Vector Analysis and

Cartesian Tensors

Vector Analysis and

Cartesian Tensors

Third edition

D.E. Bourne Department of Applied and Computational Mathematics

University of Sheffield, UK

and

P.C. Kendall Department of Electronic and Electrical Engineering

University of Sheffield, UK

SPRINGER-SCIENCE+BUSINESS MEDIA, B.Y.

First edition 1967Reprinted 1982, 1983, 1984, 1985, 1986, 1988, 1990, 1991Second edition 1977/81

1967, 1977, 1992 D.E. Bourne and P.C. KendallOriginally published by Chapman & Hall in 1992

ISBN 978-0-412-42750-3 ISBN 978-1-4899-4427-6 (eBook)DOI 10.1007/978-1-4899-4427-6

Typeset in 10/12 pt Times by Pure Tech Corporation, India

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The publisher makes no representation, express or implied, with regard to theaccuracy of the information contained in this book and cannot accept anylegal responsibility or liability for any errors or omissions that may be made.

A catalogue record for this book is available from the British Library

Library of Congress Cataloging-in-Publication dataBourne, Donald Edward.

Vector analysis and cartesian tensors / D.E. Bourne and P.C. Kendall.3rd ed.

p. cm.Includes bibliographical references and index.I. Vector analysis. 2. Calculus of tensors. I. Kendall, P.C. (Peter Calvin)II. Title.

QA433.B63 1992 91-28636515'.63dc20 CIP

Contents

Preface ix

Preface to second edition Xl

1 Rectangular cartesian coordinates and rotation of axes 1.1 Rectangular cartesian coordinates 1 1.2 Direction cosines and direction ratios 5 1.3 Angles between lines through the origin 6 1.4 The orthogonal projection of one line on another 8 1.5 Rotation of axes 9 1.6 The summation convention and its use 14 1.7 Invariance with respect to a rotation of the axes 17 1.8 Matrix notation 19

2 Scalar and vector algebra 2.1 Scalars 21 2.2 Vectors: basic notions 22 2.3 Multiplication of a vector by a scalar 28 2.4 Addition and subtraction of vectors 30 2.5 The unit vectors i, j, k 34 2.6 Scalar products 35 2.7 Vector products 40 2.8 The triple scalar product 48 2.9 The triple vector product 51 2.10 Products of four vectors 52 2.11 Bound vectors 53

3 Vector functions of a real variable. Differential geometry of curves

3.1 Vector functions and their geometrical representation 55 3.2 Differentiation of vectors 60 3.3 Differentiation rules 62 3.4 The tangent to a curve. Smooth, piecewise smooth and

simple curves 63

VI I I~ ______________________ C_O __ N_T_E_NT __ S ______________________ ~ 3.5 Arc length 69 3.6 Curvature and torsion 70 3.7 Applications in kinematics 75

4 Scalar and vector fields 4.1 Regions 89 4.2 Functions of several variables 90 4.3 Definitions of scalar and vector fields 96 4.4 Gradient of a scalar field 97 4.5 Properties of gradient 99 4.6 The divergence and curl of a v~ctor field 104 4.7 The del-operator 106 4.8 Scalar invariant operators 110 4.9 Useful identities 114 4.10 Cylindrical and spherical polar coordinates 118 4.11 General orthogonal curvilinear coordinates 122 4.12 Vector components in orthogonal curvilinear coordinates 128 4.13 Expressions for grad fl, div F, curl F, and V2 in

orthogonal curvilinear coordinates 130 4.14 Vector analysis in n-dimensional space 136 4.15 Method of Steepest Descent 139

5 Line, surface and volume integrals 5.1 Line integral of a scalar field 147 5.2 Line integrals of a vector field 153 5.3 Repeated integrals 156 5.4 Double and triple integrals 158 5.5 Surfaces 172 5.6 Surface integrals 5.7 Volume integrals

6 Integral theorems

181 189

6.1 Introduction 195 6.2 The divergence theorem (Gauss's theorem) 195 6.3 Green's theorems 204 6.4 Stokes's theorem 209 6.5 Limit definitions of div F and curl F 220 6.6 Geometrical and physical significance of divergence and curl 222

7 Applications in potential theory 7.1 Connectivity 225 7.2 The scalar potential 226 7.3 The vector potential 230 7.4 Poisson's equation 232

CONTENTS

7.5 Poisson's equation in vector form 7.6 Helmholtz's theorem 7.7 Solid angles

8 Cartesian tensors 8.1 Introduction 8.2 Cartesian tensors: basic algebra 8.3 Isotropic tensors 8.4 Tensor fields 8.5 The divergence theorem in tensor field theory

9 Representation theorems for isotropic tensor functions 9.1 Introduction 9.2 Diagonalization of second order symmetrical tensors 9.3 Invariants of second order symmetrical tensors 9.4 Representation of isotropic vector functions 9.5 Isotropic scalar functions of symmetrical second order

tensors 9.6 Representation of an isotropic tensor function

Appendix A Determinants

Appendix B Expressions for grad, diy, curl, and V2 in cylindrical and spherical polar coordinates

Appendix C The chain rule for Jacobians

Answers to exercises

Index

II Vll

237 238 239

244 245 250 259 263

265 266 272 273

275 277

282

284

286

287

299

Preface

We are very grateful to Chapman & Hall for their offer to reset this book completely. This has given us the opportunity to include small but important teaching points which have accumulated over a long period, to improve the notation and the diagrams, and to introduce some new material. The kinema-tics section has been extended to deal with the existence and nature of angular velocity, including rotating frames of reference and the concept of the Coriolis force. A new section on the application of vector analysis in optimization theory has been added, giving a simple approach to the method of steepest descent, which students have found stimulating. New examples and exercises have been added, and some deleted.

D.E. BOURNE P.e. KENDALL

Preface to second edition

The most popular textbook approach to vector analysis begins with the defini-tion of a vector as an equivalence class of directed line segments - or, more loosely, as an entity having both magnitude and direction. This approach is no doubt appealing because of its apparent conceptual simplicity, but it is fraught with logical difficulties which need careful handling if they are to be properly resolved. Consequently, students often have difficulty in understanding fully the early parts of vector algebra and many rapidly lose confidence. Another disadvantage is that subsequent developments usually make frequent appeal to geometrical intuition and much care is needed if analytical requirements are not to be obscured or overlooked. For example, it is seldom made clear that the definitions of the gradient of a scalar field and the divergence and curl of a vector field imply that these fields are continuously differentiable, and hence that the mere existence of the appropriate first order partial derivatives is insufficient.

