2.5.7) into (2.5.6) gives

x = 14, y = 5, z = 9.

(b) A side-on view of the plane is shown in Figure 2.5.7. The angle between the plane andthe straight line is and the angle between the normal to the plane and the straightline is . As the two angles add up to a right angle, = pi/2 . Given that thestraight line is in the direction of the vector a, and n is a normal to the plane, isfound using the dot product, i.e.

cos =n a|n||a| . (2.5.8)


Plane (side on)

Straight line

an

Figure 2.5.7

To find a (the direction of the line), substitute (2.5.6) into the the vector equation ofa point on the line, r = xi+ yj+ zk, which givesr = (2 + 3t)i+ (1 + t)j+ (1 + 2t)k

= (2i+ j+ k) + t(3i+ j+ 2k). (2.5.9)As the equation of the line is of the form r = c+ ta, comparison with (2.5.9) tells usthat

a = 3i+ j+ 2k. (2.5.10)As the equation of the plane is x+ y + 2z = 37, the normal is

n = i+ j+ 2k. (2.5.11)Substituting (2.5.10) and (2.5.11) into (2.5.8) gives

cos =(i+ j+ 2k) (3i+ j+ 2k)12 + 12 + 22

32 + 12 + 22

=884

= = cos1 884

291221

and

90 291221 = 604739.

Intersection of planes

Example

In each of the problems given below, find whether and how the given planes intersect.

(a) x+ y 2z = 2, 2x y + z = 9, x+ 4y + z = 1(b) x+ 3y + z = 6, x+ 2y z = 5, 3x+ 8y + z = 10(c) x+ 3y + z = 4, 2x+ 2y z = 6,(d) x+ y z = 3, 2x y z = 0, 3x+ 9y 5z = 21


Solution.

(a) The equations are:

x+ y 2z = 22x y + z = 9x+ 4y + z = 1.

We use Gaussian elimination on the matrix of coefficients

M =

1 1 2 | 22 1 1 | 91 4 1 | 1

.We give the row operations here just in this example and leave the reader to workthem out in parts (b) to (d).R2 R2 2R1, along with R3 R3 R1.1 1 2 | 20 3 5 | 13

0 3 3 | 3

.Now R3 R3/3, then swap R2 and R3:1 1 2 | 20 1 1 | 1

0 3 5 | 13

.R1 R1 R2, along with R3 R3 + 3R2:1 0 3 | 30 1 1 | 1

0 0 8 | 16

.R3 R3/8, followed by R2 R2 R3 and R1 R1 + 3R3:1 0 0 | 30 1 0 | 1

0 0 1 | 2

,from which we can read the solution x = 3, y = 1, z = 2. Hence, the planesintersect at the point (3,1, 2). Check: Substitute these coordinates into all three ofthe original equations to make sure that the point (3,1, 2) is on all three planes.

(b) The equations are

x+ 3y + z = 6,

x+ 2y z = 5,3x+ 8y + z = 10.

M =

1 3 1 | 61 2 1 | 53 8 1 | 10


Two simple row operations along with multiplication of rows by 1 bring this to theform 1 3 1 | 60 1 2 | 1

0 1 2 | 8

,from which we see that y + 2z = 1 as well as y + 2z = 8. This is impossible, so theequations are inconsistent. There is no solution, and the three planes do not intersectat a single point. This can happen for a number of reasons. The planes might be(i) all parallel and non-coincident, (ii) two of them parallel and non-coincident, withthe third intersecting both in lines or (iii) a roof-like shape consisting of two slopingsides and a base, i.e. three sides of a triangular cylinder, in which each pair intersectsin a line but there is no common intersection of all three planes.

(c) Number the equations as follows:

x+ 3y + z = 4 (2.5.12)

2x+ 2y z = 6. (2.5.13)

In this case, we have only two equations, each representing a plane in 3D space.

If (2.5.12) was exactly the same as (2.5.13) (or a multiple of (2.5.13)), then we wouldreally have only one plane, and the intersection would consist of all the points in theplane; but that is not what we have here.

If the left-hand side of (2.5.12) was a constant k times the left-hand side of (2.5.13),but the right-hand side of (2.5.12) was not k times the right-hand side of (2.5.13), wewould have two parallel but unequal planes, which would never meet, and there wouldbe no points of intersection; but, again, that is not what we have here.

So we must have two non-parallel planes. Any two non-parallel planes in 3D spacemust intersect along a straight line. We now proceed to find the parametric equationsof that line. Here

M =

(1 3 1 | 42 2 1 | 6

).

Row reduction turns this into (1 0 54 | 520 1 34 | 12

),

which can only be solved to the extent that

x =5

2+

5z

4, y =

1

2 3z

4.

We rename z as the parameter t to obtain the result that the planes intersect in thestraight line whose parametric equations are

x =5

2+

5t

4, y =

1

2 3t

4, z = t.

(d) Proceeding as before,

M =

1 1 1 | 32 1 1 | 03 9 5 | 21

.


A few row operations bring M to the form1 1 1 | 30 3 1 | 60 3 1 | 6

.Note the second and third rows contain the same information. This means that oneof our original equations was redundant, i.e. it was a combination of the other twoequations.

Further reduction gives 1 1 1 | 30 3 1 | 60 0 0 | 0

.In the language of the final section of this chapter, the rank of the matrix is only 2.By continuing the reduction as far as possible to

1 0 23 | 10 1 13 | 20 0 0 | 0

,or by immediately solving then back-substituting, we obtain a solution (with z chosenas the parameter) x = 1 + 2t3 , y = 2 +

t3 , z = t, i.e. we find that the planes intersect

in the straight line whose parametric equations are:

x = 1 +2t

3, y = 2 +

t

3, z = t.