The account of vector analysis presented in this volume is based upon the definition of a vector in terms of rectangular cartesian components which satisfy appropriate rules of transformation under changes of axes. This ap-proach has now been used successfully for ten years in courses given from the first year onwards to undergraduate mathematicians and scientists, and offers several advantages. The rules for addition and subtraction of vectors, for finding scalar and vector products and differentiation are readily grasped, and the ability to handle vectors so easily gives the student immediate confidence. The later entry into vector field theory takes place naturally with gradient, divergence and curl being defined in their cartesian forms. This avoids the alternative, more sophisticated, definitions involving limits of integrals. An-other advantage of the direct treatment of vectors by components is that introducing the student at a later stage to tensor analysis is easier. At that stage tensors are seen as a widening of the vector concept and no mental readjust-ment is necessary.

The approach to vectors through rectangular cartesian components does not obscure the intuitive idea of a vector as an entity with magnitude and direction. The notion emerges as an almost immediate consequence of the definition and

xu I L-I _______ P_R_E_F_A_C_E_T_O_S_E_C_O_ND __ E_D_IT_I_O_N _______ ---'

is more soundly based, inasmuch as both the magnitude and direction then have precise analytical interpretations. The familiar parallelogram law of ad-dition also follows easily.

The essential background ideas associated with rotations of rectangular cartesian coordinate axes are introduced in Chapter I at a level suitable for undergraduates beginning their first year. The second and third chapters deal with the basic concepts of vector algebra and differentiation of vectors, res-pectively; applications to the differential geometry of curves are also given in preparation for later work.

Vector field theory begins in Chapter 4 with the definitions of gradient, divergence and curl. We also show in this chapter how orthogonal curvilinear coordinate systems can be handled within the framework of rectangular car-tesian theory.

An account of line, surface and volume integrals is given in the fifth chapter in preparation for the integral theorems of Gauss, Stokes and Green which are discussed in Chapter 6. The basic approach to vectors that we have adopted enables rigorous proofs to be given which are nevertheless within the grasp of the average student. Chapter 7 deals with some applications of vector analysis in potential theory and presents proofs of the principal theorems.

Chapters g and 9, on cartesian tensors, have been added to this second edition in response to the suggestion that it would be useful to have between two covers most of the vector and tensor analysis that undergraduates require. The case for adding this material is strengthened by the fact that the approach to vectors in the early chapters makes the transition to tensors quite straight-forward. Chapter 8 deals with the basic algebra and calculus of cartesian tensors, including an account of isotropic tensors of second, third and fourth order. Chapter 9 briefly discusses those properties of second order tensors which have risen to importance in continuum mechanics over the last twenty years. Some theorems on invariants and the representation of isotropic tensor functions are proved.

We warmly acknowledge the many useful comments from students and colleagues who have worked with the first edition. They have enabled us to make improvements to the original text. We particularly thank the following: Dr G.T. Kneebone and Professor L. Mirsky for their early interest in the first edition; Professor A. Jeffrey and Thomas Nelson and Sons Ltd without whom this new edition would not have appeared.

D.E. BOURNE P.e. KENDALL

Rectangular cartesian 1 coordinates and rotation of

axes

1.1 RECTANGULAR CARTESIAN COORDINATES

From a fixed point 0, which we shall call the origin of coordinates, draw three fixed lines Ox, Oy, Oz at right angles to each other as in Fig. 1.1. These will be called the x-axis, y-axis, z-axis respectively and they will be referred to collectively as rectangular cartesian axes Oxyz. The planes Oyz, Ozx and Oxy are called the coordinate planes and they may be referred to as the yz-plane, zx-plane and xy-plane, respectively.

It is customary to choose the axes in such a way that Ox, Oy and Oz form a right-handed set, in that order. This means that to an observer looking along Oz, the sense of the smaller arc from a point on Ox to a point on Oy is

z

~----I------- Y

x

Fig. 1.1 Rectangular cartesian coordinates.

~_2 __ ~1 ~I _______________ C_A_R_T_E_S_IA_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~ clockwise. Figure 1.2(a) illustrates this and Fig. 1.2(b) shows the relationship of Fig. 1.2(a) to a right hand. Notice that to an observer looking along Ox, the sense of the srnatlerarc-fromOyro-Ozis clockwise; and to' an observer looking along Oy, the sense of the smaller arc from Oz to Ox is clockwise.

The three statements concerning observers looking along the respective axes exhibit cyclic symmetry in x, y, z, i.e. if, in anyone of the three statements, we replace x by y, y by z and z by x, then one of the other two statements is produced. The operation of replacing x by y, y by z and z by x is called a cyclic interchange of x, y, z.

The position of a point P relative to a given set of rectangular cartesian axes may be specified in the following way. Draw the perpendiculars PL, PM, PN from P to the YZ-, zX-, xy-planes respectively, as in Fig. 1.1. Let

x = length of PL

taking the positive sign if P lies on the same side of the yz-plane as Ox and the negative sign otherwise. Similarly, define

y = length of PM, z = length of PN,

the positive or negative sign for y being taken according as P lies on the same or on the opposite side of the zx-plane as Oy, and the positive or negative sign for z being taken according as P lies on the same or on the opposite side of the xy-plane as Oz. The numbers x, y, z are called the x-coordinate, y-coordinate, z-coordinate of P. We may refer to P as the point (x, y, z).

:Ie

It is an elementary observation that, when x, y, z are given, the position of

J---r---Y

to y-axis

(a) (b)

t Thumb pointing along z-axis

Fig. 1.2 (a) The arrow turns in a clockwise sense as seen by an observer looking along Oz and (b) relation to a right hand.

L-________ R_E_C_T_A_N_G_U_L_A_R __ C_A_R_T_E_SI_A_N_C_O __ O_R_D_IN_A_T_E_S ________ ~I 1~ __ 3 __ ~ P relative to the given axes is determined uniquely. Conversely, a given point P detennines a unique triad of coordinates. In other words, there is a one-one correspondence between points P and triads of real numbers (x, y, z).

Distance from origin

To find the distance of P from the origin 0, construct the rectangular paralle-lepiped which has PL, PM, PN as three edges (Fig. 1.3). Using Pythagoras's theorem, we have

Op2 = ON2 + PN2 = PL2 + PM2 + PN2

z

J-____ ~L

y

x

Fig. 1.3 Construction to find the distance OP.

Since the perpendicular distances of P from the coordinate planes are lxi, I y I, I z I, it follows that

OP=...j(x2+l+i). (1.1)

Distance between points

The distance between the points P(x, y, z) and P'(x, y', z') may be found in the following way. Through P construct three new coordinate axes PX, PY, PZ parallel to the original axes Ox, Oy, Oz as shown in Fig. 1.4. Let the coordinates of p' relative to these new axes be X, Y, Z. Then it is easily seen that

X = x' - x, Y = y' - y, Z = Z' - z. Applying result (1.1)

~_4 __ ~1 ~I _______________ C_A_R_T_E_S_IA_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~

and so, in tenns of coordinates relative to the original axes,

PP' =.j {(x' _X)2+ (y' - y)2 + (z' - Z)2}.

~------------~----7Y

x

Fig. 1.4 Construction to find PP'.

EXERCISES

(1.2)

1.1 Show that the distance between the points (5, 4, 2) and (0, 3, 1) is 3.j3.

1.2 Show that the distance of the point (a - b, a + b, c) from the origin is .j(2a2 + 2b2 + c2).

1.3 Find the points in the xy-plane which are at unit distance from the origin and equidistant from the x-axis and y-axis.

1.4 Find the points which are at a distance of 5 units from the origin and whose distances from both the xy- and zx-planes are 2.j2 units.

1.5 Find the points which are at a distance of H2 from every axis. 1.6 Find the distance between (i) the points (1, - 1,0) and (1, 2, 4), (ii) the

points (3, - 1,2) and (- 1,5, - 1).

1.7 The coordinates of a point 0' relative to rectangular cartesian axes Ox, Oy, Oz are (1,1, - 1). Through 0' new axes O'x', O'y', O'z' are taken such that they are respectively parallel to the original axes. Find the coordi-nates of 0 relative to the new axes. If a point P has coordinates (- 1, 2, 0) relative to the new axes, find its perpendicular distances from the xy-, xz-, yz-planes.

~ _______ D_I_R_E_C_T_IO_N __ C_O_S_IN_E_S_A_N __ D_D_I_R_E_C_T_IO_N __ R_A_T_IO_S ________ ~I ~I __ 5 __ ~ 1.8 Find the perimeter of the triangle whose vertices lie at the points (1, 0, 0),

(0, I, 0) and (0, 0, I).

1.2 DIRECTION COSINES AND DIRECTION RATIOS

Direction cosines

Let OP be a line described in the sense from 0 (the origin) to a point P, and denote by ex, p, y the angles that OP makes with Ox, Oy, Oz (Fig. 1.5). We define the direction cosines of OP to be cos ex, cos p, cos y. For convenience we write

I = cos ex, m = cos p, n = cos y. (1.3)

p (x. y.:)

""--....L...----N":------ x

Fig. 1.5 The line OP makes angles a, p, y with the axes.

The direction cosines of the x-axis, for example, are 1, 0, 0. Denote the foot of the perpendicular from P to the x-axis by N, let OP = r

and suppose that the coordinates of P are (x, y, z). From the triangle OPN we have ON = 1 xl = r 1 cos ex I. Also, if ex is an acute angle, cos ex and x are both positive, whilst if ex is an obtuse angle, cos ex and x are both negative. It follows that x = r cos ex, and similarly we may show that y = r cos p, z = r cos y. The direction cosines of OP are therefore

1= x/r, m = y/r, n = z/r. (1.4) Since r2 = x 2 + l + Z2, we have

[2 + m2 + n2 = 1. ( 1.5) This shows that the direction cosines of a line are not independent - they must satisfy (1.5).

L-_6 __ ~1 LI ______________ C_A_R_T_E_s_IA_N __ C_O_O_R_D_I_N_~_T_ES ______________ ~

The direction cosines of a line 5 not passing through the origin are defined to be the same as those of the parallel line drawn from the origin in the same sense as 5.

Direction ratios.

Any three numbers a, b, c such that

a:b:c=/:m:n

are referred to as direction ratios of OP. If (1.6) holds we have

1 = aId, m = bId, n = c/d where, by substituting into equations (1.5),

. d = --J(a2 + b2 + c2).

(1.6)

(1.7)

(1.8)

The choice of sign in (1.8) indicates that there are two possible sets of direction cosines corresponding to a given set of direction ratios. These sets of direction cosines refer to oppositely directed parallel lines.

EXERCISES

1.9 Show that it is impossible for a line through the origin to be inclined at angles of 60, 120 and 30 to the x-axis, y-axis and z-axis, respectively, but that angles of 60, 120 and 135 are possible.

1.10 Find the direction cosines of the line joining the origin to the point (6,2,5).

1.11 A line makes angles of 60 with both the x-axis and y-axis and is inclined at an obtuse angle to the z-axis. Show that its direction cosines are ~, ~, -H2 and write down the angle it makes with the z-axis.

1.12 Find the direction cosines of the line which is equidistant from all three axes and is in the positive octant x ~ 0, y ~ 0, z ~ o.

1.13 Find direction ratios for the line which makes an angle of 45 with the x-axis and an angle of 45 with the y-axis and which lies in the positive octant.

1.3 ANGLES BETWEEN LINES THROUGH THE ORIGIN

Consider two lines OA and OA' with direction cosines I, m, n and r, m', n'. To find the angle 0 between them, denote by B, B' the points on OA, OA'

~ ______ A_N_G_L_E_S_B_ETW ___ EE_N __ L_IN_E_S_T_H_R_O_U_G_H __ T_H_E_O_R_I_G_IN ______ ~I ~I __ 7 __ ~

Fig. 1.6

A'

O~--~-----.----~--------A

""(-- 1 --+) B

(produced if necessary) such that OB = OB' = 1 (Fig. 1.6). Then the coordinates of B, B' are (I,m, n), (t, m', n'), using equation (1.4) with r= 1. Applying the cosine rule to the triangle OBB' gives

cos ()= Olf+OB,2_ BB,2 = 1-'!BB,2. 20B.OB' 2

But from (1.2) BB,2 = ([' _/)2 + (m' - m)2 + (n' _ n)2

= (['2 + m,2 + n,2) + (12 + m2 + n2) - 2([t + mm' + nn').