Exercises

1. Find the equation of the plane which passes through (4, 7, 8) and is normal to 2i+3j+k;hence find the planes (perpendicular) distance from the origin.2. Find the equation of the plane normal to the vector 2i+ j 3k and a distance 5 fromthe origin.3. For the plane defined by 4x+ 2y + 4z = 9, find

(a) a unit vector normal to the plane,

(b) the perpendicular distance from the plane to the origin, and

(c) the intercepts of the plane with the axes.

(d) the distance between the plane in (a) and the plane 4x+ 2y + 4z = 6.4. Repeat Question 3(a)(c) for the plane x y + z = 0.5. Which of the planes P1 and P2 below passes closer to the origin?

P1 = {(x, y, z) : 4x+ y 3z = 9};P2 = {(x, y, z) : x+ y + z = 3}.

6. Find the angle between the planes 5x+ y + z = 1 and 3x+ 3y + 3z = 5.

7. Find the equation of the plane which passes through the points (1, 6, 1), (3, 5, 2), and(3, 2, 1).


8. Find the equation of the plane which passes through (3, 0, 4) and is normal to 2i+j3k.9. (a) Find the equation of the plane which passes through the points (1, 1, 0), (1, 0, 1),

and (0, 0, 1).

(b) Graph the plane x+ 2y + z = 4.

10. For the line x = 2 + t, y = 3 t, z = 1 + 2t, and the plane x+ y + 3z = 4, find(a) the point of intersection, and

(b) the angle between the line and the plane.

11. In each of the problems given below, find where the given planes intersect.

(a) x 3y + z = 2 2x+ y + z = 7 x 4y + 2z = 1(b) x+ 4y 3z = 2 x+ 2y + z = 5 x+ 3y z = 4(c) x+ 2y + z = 3 2x+ y z = 4(d) 2x y z = 1 x+ 3y + z = 2

Answers

1. 2x+ 3y + z = 37,3714

2. 2x+ y 3z = 514 or 2x+ y 3z = 5143. (a) 13(2i+ j+ 2k)

(b) 32(c) 94 ,

92 ,

94

(d) 52

4. (a)13

(i j+ k)(b) 0

(c) (0, 0, 0)

5. The distance of P1 is about 1.77. The distance of P2 is about 1.73. Hence, P2 passescloser to the origin.

6. 385633

7. 2x+ y 3z = 58. 2x+ y 3z = 69. (a) y + z = 1

(b) See Figure 2.5.8

z

y

x

O2

4

4

Figure 2.5.8


10. (a) When t = 23 , (43 , 113 ,13) (b) 473629

11. (a) (1, 2, 3)

(b) do not intersect

(c) x = 13(5 + 3t), y =13(2 3t), z = t

(d) x = 17(2t 1), y = 17(5 3t), z = t

2.6 Vector spaces, linear independence, basis and rank

Up to now, we have talked loosely about vectors in 2 and 3 dimensions. When introducinga new entity (such as vectors) in mathematics, it is important to set up a rigorous formalstructure in which to embed that entity. For the recently introduced concept of vectors, weintroduce the idea of a vector space. In engineering mathematics and most areas of physicsand economics, we use the notion of a vector space defined over the real (or possiblycomplex) (see below) numbers. Quantum mechanics in particular, requires use of vectorspaces over the complex numbers. (We will stick to the reals here however.)

We will not go very deeply into the formal description, but rather, we give some briefsketches of how things work. For a deeper treatment the reader is referred to any goodbook on linear algebra.

A vector space over the real numbers is a set V containing elements known as vectors, u,v, w etc., along with rules for combining those vectors with each other using also elementsfrom the real number set. The space is closed under application of the rules in use. Thatmeans that if we use the rules to combine vectors in the space, we just get other vectors inthe space and not something new.

Given u, v V and k R (this is what we mean by over the real numbers), the allowedrules are the sum (e.g. u+ v) and the multiplication by a scalar (e.g. kv).For V to be a vector space, the means by which these rules act must satisfy a number ofrestrictions. For all u, v, w V , all k, l R we require the following eight rules.Four are connected with the addition:

1. Associativity: u+ (v+ w) = (u+ v) + w .2. Existence of null vector: there exists 0 V such that u+ 0 = 0+ u = u.3. Existence of inverse for addition: for every u V , there exists another vector u Vsuch that u+ (u) = u+ u = 0.4. Commutativity: u+ v = v+ u.

Four more are connected with the scalar multiplication:

5. Distributivity1: for u, v V and k R, k(u+ v) = ku+ kv.6. Distribitivity2: (k + l)u = ku+ lu.7. Associativity: (kl)u = k(lu).8. Existence of scalar multiplicative unit (1 R), such that 1u = u1 = u.

2.6. Vector spaces, linear independence, basis and rank 43

You should verify that the 2- and 3-dimensional sets of vectors we have been using to studygeometry do indeed satisfy the conditions to belong to a vector space.