Using the results [2 + m2 + n2 = 1, ['2 + m'2 + n'2 = 1, we obtain cos () = 1/' + mm' + nn'. (1.9)

Note that, because cos (2n - ()) = cos (), we still obtain equation (1.9) when the angle between OA and OA' is taken as 2n - ().

Condition for perpendicular lines

Two lines through the origin are perpendicular if and only if

1/' + mm' + nn' = O.

Proof

(1.10)

The two lines OA, OA' are at right angles if and only if () = t n or () = ~ n, i.e. if and only if cos () = O. The result now follows from (1.9).

EXERCISES

1.14 Show that the angle between the lines whose direction cosines are .!-I2 .!-I2 0 and.!-l3 .!-I3 .!-I3 is cos - I .!-I6 2"" , 2"" , 3"" , 3"" , 3"" 3"" .

1.15 Show that the lines whose direction cosines are ~..J2, 0, ~..J2 and H2, 0, - H2 are perpendicular.

~_8 __ ~1 ~I _______________ C_A_R_T_E_S_IA_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~ 1.16 Find the angle between any two of the diagonals of a cube. [Hint. Choose

axes suitably with origin at the centre of the cube.]

1.4 THE ORTHOGONAL PROJECTION OF ONE LINE ON ANOTHER

Let two lines OP, OA meet at an angle e. Then we define the orthogonal projection of OP on OA to be OP cos e (Fig. 1.7).

Note that if N is the foot of the perpendicular from P to OA (produced if necessary beyond 0 or beyond A), then ON = OP 1 cos e I.

The work in Section 1.2 shows that the orthogonal projections of OP on rectangular cartesian axes with origin 0 are the x, y, z coordinates of P relative to these axes. We now extend this result to find the orthogonal projection of OP on a line OA which is not necessarily part of one of the coordinate axes.

Let the direction cosines of OA be t, m, n and let P be the point (x, y, z). Then the orthogonal projection of OP on OA is

tx+my+nz. (1.11)

Proof

By equations (1.4) the direction cosines of OP are x/r, y/r, z/r, where r = OP. Hence, by formula (1.9), the angle e between OP and OA is given by

cos 8= (lx+ my + nz)/r.

z

~~------------------- y

x

Fig. 1.7 ON is the orthogonal projection of OP on the line OA.

L-_________________ R_O_T_AT_I_O_N __ O_F_A_X_E_S ________________ ~II L __ 9 __ ~

From the definition of the orthogonal projection of OP on OA, expression (1.11) follows at once.

EXERCISES

1.17 Points A, B have coordinates (1,4, - 1), (- 1,3,2) respectively. If 0 is the origin, find the point P on OA produced which is such that the orthogonal projection of OP on OB is of length 9.)14/7.

1.18 A line OP joins the origin 0 to the point P (3, 1,5). Show that the orthogonal projection of OP on the line in the positive octant making equal angles with all three axes is 3.)3.

1.19 The feet of the perpendiculars from the point (4, - 4,0) to the lines through the origin whose direction cosines are Q.)2, H2, 0), q,~, j) are denoted by N, N'. Find the lengths of ON, ON', where 0 is the origin, and explain why one of these lengths is zero.

1.5 ROTATION OF AXES

The transformation matrix and its properties

Consider two sets of right-handed rectangular cartesian axes Oxyz, Ox'y'z'. It is easily seen that, by a suitable continuous movement about 0, the set of axes Oxyz (with Ox, Oy, Oz always fixed relative to each other) may be brought into coincidence with the set Ox'y'z'. Such a movement will be called a rotation of the axes.

Note that if one set of axes is right-handed and the other left-handed, it is impossible to bring them into coincidence by a rotation. It will be convenient to refer to Oxyz as the original axes and Ox'y'z' as the new axes.

Let the direction cosines of Ox' relative to the axes Oxyz be II" 112,113 , Further, denote the direction cosines of Oy' and Oz' by 121 ,122,123 and 13" 132, 133 , We may conveniently summarize this by the array

0 x y z

x ,

I" 112 113 (1.12) y ,

121 In b z ,

131 132 133

In this array, the direction cosines of Ox' relative to the axes Oxyz occur in the first row, the direction cosines of Oy' occur in the second row and those of Oz' in the third row. Furthermore, reading down the three columns in turn, it is easily seen that we obtain the direction cosines of the axes Ox, Oy, Oz

10 I LI ________ CA_R_TE_S_IA_N_C_O_O_R_D_I_N_A_T_ES _______ -----'

Z'_

---

x

Fig. 1.8 A rotation of axes.

z

" " ", "

", ",

",

/y' ",

"*-------y

x'

relative to the axes Ox'y' z'. The array of direction cosines in (1.12) is called the transformation matrix.

Since the axes Ox', Oy', ot are mutually perpendicular,

111/21 + 112/22 + 113b = 0, 121/31 + 122/32 + 123/33 = 0, 131/11 + 132/12 + 133/13 = 0.

(1.13)

Also, from Section 1.2, the sums of the squares of direction cosines are all unity and so

1121 + 1122 + 1123 = 1,

Iii + 1222 + li3 = 1,

1321 + Il2 + Il3 = 1. (1.14)

The six equations in (1.13), (1.14) are called the orthonormality conditions: it should be observed how they are formed from the array (1.12).

Since the elements of the columns form the direction cosines of the axes Ox, Oy, Oz relative to the axes Ox'y'z', it follows by a similar argument that

111112 + 121122 + 131/32 = 0, 112 /13 + 122 123 + 132133 = 0, 113/11 + 123121 + 133/31 = 0,

(1.15)

~ ________________ R_O_T_~_T_IO_N __ O_F_A_X_E_S ________________ ~I I 11 and

1121 + IiI + If I = 1, 1122 + li2 + 1322 = 1,

1123 + IA + Il3 = 1. (1.16)

Equations (1.15) and (1.16) are an important alternative form ofthe orthonor-mality conditions. They may be derived from equations (1.13) and (1.14) by a purely algebraic argument.

The transformation matrix satisfies one further condition which arises from the fact that the axes Oxyz, Ox'y'z' are both right-handed. Consider the deter-minant

III 112 113 T= 121 122 123

131 132 133

(For the reader unacquainted with determinants, an account of all the theory needed in this book is included in Appendix 1.)

Denoting the transpose of T by T' we have

III 112 113 T2 = TT' = 121 122 123 X

131 132 133

111 121 131

112 122 132 113 123 133

Hence, multiplying the two determinants and using the orthonormality condi-tions (1.13) and (1.14)

Thus T= 1.

100 T2 = 0 1 0 = 1.

o 0

Now, when the axes Oxyz, Ox'y'z' coincide, it is easily seen that the appro-priate values of the direction cosines in the array (1.12) are lij= 1 when i = j, lij = 0 when i:F- j , and so for this particular case

100 T= 0 1 0 = 1.

001

If the axes are rotated out of coincidence, the direction cosines lij will vary during the rotation in a continuous manner (i.e. with no 'sudden jumps' in value) and as the determinant T is the sum of products of the direction cosines its value will also vary continuously. But at all stages of the rotation T = 1 or - 1 and so, for no discontinuity in value to occur, T must take the value 1

12 I IL ________________ C_A_R_T_E_SI_A_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~ throughout the rotation, or else take the value - 1. Since T = 1 when the two coordinate systems coincide, it follows that in all positions

III 112 113 T = 121 122 123 = 1.

131 132 133

(1.17)

This is the additional condition to be satisfied by the transformation matrix. We have shown that, if the components of the array (1.12) are the direction

cosines of the new axes relative to the original axes, conditions (1.13), (1.14) and (1.17) are necessarily satisfied. These conditions are also sufficient for the array to represent a rotation of right-handed axes Oxyz. For, firstly, if equations (1.13) are satisfied the axes Ox', Oy', Oz' are mutually perpendicular; second-ly, if equations (1.14) are satisfied the rows in the transformation matrix represent direction cosines of Ox', Oy', Oz'; and finally, if (1.17) is satisfied the system Ox'y'z' is right-handed.

Transformation of coordinates

Let a point P have coordinates (x, y, z) and (x', y', z') relative to the axes Oxyz and Ox'y'z' respectively. The x'-, y'-, z'-coordinates of P are the orthogonal projections of OP on Ox', Oy', Oz'. Hence, using (1.11) to calculate these, we obtain

x' = III X + 112 y + 113 Z, y' = 121 X + 122 y + 123 Z, z' = 131 x+ 132 y+ 133 z.

(1.18)

Equations (1.18) show how the coordinates of P transform under a rotation of axes. It should be noted how these expressions are formed from the array ( 1.12).

We could, of course, regard the axes Ox'y'z' as the original set and the axes Oxyz as the new set, and determine the coordinates (x, y, z) in terms of (x',y',z'). Remembering that the elements of the columns in (1.12) are the direction cosines of the X-, y-, z-axes relative to the axes Ox'y'z', it follows, by using (1.11) again, that

x = III x' + 121 y' + 131 z', Y = 112 x' + 122 y' + 132 Z' , Z = 113 x' + 123 y' + 133 Z'

(1.19)

The reader may verify as an exercise that equations (1.19) also follow alge-braically from equations (1.18) by solving for x, y, z.

L-________________ R_O __ TA_TI __ O_N __ O_F_A_X_E_S ________________ ~I I 13 EXERCISES

1.20 Two sets of axes Oxyz, Ox'y'z' are such that the first set may be placed in the position of the second set by a rotation of 180 about the x-axis. Write down in the form of array (1.12) the set of direction cosines which corresponds to this rotation. If a point has coordinates (1,1,1) relative to the axes Oxyz, find its coordinates relative to the axes Ox'y'z'.

1.21 A set of axes Ox'y'z' is initially coincident with a set Oxyz. The set Ox'y'z' is then rotated through an angle 0 about the z-axis, the direction of rotation being from the x-axis to the y-axis. Show that

x' = x cos 0 + y sin 0, y' = - x sin 0 + y sin 0, z' = z.

[Hint. Consider the direction cosines of the new axes and use equations (1.18).]

1.22 Show that the following equations represent a rotation of a set of axes about a fixed point:

x' = x sin 0 cos qJ + y sin 0 sin qJ + z cos e, y' = x cos e cos qJ + Y cos e sin qJ - z sin 0, z' = - x sin qJ + y cos qJ.

[Hint. Show that the coefficients of x, y, z satisfy (1.17) and the ortho-normality conditions.]

1.23 Solve the equations of Exercise 1.22 for x, y, z in terms of x', y', z'. [Hint. Multiply the first equation by sin 0 cos qJ, the second by cos 0 cos qJ, the third by - sin qJ and add to obtain

x = x' sin 0 cos qJ + y' cos 0 cos qJ - z' sin qJ.

Similarly for y and z.]

1.24 With reference to the transformation array (1.12) in the text, show that

III = b 133 -/23 / 32, 112 = 123/31 -/21 / 33 ,

113 = /21 132 -/22 / 31

Write down two sets of three similar relations. [Hint. By using (1.17) show that if 1110 112, 113 have the values given above then If 1 + If2 + If3 = 1. The solution to this exercise is given with the answers at the end of the book.]

14 I LI _______________ C_A_R_T_E_S_IA_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~

1.6 THE SUMMATION CONVENTION AND ITS USE

It is possible to simplify the statement of equations (1.18) and (1.19) by relabelling the coordinates (x, y, z) as (Xh X2, X3) and the coordinates (.I, y', z') as (x;, x~, .x;). With this change of notation equations (1.18) become

and

3

X; = luxi + 112X2 + Il3X3 = L Ilj Xj j=1

3

~= L 12j Xj, j=1

3

x;= L 13jXj. j=1

(1.20)

(1.21)

(1.22)

Even more briefly, we may write equations (1.20) to (1.22) in the form 3

xi= L lijXj, i= 1,2,3. j=1

Similarly, equations (1.19) may be reduced to the form 3

Xi= L Ijix}, i=1,2,3. j=1

(1.23)

(1.24)

In equations (1.23), (1.24) the suffix j appears twice in the sums on the right-hand sides and we sum over all three possible values of j. This situation occurs so frequently that it is convenient to adopt a convention which often avoids the necessity of writing summation signs.