Let u1, u2, u3, u4 V , k1, k2, k3, k4 R.A linear combination of the vectors ui, i = 1 n consists of the sum:

k1u1 + k2u2 + k3u3 + k4u4 + + knun.The set of all vectors that can be constructed using all possible values of the ks is a subspace(we really ought to define that but you can guess roughly what it means) of V and is calledthe space spanned by the set of n vectors um, m = 1 n. It is possible (in fact essential)that there are some sets of vectors that span the whole of the space V .You are already familiar with this spanning concept. Just think of the vectors i and jin 2 dimensions. Any vector in 2 dimensions can be constructed with combinations ofmultiples of these two vectors, so i and j span 2-dimensional space. Similarly, i, j and kspan 3-dimensional space.Back in 2 dimensions, it is clear that we could choose more vectors in addition to i and j,for example i, j and i+ j and these would still span the space. One of the useful thingsabout i and j is that they amount to a minimal number of vectors needed to span thespace. Similar considerations apply to i, j and k in 3 dimensions.In fact we can define the dimension of a vector space to be precisely the number which isthe smallest number of vectors needed to span the space.

There is something else useful about the vectors i, j and k that makes us want to use them.In 2 dimensions again, consider the vector 3i+ 4j.We do not have to express such a vector in terms of i, and j.Lets define a = i+ j, b= i+ 2j. Then 3i+ 4j can be written correctly as 3i+ 4j = 2a+ b(check it). You have to stop and think for a moment to see this though. It is not convenientto use a and b in this way. i and j are much better. The reason for that is that i and j areat right-angles to each other. In vector space language, we say they are orthogonal.In fact they form an orthogonal basis for the 2-dimensional vector space.

The concept of orthogonality merits further investigation, but here we just take a visualapproach and understand it to mean that the vectors are at right angles. Technically,this means that once we have defined the dot product (known as an inner product, a newstructure associated with the vector space), then we have orthogonality of u and v definedby u v = 0. We will not explore this further here.In 3 dimensions, i, j and k are all mutually orthogonal and form an orthogonal basis forthe space.So what do we mean exactly by a basis? Well first we need one more concept: that of linearindependence.

Clearly (by definition of the dimension) we require at least n vectors to span an n-dimensionalvector space. It wont do just to choose any old set of n vectors though.


In 3 dimensions for example, we need 3 vectors. We know that i, j and k work, but whatabout u = i+ j, v = i j, and w = i+ 2j? Clearly these are not good enough to reacheverywhere in 3-dimensional space because they have no k component, i.e. no componentperpendicular to the vectors i and j. We cannot generate any height with u, v and w .The point here is that u, v and w might be 3 vectors but they are not 3 linearly independentvectors. It so happens that 32u 12v = w . (Check it!) w depends on u and v. It can beconstructed from u and v and so is not really a third new vector. Obviously k cannot beconstructed from i and j (as it is orthogonal to both) and it is this extra property thatmakes i, j and k a suitable choice for constructing all vectors in 3 dimensions.Another way of writing the relation between u, v and w above is to say that

32u 12v w = 0.

It is in this form that we define linear independence.

In an n-dimensional vector space V , a set of n vectors v1, v2, v3 vn is linearly independentif and only if when we try to combine them to get 0, the only way is to have all zerocoefficients.That is, the only way c1v1 + c2v2 + c3v3 + c4v4 can equal 0 is to have all the cs equal to 0.If there is any combination that works, in which two or more of the cs are not zero, thenthe vectors are linearly dependent.

Our vectors u, v, w above are linearly dependent while i, j and k are linearly independent.It is this independence property that allows i, j and k to span the 3-dimensional space andbeing 3 vectors in 3 dimensions, this is the minimum number required. The vectors i, jand k form a minimal spanning set for 3-dimensional space and since they are mutuallyorthogonal, they amount to an orthogonal basis for the space.If V is an n-dimensional vector space, then a basis for V isa set of n linearly independent vectors in V which span V .

Note that in n dimensions, a set of more than n vectors is necessarily linearly dependentso its use as a basis would contain redundancy, while any less than n vectors could notspan the space. The basis is best chosen to be exactly n-dimensional and preferably tobe orthogonal too. Given n linearly independent vectors in n dimensions, we can alwaysconstruct an orthogonal set from them. The process is not explored here but if you areinterested then look up the GramSchmidt process. It can get quite complicated if n islarge.

Given a set of vectors, how can we check whether they are linearly independent or not?Sometimes this can be done by inspection.

Example 2.6.1. In 2 dimensions let u =(

11

), v =

(55

).


Solution. Clearly 5u v = 0 and so u and v are linearly dependent.In this notation we can represent i as

(10

)and j as

(01

)from which it can clearly be seen

that the only solution of ai+ bj = a(

10

)+ b

(01

)=

(ab

)=

(00

), is to have a = b = 0.

Hence i and j are linearly independent.In 3 dimensions we can represent i, j and k as i=

100

, j =01

0

, k =00

1

.Example 2.6.2. Decide on the linear (in)dependence of u = 2i+ j k, v = 6i+ 2j andw = i+ k.Solution. Write u, v and w as column vectors, u =

211

, v =62

0

, w =10

1

.Clearly if some combination of these is to make

000

the third entry can only disappearby having equal amounts of u and w .For example

u+ w = 2 + 11 + 01 + 1

=31

0

.Seeing this, it is then obvious that 2u+ 2w v = 0 and the vectors are linearly dependent.As the vector spaces get bigger or we increase the number of vectors in the combination, itbecomes harder to simply see the result. We need a systematic way of going about things.

The process involves solving sets of linear equations, usually by Gaussian elimination.