Summation convention

Whenever a suffix appears twice in the same expression, that expression is to be summed over all possible values of the suffix.

Using the summation convention, equations (1.23), 0.24) become simply

xi = lijxj, Xi= ljix} .

(1.25)

(1.26)

It is understood here that when a suffix is used alone in an equation (such as i on the left-hand and right-hand sides of (1.25) and (1.26)) the equation under consideration holds for each value of that suffix. The reader should notice that

'-----____ T_H_E_S_U_M_M_A_TI_O_N_C_O_NVE __ N_TI_O_N_A_N_D_I_T_S_U_S_E ___ ---l\\ 15 equations (1.25), (1.26) are much more elegant and more convenient than the original forms (1.18), (1.19).

Kronecker delta

The Kronecker delta is defined by

Jij ={~ when i:f:.j when i=j. (1.27) By introducing this symbol and using the summation convention the orthonor-mality conditions (1.13), (1.14) are embodied in the single equation

lik Ijk = Jjj. For example, taking i = l,j = 2, this becomes

Ilk 12k = 0;

that is

111/21 + 1\2 /22 + l\3b = 0,

which is the first of equations (1.13). Again, taking i = j = 1 in (1.28), Ilk l lk=l;

that is

(1.28)

which is the first of equations (1.14). Similarly, taking the other possible combinations of the suffixes i, j we may obtain the remaining four orthonor-mality conditions.

The alternative form of the orthonormality conditions as expressed by (1.15), (1.16) are embodied in the equation

(1.29)

The Kronecker delta is a useful symbol in many contexts other than vector analysis.

Further remarks on the summation convention

1. A repeated suffix is known as a dummy suffix because it may be replaced by any other suitable symbol. For example,

Ilj 12j = Ilk 12k = Ila/2a, since in each expression summation over the repeated suffix is implied.

2. When the summation convention is in use, care must be taken to avoid using any suffix more than twice in the same expression. (The meaning of llj ljj, for example, is not clear.)

16 I LI _______________ C_A_R_T_E_S_IA_N __ C_O_O_R_D_I_N_A_T_E_S ______________ ~

3. As far as we are concerned in this book there are only three possible values for a suffix, namely 1, 2 and 3. The reader will appreciate, though, that elsewhere it might be convenient to increase or decrease the range of a suffix.

Some parts of vector analysis can be shortened considerably by using the convention. We shall usually warn the reader when the summation convention is in use.

EXERCISES

1.25 If

all ::;: 1, a12::;:-I, aI3::;: 0, a21::;: - 2, a22::;: 3, a23::;: 1, a31::;: 2, a32::;: 0, a33 ::;:4,

show that

aii::;: 8, ail ai2::;: -7, ai2 ai3::;: 3, ali a2i::;: - 5, a2i a3i::;: 0, ail a2i::;: - 6.

1.26 If the numbers aij are as given in Exercise 1.25 and if

bl ::;: 1, b2::;: - 1, b3::;: 4,

show that

alibi::;: 2, ajlbj ::;: 11, aji ailbj ::;: 49.

[Hint. For the last part, first evaluate ajl bj , aj2 bj and aj3 bj .]

1.27 Show that

Oij bj ::;: Oji bj ::;: bi.

1.28 If the numbers aij are as given in Exercise 1.25, evaluate

(i) alj Olj, (ii) al2 0;;, (iii) ali a2k Oik.

1.29 The suffix i may assume all integral values from to 00. If a;, bi are defined by

1 b;::;: -=t '

l.

where x is a constant and, by definition, o!::;: 1, show that

1.30 If the quantities eij, efj satisfy the relation

,--_I_NV_A_R_IA_N_C_E_W_I_T_H_R_E_S_P_E_C_T_T_O_A_R_O_T_A_T_IO_N_O_F_TH_E_A_X_E_S_-----'II 17

and if

show that

[Hint. Multiply the first equation by Imp Inq.]

1.7 INVARIANCE WITH RESPECT TO A ROTATION OF THE AXES

Consider a function f( 0.1 , ~, ... ) of several elements 0.1, ~, ... such that given any set of rectangular cartesian axes Oxyz, the elements 0.1, 0.2, ... are determined by a definite rule. Denote by a.~, a.~, ... the elements corresponding to any other set of rectangular cartesian axes Ox'y'z' with the same origin O. Then, if

f(a.~, a.;, ... ) = f(a.l, 0.2, ... )

the function f is said to be invariant with respect to a rotation of the axes. The examples which follow should clarify the idea of invariance.

Examples of invariants

1. The function ...j(x2 + l + i) is invariant, since (1.19) gives ...j(x2 + l + i) = ...j {(l121 + 1122 + 1123)x'2 + (Iii + IA + li3)y,2

+ (Iii + li2 + 1323)z,2 + 2(111 121 + 112/22 + 113/23)x' Y' + 2(121 131 + 122 /32 + 123/33)y' z' + 2(131/11 + 132112 + 133/13)z' x').

Using the orthonormality conditions (1.13) and (1.14), this reduces to

...j(r + l + i) = ...j(X,2 + y'2 + z'2). This result has an immediate geometrical interpretation - it expresses the fact that the distance between the origin 0 and a point P does not depend upon the system of coordinates used in calculating the distance.

The proof given above can be shortened considerably by using the sum-mation convention (see Exercise 1.32 at the end of this section).

2. If OA and OB are two lines through the origin, the expression representing the cosine of the angle between them is clearly invariant with respect to a rotation of the axes. To verify this algebraically let (ai, a2, a3), (blo b2, b3) be the coordinates of A, B relative to the axes Oxyz. If OA = a, OB = b, the direction cosines of OA and OB are

18 I ~I _______________ CA_R_T_E_S_IA_N __ C_O_O_R_D_I_N_X_T_ES ______________ ~

If () is the angle between OA, OB, formula (1.9) gives

cos () = a(b( + a2 b2 + a3 b3 = aibi, ab ab

using the summation convention. From Example 1 we see that a and b are invariant with respect to a rotation of the axes. Thus, to show that cos () is invariant, it only remains to show that ai bi is invariant.

Using (1.25) we see that on transforming to new axes Oxy'z', the coor-dinates of A and B become (a~, a;, a;) and (b~, b;, b;), where

Thus

(1.30)

(Notice that before forming the expression for at hi different dummy suf-fixes must be used in the formulae for ai and bi; otherwise a suffix would appear more than twice in the right-hand side of (1.30).) By using the orthonormality conditions in the form (1.29) (with the suffixes changed to those required here), (1.30) becomes

ai bi =Ojkajbk = akbk = aibi.

It follows that the quantity aibi (= a(b( + a2b2 + a3b3) is invariant under a rotation of the axes, as required, and that cos () is invariant also.

The concept of invariance with respect to rotation of the axes is important because the recognizable aspects of a physical system are usually invariant in this way.

For example, the distance between two points, the volume of a specified region and the resolute of a force along a given line are all independent of any special coordinate system, and the expressions which represent them are in-variant with respect to a rotation of the axes.

EXERCISES

1.31 Find the x, y', z' coordinates of the points x = I, Y = I, z = 0 and x = 0, y = I, z = 1 for the rotation of axes given in Exercise 1.21, namely

x = x cos () + y sin (),

L-________________ M_~_T_R_I_X_N_O_T_~_TI_O_N ________________ ~I I 19 y'=-xsinO + y cos 0, z' = z.

Verify that the angle between the lines joining the origin to these two points works out as 60 with either set of axes.

1.32 Show that the quantity xi + x~ + x~ = Xi Xi is invariant under a rotation of axes.

1.8 MATRIX NOTATION

Another way of expressing some of the results obtained in this chapter is afforded by the use of matrices. It is not our intention to make much use of matrix notation in this book, but those readers familiar with matrices may welcome the following brief remarks.

The matrix of direction cosines in (1.12) may be denoted by

L=[:~: :~~ :~:]. (1.31) 131 132 133

Its transpose is

[111 121 131]

LT = 112 122 132 . 113 123 133

(1.32)

With this notation the orthonormality conditions (1.13) and (1.14) may be expressed as

(1.33)

where

[1 0 0] 1= 0 1 0 001

(1.34)

is the unit matrix. By pre-multiplying (1.33) by C 1, which is the inverse of L, the result that

is obtained. Writing

(1.35)

(1.36)

20 I ,-I ________ CA_R_T_E_S_IA_N_C_O_O_R_D_I_N_Pl._T_ES _______ --'

the transformation rules (1.25) and (1.26) may be expressed as

x'=Lx

and

(1.37)

(1.38)

respectively. Expression (1.38) may be derived at once from (1.37) by pre-multiplying by eland using (1.35).

L...--S_c_a_l_af_a_D_d_v_e_c_to_f_al_g_eb_f_a-----l10

2.1 SCALARS

Any mathematical entity or any property of a physical system which can be represented by a single real number is called a scalar. In the case of a physical property, the single real number will be a measure of the property in some chosen units (e.g. kilogrammes, metres, seconds).

Particular examples of scalars are: (i) the modulus of a complex number; (ii) mass; (iii) volume; (iv) temperature. Note that real numbers are themselves scalars.

Single letters printed in italics (such as a, b, c, etc.) will be used to denote real numbers representing scalars. For convenience statements like 'let a be a real number representing a scalar' will be abbreviated to 'let a be a scalar'.

Equality of scalars

Two scalars (measured in the same units if they are physical properties) are said to be equal if the real numbers representing them are equal.

It will be assumed throughout this book that in the case of physical entities the same units are used on both sides of any equality sign.

Scalar addition, subtraction, multiplication and division

The sum of two scalars is defined as the sum of the real numbers representing them. Similarly, scalar subtraction, multiplication and division are defined by the corresponding operations on the representative numbers. In the case of physical scalars, the operations of addition and subtraction are physically meaningful only for similar scalars such as two lengths or two masses.

Some care is necessary in the matter of units. For example, if a, b are two physical scalars it is meaningful to say their sum is a + b only if the units of measurement are the same.

22 I LI _____________ S_C_A_L_A_R_A_N_D __ V_E_C_T_O_R_A __ LG __ EB_R_A ____________ ~

Again, consider the equation T = i mv2 giving the kinetic energy T of a particle of mass m travelling with speed v. If T has the value 30 kg m2 s- 2 and v has the value 0.1 km S-I, then to calculate m = 2T Iv 2 consistent units for length and time must first be introduced. Thus, converting the given speed to m S-I we find v has the value 100m S-I. Hence the value of m is 2 x 30/10000 = 0.006 kg.

Henceforth it is to be understood that consistent units of measurement are used in operations involving physical properties.

2.2 VECTORS: BASIC NOTIONS

From an elementary standpoint the reader will probably have already en-countered properties of physical systems which require for their complete specification a scalar magnitude and a direction - the velocity of a moving point and the force on a body are particular examples. Such properties are called vectors. We define a vector formally below, but as the definition might otherwise seem strange we shall try to put it into perspective by continuing for the moment on an intuitive basis.

Consider the velocity of a point P moving relative to fixed rectangular cartesian coordinate axes Oxyz. Denote by Vh V2, V3 the rate at which P is travelling away from the YZ-, ZX-, xy-planes in the directions X-, y-, z-increasing respectively. Then v h V2, V3 are called the components of the velocity; together they completely describe the instantaneous motion of P.

If the axes Oxyz are moved to new fixed positions Ox'y'z', without rotation, the velocity components of P relative to the new axes will be Vh V2, V3 as before: for VI is the rate at which P moves away from the yz-plane and this will be the same as the rate at which it moves away from the y'z'-plane since the two planes are parallel; and similarly for the other two directions.

If the axes are rotated about 0 to new positions Ox'y'z' the velocity compo-nents, v~ , v~, vi say, of P relative to the new axes must be related in some way to the original components, for the two sets of components are each sufficient to define the motion of P. The relationship will depend upon the relative position of the two sets of axes (which is defined by the direction cosines of one set relative to the other), and by continuing this reasoning on an intuitive basis it could be obtained explicitly.

The formal definition which follows will be seen to fit into the pattern which the brief remarks above suggest. That vectors do indeed have an associated magnitude and direction will follow later in this section as a consequence of the definition and so the consistency of this with intuitive ideas will be confirmed.