Example 2.6.3. Determine whether the following set of vectors is linearly independent.

u =13

2

, v = 6110

, w =21

3

.Solution. Combine the vectors with coefficients a, b and c, making the combination zero.

a

132

+ b 6110

+ c21

3

=00

0

.This leads to three linear equations for a, b and c:

a+ 6b 2c = 0,3a+ b c = 0,

2a 10b+ 3c = 0.


These are solved by Gaussian elimination. Notice that the left-hand side of the augmentedmatrix is just formed from the three column vectors. Thus we did not really need to writeout the equations but could have gone straight to the augmented matrix:1 6 2 | 03 1 1 | 0

2 10 3 | 0

Multiply row 1 by 1. Then apply R2 R2 3R1, R3 R3 2R1 to get1 6 2 | 00 19 7 | 0

0 2 1 | 0

.Now multiply row 3 by 12 and swap it with row 2:1 6 2 | 00 1 1

2| 0

0 19 7 | 0

then R1 R1 + 6R2, R3 R3 19R2 to get

1 0 1 | 00 1 1

2| 0

0 05

2| 0

which tells us that c = 0, b = 0 and a = 0.

Thus the only combination of these vectors that comes to zero is the trivial one and thevectors are linearly independent. In general, if one makes up a set of vectors at random (ashappened here), the chances are that they will be independent.

Example 2.6.4. Determine whether the following set of vectors is linearly independent.

u =13

2

, v = 6110

, w = 620

8

.Solution. Perform Gaussian elimination on1 6 6 | 03 1 20 | 0

2 10 8 | 0

.The first four operations are the same as in Example 2.6.3. to get1 6 6 | 00 1 2 | 0

0 19 38 | 0

.


Next take R3 R3 19R1. 1 6 6 | 00 1 2 | 00 0 0 | 0

,followed by R1 R1 + 6R2 to finish with1 0 6 | 00 1 2 | 0

0 0 0 | 0

.This gives us a non-trivial solution for a, b and c as a = 6c, b = 2c, where c can take anyvalue.

This makes sense because if some combination of the vectors combines to zero, then so doesany multiple of that combination.

If we choose, e.g. c = 1, then a = 6 and b = 2 so 6u+ 2v+w = 0. (You should check this.)To conclude the discussion of linear independence, we will view the solution of our equationsa slightly different way.

The working for Examples 2.6.3. and 2.6.4. just considered could instead be written as amatrix equation Mix = 0 (i = 3 or 4), where

x =abc

and M3 =1 6 23 1 1

2 10 3

(Example 2.6.3.) or

M4 =

1 6 63 1 202 10 8

(Example 2.6.4.).

If M has an inverse then in principle we can find it and apply it to these equations to getx = M10 = 0. This says that a, b and c are all 0 and we have linear independence.If M is singular (i.e. no inverse) then the matrix equation cannot be solved this way. Itmeans that there are non-trivial values for a, b and c and the vectors are dependent. To findout what particular combination of them disappears, youd still have to solve the equationsby Gaussian elimination or an equivalent process.

So how do we decide if M is singular or not? It comes down to the determinant. IfDet(M) = 0 then M is singular and has no inverse. Then the vectors are dependent. IfDet(M) 6= 0 then M1 exists and can be used to show that a, b and c are 0 and the vectorsare independent as described above. You should check that the M in Example 2.6.3. issingular and that for Example 2.6.4. is not, by calculating the determinants. This is enoughto decide on the (in)dependence of the vectors, if you dont care about which particularcombination disappears.


Note also that the property of singularity of the matrix is preserved by the valid rowoperations. Thus in Example 2.6.4. we could instead calculate

Det

1 0 60 1 20 0 0

.This is clearly 0 on account of the row of zero entries and so the vectors must be dependent.On the other hand, in Example 2.6.3.,

Det

1 0 10 1 1

2

0 05

2

is easily seen to evaluate to 52 6= 0 and so these vectors are independent.This leads us finally to introduce one more concept that of the rank of a matrix.

The precise definition of rank involves setting up the concepts of row and column subspacesof a matrix. We sidestep this and simply point out that the rank turns out to be a measure ofthe number of linearly independent rows (regarded as vectors) of the matrix. This propertyis preserved through the Gaussian elimination process and so the rank can be decided afterapplying row operations.

Again referring to our Examples 2.6.3. and 2.6.4. we have

Rank(M3) = Rank

1 6 23 1 12 10 3

= Rank

1 0 10 1 1

2

0 05

2

= 3because no row can be completely eliminated.

Here the rank is equal to the dimension of the vectors (they had three components) beingcombined (which is the maximum that it can be). When that happens, the vectors areindependent.

On the other hand

Rank(M4) = Rank

1 6 63 1 202 10 8

= Rank1 0 60 1 2

0 0 0

= 2.This time, the rank is less than the dimension of the vectors and so they are dependent.

Furthermore, the fact that the rank is only one less than the dimension, tells us that thereis only one combination (and its multiples) which disappears.

Example 2.6.5. This is a trivial-looking example but it makes a point.

Consider the vectors

u =11

1

, v =22

2

, w =111

.


Its obvious by looking that they are dependent. If we construct

M5 =

1 2 11 2 11 2 1

then Gaussian elimination brings it to the form1 2 10 0 0

0 0 0

which has rank 1. This is two less than the dimension of the vectors which tells us thatthere is more than one way to combine the vectors to make them disappear. For examplev + w u or just u + 0v + w etc. (Note that linear dependence can allow some of thecoefficients to be 0 as long as not all are.)The rank is a well-defined quantity for all matrices, not only square ones.

If we are testing the independence ofm vectors in n dimensions, then we will be investigatingthe rank of an n m matrix. If m > n (more vectors than their dimension) then theGaussian elimination cannot possibly reach row echelon form. In that case the rank is nand the vectors are dependent. As an example consider the 3 vectors in 2 dimensions, i, jand i+ j. The rank of

(1 0 10 1 1

)is clearly 2. The vectors are dependent.

So we can state the following. If the number of vectors is greater than their dimension,then they are linearly dependent. If m n on the other hand, there is no quick way todecide.

Exercises

1. Determine which sets of vectors of the following form a linearly independent set.Do this by inspection where possible but also, in all cases investigate the appropriateGaussian elimination.When the vectors are dependent, write down a combination of them that simplifies tothe zero vector.

(a) u =10

0

, v =12

0

, w =12

3

,(b) u =

101

, v =12

3

, w =32

5

,(c) u =

100

, v =11

0

, w =21

0

,(d) u =

412

, v =410

2

, w =87

8

,


(e) u =30

4

, v = 51

2

, w =11

3

,(f) u =

201

, v =32

5

, w = 61

1

, t= 702

,

(g) u =

3873

, v

1531

, w =

2126

, t=

1403

,

(h) u =

0022

, v =

3300

, w =

1101

.Based on your results above, write down in terms of i j and k some possible non-standard bases for 2- and 3-dimensional Euclidean space (i.e. the usual 2- and 3-dimensional spaces spanned by i, j or i, j and k).

2. Find the equations of the planes spanned by the following pairs of vectors.

(a) u = 2i j+ k, v = i+ 2j+ 3k,(b) u = i 3j+ 2k, v = i+ 2j+ 2k,(c) u = i+ 2k, v = i j+ 3k.

3. Given u = i+ 2k, v = i j+ 3k, w = i+ 4j and x = i+ j k, show that u, v andw could be used as a basis for 3-dimensional space and express x in terms of u, v andw . Show that this new basis is not orthogonal.Answers

1. We take the coefficients of the vectors in the linear combinations to be as many asneeded of a, b, c and d respectively.Probably the only ones that are easy to see by inspection are (a), (c), (f) and (h).For part (a)

au+ bv+ cw =a+ b+ c2b+ 2c

3c

.This can only be 0 if c = 0. But then that requires b = 0 also. Finally, both of theseresults force a = 0. Hence the vectors are linearly independent as a, b and c are all 0.In part (c), we have three vectors in 3 dimensions, however, none of them has any kcomponent. That means that effectively, they are 3 vectors in 2 dimensions and somust be linearly dependent as 3 > 2.In (f) the vectors are dependent for an analogous reason, 4 vectors in 3 dimensions.Part (h) is similar to (a). The vectors are independent. The coefficient of u must be0 because of the 2 in the 3rd position. But that in turn means that w must have zero


coefficient because of the 1 in the 4th position. That just leaves v which must havezero coefficient as a consequence.So now lets look at the Gaussian elimination and rank.

(a) M =(1 1 10 2 20 0 3

). This is already in row echelon form (thats why the answer was

obvious at sight) but we could reduce it further to(1 0 00 1 00 0 1

). In both forms it is

clear that the rank is 3 and the vectors must be independent.

(b) M =(1 1 30 2 21 3 5

). Row operations can bring this to either of the forms

(1 1 30 1 10 0 0

)or(

1 0 20 1 10 0 0

). Either way, the rank is 2 and the vectors are linearly dependent with

any multiple of one relation between them. The 2nd form tells us easily thatb = c and a = 2c. So taking c = 1 (any choice is equally valid), we see that2u+ v w = 0.

(c) M =(1 1 20 1 10 0 0

). Its already reduced as far as we need though we could go one step

further:(1 0 10 1 10 0 0

). The rank is 2. The vectors are dependent and a = c, b = c.

Choose c = 1 so u+ v w = 0.(d) M =

(4 4 81 10 72 2 8

). M can be row reduced to

(1 0 30 1 10 0 0

). The rank is 2 so the vectors

are dependent. a = 3c, b = c. Take c = 1 so 3u+ v w = 0.(e) M =

(3 5 10 1 14 2 3

). M can be reduced all the way to the unit matrix I3 so the rank

is 3 and the vectors are linearly independent.

(f) M =(2 3 6 7

0 2 1 01 5 1 2

). A little less pleasant than the earlier cases. Reduce to the

rank 3 matrix 1 0 0 79

29

0 1 03

29

0 0 16

29

.Hence these four 3-dimensional vectors are dependent (as we knew they had tobe) and by choosing d = 29 we get a = 79, b = 3, c = 6 and so 79u 3v6w + 29t= 0. It works. Check it!

(g) M =

(3 1 2 18 5 1 47 3 2 03 1 6 3

). It reduces to I4 and so the rank is 4. The vectors are

independent.

(h) M =

(0 3 10 3 12 0 02 0 1

). M reduces to

(1 0 00 1 00 0 10 0 0

)which has rank 3 and tells us that

a = b = c = 0.

2. Each of these problems can be solved in two ways, one of which involves row reducinga matrix. We will address (a) in both ways then use only one way for the otherproblems.

(a) The method that should already be known to you is to recognise that the plane


must contain both the vectors and so their cross-product is normal to the plane.We have

u v =i j k2 1 11 2 3

= 5i 5j+ 5k.Hence the plane has equation 5x 5y + 5z = d.The constant d is fixed by identifying a point in the plane. Since the vectorsspan the plane, the vector 0u + 0v is in the plane, i.e. the origin is in the planeand so d = 0.The equation is 5x 5y + 5z = 0 and after dividing by -5 it is x+ y z = 0.Alternatively: all points in the plane take the form

a

211

+ b12

3

.Thus given any a and b the coordinates are x = 2a+ b, y = a+ 2b, z = 1 + 3b.The plane has an equation formed from a linear combination of x, y and z setto 0 (because the origin is a point in this plane). Take the combination to belx+my + nz = 0 and find l, m and n as follows:

l(2a+ b) +m(a+ 2b) + n(a+ 3b) = 0,for any a and b. Hence taking coefficients of a and b we find 2lm+ n = 0 andl + 2m+ 3n = 0.Form the augmented matrix and reduce it. M =

{2 1 1 | 01 2 3 | 0

}which reduces to{

1 0 1 | 00 1 1 | 0

}and solution

l = n, m = n.Taking n to be 1 gives l = 1, m = 1 and so x+ y z = 0 as before. Notice thatthe matrix M that was reduced is the transpose of the matrix whose columnsare the vectors u and v. We can use this observation to hasten the calculationsin the other parts of the question.

(b) Following what we learned in part (a) put the vector components into rows androw reduce the resulting matrix, M =

(1 3 21 2 2

)to the form ( 1 0 20 1 0 ). Using the

variables l, m and n as in part (a), we get m = 0 and l = 2n. Choosing n = 1gives us the plane equation 2x + z = 0. The variable y does not appear sinceits coefficient m = 0.

(c) M =(1 0 21 1 3

)reduces to

(1 0 20 1 1

)resulting in l = 2n, m = n.

And a plane equation 2x+ y + z = 0.3. We are asked to show that u, v and w are linearly independent. It will be enough totake the matrix of components and check its determinant.

M =

1 1 10 1 42 3 0

, Det(M) = 12 + 10 = 22 6= 0.


Hence the vectors are linearly independent. They are therefore 3 linearly independentvectors in 3 dimensions and so could be used as a basis for the space. To express x interms of u, v and w write x = au+ bv+ cwso 11

1

= a10

2

+ b 11

3

+ c14

0

=a+ b+ cb+ 4c

2a+ 3b

to be solved for a, b and c. Row reduce

(1 1 1 | 10 1 4 | 12 3 0 | 1

)to get

(1 0 0 | 70 1 0 | 50 0 1 | 1

)so a = 7,

b = 5, c = 1 and x = 7u+5v w . (Check it.)For orthogonality we need the dot products of u, v and w with each other all to bezero. It only needs one of these to fail and the basis is not orthogonal.Now u w = 1 + 0 + 0 = 1 6= 0 so the basis is not orthogonal.

Epilogue

This has been just a very brief and cursory summary of basic vector space material. Forfurther development see below. If you study Fourier series then you will be dealing withan infinite dimensional vector space in which the vectors are periodic functions and thebasis vectors (in terms of which all vectors in the space can be expanded) are sin(nx) andcos(nx). The dot (so called inner) product of two vectors in this space is just the inte-gration over a period of the product of two functions. (Note the analogy: finite infinite,sum integral).Taylor series is also an application of infinite dimensional vector spaces in which the vec-tors are also functions and the basis vectors are integer powers of x. To progress fromfinite to infinite dimensional vector spaces, in effect the labels 1, 2, 3, . . . etc. labelling thecomponents of the vector have become continuous values, x.

Finite degree polynomials of degree up to n also form a vector space of dimension n+ 1 inwhich the basis vectors are powers of x from 0 to n.

We have stood at the edge of a very large vista. To explore further if interested look up:change of basis, GramSchmidt process, inner product, infinite dimensional vector spaces,complex vector spaces, Hilbert space.

Module 3Complex Numbers

The equationx2 + 1 = 0

is not satisfied by any real number. For this reason mathematicians have extended the setof real numbers by introducing a new quantity, written j which does satisfy this equation.The defining property of j is that

j2 + 1 = 0,

that is, j2 = 1. Given this new quantity, we can develop the algebra of quantities involvingj by applying all of the usual rules of algebra with the single additional rule j2 = 1.We call any number of the form x+ jy, where x and y are real numbers, a complex number.We call x the real part of the complex number and y the imaginary part of the complexnumber, and if z = x + jy, we write Re(z) = x and Im(z) = y. Here are examples ofcomplex numbers with their real and imaginary parts:

z1 = 2 + 3j, Re(z1) = 2, Im(z1) = 3z2 = 5.324 j, Re(z2) = 5.324, Im(z2) = 1z3 =

3 + 47 j, Re(z3) =

3, Im(z3) =

47

z4 = 8e+ pij, Re(z4) = 8e, Im(z4) = pi

z5 = 457.92 + 106.1j, Re(z5) = 457.92, Im(z5) = 106.1

54

3.1. Arithmetic operations with complex numbers 55

3.1 Arithmetic operations with complex numbers

To state the rules for operating with complex numbers we will use the convention thatz = x + jy is a complex number and if z1 and z2 are complex then z1 = x1 + jy1 andz2 = x2 + jy2.

Equality

Two complex numbers z1 and z2 are equal if and only if Re(z1) = Re(z2) and Im(z1) =Im(z2). z1 = z2 if and only if x1 = x2 and y1 = y2.

Example 3.1.1. Solve the equation 3z = 6 9j.

Solution. Letting z = x+ jy we have 3(x+ jy) = 6 9j so 3x+ j3y = 6 9j.Equating real and imaginary parts gives 3x = 6 and 3y = 9. Thus, x = 2 and y = 3 soz = 2 3j.

Addition and subtraction

z1 + z2 = (x1 + jy1) + (x2 + jy2) = (x1 + x2) + j(y1 + y2)z1 z2 = (x1 + jy1) (x2 + jy2) = (x1 x2) + j(y1 y2)

Examples

(a) (2 5j) + (4 + j) = (2 4) + j(5 + 1) = 2 4j(b) (3 + 0.7j) (4 2j) = (3 4) + j(0.7 (2)) = 7 + 2.7j

Multiplication

The important new element here is j2 = 1. When we apply this to multiplication ofcomplex numbers, we obtain

z1z2 = (x1 + jy1)(x2 + jy2) = x1x2 + jx1y2 + jy1x2 + j2y1y2

z1z2 = x1x2 y1y2 + j(x1y2 + x2y1).

Examples

(a) (3 + 2j)(2 5j) = 6 10j2 + j(4 15) = 6 + 10 11j = 16 11j(b) (

2 + j)(

2 j) = (2)2 j2 + j(22) = 2 + 1 = 3

Note that we can compute the successive powers of j:

j3 = j2j = 1 j = j,j4 = j2j2 = 1 1 = 1,j5 = j,

and so on.

56 Module 3. Complex Numbers

Note also that(x+ jy)(x jy) = x2 j2y2 = x2 + y2.

From this we have such remarkable results as

(3 + 2j)(3 2j) = 32 + 22 = 9 + 4 = 13.

Thus the number 13 can be factorised if we use complex numbers.

Conjugation

If z = x+jy then we call the number z = x jy the (complex) conjugate of z. For example,3 2j = 3+2j. We always have zz = x2+y2, i.e. the product of a number and its conjugateis a real number. Note that sometimes the conjugate of z is written z, i.e. (x+jy) = xjy.It is clear that (z) = z.

Division

To expressz1z2

=x1 + jy1x2 + jy2

in the form z = x+ jy, we multiply both numerator and denom-

inator by z2

x1 + jy1x2 + jy2

=(x1 + jy1)

(x2 + jy2)

(x2 jy2)(x2 jy2)

=(x1x2 + y1y2) + j(x2y1 x1y2)

(x22 + y22)

.

Example 3.1.2.

(5 + 3j)

(4 j) =(5 + 3j)

(4 j)(4 + j)

(4 + j)=

20 + 5j + 12j + 3j2

42 + 12=

17 + 17j

17= 1 + j.

Example 3.1.3.

1

2 3j =1

(2 3j)(2 + 3j)

(2 + 3j)=

2 + 3j

22 + 32=

2

13+

3j

13.

3.2. Geometrical representation of complex numbers 57

3.2 Geometrical representation of complex numbers

A complex number z = x + jy can be represented by the point P with coordinates (x, y)in the Cartesian plane. We then call the x-axis the real axis and the y-axis the imaginaryaxis. A representation of complex numbers in this way is called an Argand diagram. TheCartesian plane is referred to as the complex plane, when used to represent the set ofcomplex numbers. Addition of complex numbers is analogous to head-to-tail addition ofvectors with components x and y.

O

1

112 2 3

2

1

2

x

y

2+2j 1+2j

3+j

12j

Figure 3.2.1

Polar form

Given a complex number z = x+jy we can convert to polar coordinates in the plane using therelations x = r cos , y = r sin . Then we have z = r cos + jr sin or z = r(cos + j sin ).

O x

y

x= rcos

y= rsin

r

x+jy

Figure 3.2.2

The quantity r =x2 + y2 is called the absolute value, the magnitude or modulus of z. It

is written as |z|,r = |z| = |x+ jy| =

x2 + y2.


Note the relation zz = |z|2 = r2

It is usual to abbreviate cos + j sin by cis , i.e.

cis = cos + j sin .

The angle is called the argument or phase of the complex number z.

Example

|3 + 4j| = 32 + 42 = 9 + 16 = 25 = 5|1 j| = 1 + 1 = 2| 1 +3j| = 1 + 3 = 4 = 2

Now there are infinitely many possible values for , for if z = r(cos + j sin ) then

z = r(cos( + 2npi) + j sin( + 2npi))

for any integer n. For this reason we define the principal argument of z to be that suchthat z = r cis where r is positive and satisfies the condition pi < pi. This is denotedby arg(z).

Example 3.2.1. Express 1 + j in polar form.

Solution. |1 + j| = 2, and arg(1 + j) = arctan(1) = pi4

, so 1 + j =

2 cispi

4.

Example 3.2.2. Express 2

3 2j in polar form.

Solution. |23 2j| =

(2

3)2(2)2 =

(2

3)2 + (2)2 = 16 = 4.

To calculate the argument, note that 2

3 2j lies in the fourth quadrant so

arg(2

3 2j) = arctan( 2

2

3

)= arctan 2

2

3= pi

6.

Thus, 2

3 2j = 4 cis(pi

6

).

We have, for z = x+ jy,

arg(z) = arctany

xif x > 0 (1st and 4th quadrants)

arg(z) = arctany

x+ pi if x < 0, y 0 (2nd quadrant)

arg(z) = arctany

x pi if x < 0, y < 0 (3rd quadrant).

In fact, when calculating the polar form of a complex number, it is simplest to draw adiagram locating the complex number on an Argand diagram and then it should be clearwhich value to give to the argument.

3.2. Geometrical representation of complex numbers 59

Example 3.2.3. Express 1 j in polar form.

Solution. | 1 j| = 2, and

arg(1 j) = arctan(11) pi

= arctan(1) pi=pi

4 pi

= 3pi4,

so

1 j =

2 cis

(3pi

4

).

(Compare this with the earlier example of the polar form of 1 + j.)

Example 3.2.4. Express 3 cis2pi

3in Cartesian form.

Solution. 3 cis2pi

3= 3

(cos

2pi

3+ j sin

2pi

3

)= 3

(12

+

3j

2

)=32

+3

3j

2.

Example 3.2.5. Express z = 2 3j in polar form.

Solution. We have

|z| =

22 + 32 =

13 and arg z = arctan

(32) pi = arctan

(3

2

) pi = 2.1588,

since Re(z) < 0 and Im(z) < 0. Thus, z =

13 cis (2.1588).Note also the following polar forms

1 = cis 0, j = cispi

2, 1 = cispi, j = cis pi

2.

Multiplication in polar form

Consider first those complex numbers which have the form z = cis ; we have, for these

numbers, |z| =

cos2 + sin2 = 1. Thus, these numbers are represented by the points onthe circle x2 + y2 = 1.

If we multiply two such numbers, cis and cis, we have

cis cis = (cos + j sin )(cos+ j sin)

= (cos cos sin sin) + j(sin cos+ sin cos )= cos( + ) + j sin( + )

= cis ( + ).

Thus, when we multiply these numbers, we simply add their arguments. Now to multiplyany two complex numbers z1, z2, we can bring them into polar form z1 = r1 cis 1, z2 =r2 cis 2. Then we can multiply them as follows

z1z2 = (r1 cis 1)(r2 cis 2) = r1r2 cis 1 cis 2 = r1r2 cis (1 + 2).

That is, we multiply the magnitudes and add the arguments. (We may need to add orsubtract pi to get the principal argument of the product.)


Example

Let z1 = 3j, z2 = 4 + 4j. Then in Cartesian form we have

z = z1z2 = 3j(4 + 4j) = 12j + 12j2 = 12 + 12j;

in polar form we have

|z1| = 3 and arg(z1) = pi2

and|z2| =

42 + 42 = 4

2 and arg(z2) =

pi

4,

soz1 = 3 cis

pi

2and z2 = 4

2 cis

pi

4,

and

z1z2 = 3 cispi

24

2 cispi

4

= 12

2 cis3pi

4

= 12

2

(cos

3pi

4+ j sin

3pi

4

)= 12

2

(

2

2+

2

2j

)= 12 + 12j.

Division in polar form

We can treat division of complex numbers as multiplication by the reciprocal:

z1z2

= z11

z2.

Now if we calculate1

cis , we have

1

cis =

1

cos + j sin

=1

cos + j sin

cos j sin cos j sin

=cos j sin cos2 + sin2

= cos j sin = cis (),

i.e.1

cis = cis () = ( cis ()). Thus we have

z1z2

=r1 cis 1r2 cis 2

=r1r2

cis 1cis 2

=r1r2

cis 1 cis (2) = r1r2

cis (1 2).

Thus, to divide complex numbers, we divide their magnitudes and subtract their arguments.

3.3. Powers and roots of complex numbers 61

3.3 Powers and roots of complex numbers

We now consider powers of numbers of the form cis . We have

cis 2 = cis cis = cis ( + ) = cis 2.

In the same way we have

cis 3 = cis 2 cis = cis 2 cis = cis (2 + ) = cis 3,

and we can continue in this way to conclude

cis 4 = cis 4,

and in general, for any integer n,

cis n = cisn.

This result is known as De Moivres Theorem.

Example 3.3.1. Use De Moivres Theorem to obtain expressions for sin 3 and cos 3.

Solution. We have(cos + j sin )3 = cos 3 + j sin 3.

Expanding the left-hand side, we have

(cos3 3 cos sin2 ) + j(3 sin cos2 sin3 ) = cos 3 + j sin 3.

Equating real and imaginary parts, we have

cos 3 = cos3 3 cos sin2

andsin 3 = 3 sin cos2 sin3 .

Powers of any complex number

Let z = r cis be any complex number; then

zn = (r cis )n = rn cis (n).

Thus, to raise any complex number to an integral power n, say, we express it in polar formand then raise the magnitude of z to the power n and multiply the argument by n. (Toobtain the principal argument we may need to add or subtract some multiple of pi.)

This holds for all values of n, not just integer values. It allows us to calculate powers ofcomplex numbers much more easily. It is also essential for calculating roots of complexnumbers.


Example 3.3.2. By first converting to polar form, find (1 + j)10.

Solution. We have|z| = 1 + 1 =

2 and arg(z) =

pi

4,

thus

z =

2 cis

(pi

4

)so

z10 = (

2)10 cis

(10pi

4

)= 25 cis

5pi

2

= 32 cispi

2

= 32

(cos

pi

2+ j sin

pi

2

)= 32j.

Example 3.3.3. By first converting to polar form, find (3 + j)12.

Solution. We have

|z| = 3 + 1 = 2 and arg(z) = 5pi6

;

thus

z = 2 cis

(5pi

6

),

so

z12 = 212 cis

(12 5pi

6

)= 4

Documents

MTH10007 notes s1-2015