L-_______ V_EC_T_O_R_S_:_B_A_SI_C_N_O_T_I_O_N_S _______ -'II 23 Definition

A vector is any mathematical or physical entity which is such that: 1. when it is associated with a set of rectangular cartesian axes Oxyz it can

be represented completely by three scalars ai, a2, a3 related, in turn, to the axes of x, y, and z;

2. the triad of scalars in 1 is invariant under a translation of the axes, i.e. if Oxyz, O'xy'z' are rectangular cartesian axes (with different origins 0, 0') such that Ox is parallel to O'x, Oy is parallel to O'y' and Oz is parallel to O'z', and if the triads associated with the two coordinate systems are (a[o a2, a3),(a~, a;, a;)respectively,thenal=a~, a2=a;, a3=a;;

3. if the triads of scalars associated with two sets of axes Oxyz, Ox'y'z' (with the same origin 0) are (a[o a2, a3), (a~, a;, a;) respectively, and if the direction cosines of Ox, Oy', Oz' relative to the axes Oxyz are given by the transformation matrix (1.12), then

a~ = III al + 112 a2 + 113 a3, a; = 121 al + 122 a2 + 123 a3, a; = 131 al + 132 a2 + 133 a3.

Introducing the summation convention, equations (2.1) reduce to

aj = lij aj (i = 1,2,3).

Vectors will be denoted by letters printed in heavy type. Thus

a = (a[o a2, a3)'

(2.1)

(2.2)

(2.3)

The scalars alo a2, a3 are called respectively the x-component, y-component, z-component of the vector a.

To avoid repetition, components of all vectors are henceforth to be taken as referred to axes Oxyz unless stated otherwise.

Equality of vectors

Two vectors a = (a[o a2' a3) and b = (blo b2, b3), referred to the same coordinate system, are defined to be equal if and only if their respective components are equal, i.e.

(2.4)

The zero vector

The vector whose components are all zero is called the zero vector or null vector and is written as

0= (0, 0, 0). (2.5)

24 1 ,---I _______ SC_A_L_A_R_AN_ D_V_E_C_T_O_R_A_L_G_E_B_R_A _ _____ ---'

The position vector

Let A and B be points whose coordinates relative to axes Oxyz are (a" a2, a3) and (b .. b2 , b3), respectively. The position of B relative to A is a vector written

-7 AB = (b l - aI , b2 - a2, b3 - a3)'

It is called the position vector of B relative to A.

Proof -7

To prove that AB is a vector it is necessary to establish that conditions 1 to 3 of the definition are satisfied. 1. The scalars bl - aI' b2 - a2, b3 - a3 are the coordinates of B relative to axes

Ax'y'z' drawn through A and parallel to the original axes Oxyz (Fig. 2.1). These scalars are clearly related to the axes Ox, Oy, Oz respectively and they define completely the position of B relative to A (within the coordinate system Oxyz).

2. Suppose the axes Oxyz are moved parallel to themselves so that they pass through a new origin whose coordinates are (- Xo, - Yo, - zo) referred to the axes in their original position. The coordinates of A become (xo + aI, Yo + a2,

-7, zo + a3) and those of B become (xo + bl, Yo + b2, Zo + b3). Thus, if (AB) denotes the position vector referred to the new axes,

-7 (AB), = [(xo + bl) - (xo + al), (Yo + b2) - (Yo + a2), (zo + b3) - (zo + a3)]

= (b l - aI, b2 - a2, b3 - a3)'

z

x

Fig. 2.1

~ ______________ V_E_C_T_O_R_S_:_B_A_SI_C_N __ O_T_IO_N_S ______________ ~I I 25 -7

It follows that the components of AB are invariant with respect to a translation of the axes.

3. Let Ox', Oy', Oz' be rectangular coordinate axes whose direction cosines relative to the axes Oxyz are given by (1.12). Using (1.23), the coordinates (a~, a~, a;) and (b~, b~, b;) of A and B referred to the axes ox' y' z' are

giving

3

ai = L lijaj, j=1

3

3

hi = L lij bj j=1

(i= 1,2,3)

bi - at = L lij (bj - aj) (i = 1,2, 3). j=1

-7

(2.6)

It follows that the components of AB obey the vector transformation law

(2.2). As all three conditions of the definition are satisfied, A1 is a vector. -7

The vector OP giving the position of a point P(x, y, z) relative to an origin o is called the position vector of P. Later it will also be convenient to denote this vector by r = (x, y, z).

Notice that the position vector of a point 0 relative to itself is the zero vector (0,0,0).

Examples of vectors occurring in physical systems

Many properties of physical systems are vectors. For example, velocities and accelerations of moving points, forces, angular velocities, angular accelera-tions and couples are all vectors. In the theory of electricity and magnetism, electric field strength, magnetic field strength and electric current density are also vectors.

In the next chapter (section 3.7), velocity, acceleration and angular velocity will be defined. We shall also refer occasionally to some of the other vectors mentioned above; their definitions may be found in appropriate reference books.

Geometrical representation of vectors

Let a = (al> a2' a3) be any non-zero vector and let A be the point whose X-, y-,

z-coordinates are al> a2, a3. Then oA = (al> a2, a3), showing that a and oA have the same components. Thus, the directed straight line segment OA may be taken as a geometrical representation of a. When a directed line segment such as OA represents a vector this may be shown in a diagram by drawing an arrow on the line pointing from 0 to A (Fig. 2.2). Any zero vector is appropriately represented geometrically by a single point.

26 I LI _____________ S_C_A_L_A_R_A_N_D __ V_E_CT __ O_R_A __ LG __ EB_R_A ____________ ~

z

x'

~ ~

Fig. 2.2 O'A' = OA

Let O'A' be a line segment equal in length to OA and drawn parallel to and in the same sense as OA. If O'x', O'y', O'z' are axes through 0' such that they are parallel to Ox, Oy, Oz respectively, the coordinates of A' referred to these new axes will be (aI, a2, a3)' Hence 0;;4, is a second geometrical representation of the vector a. It follows that the geometrical representation of a as a directed line segment is not unique. Any other directed line seg-ment O'A' which is parallel to OA and of equal length may also be used to represent a.

Direction of a vector

Because a non-zero vector can be represented as a directed line segment, a vector is said to have (or be associated with) a direction. Naturally this is taken to be the same as the direction of the directed line segments which represent the vector. Thus the direction of the vector a is the direction in which OA points. The direction of the null vector is not defined.

Two vectors are said to be parallel if they are in the same direction and anti-parallel if they are in opposite directions.

Magnitude of a vector

The magnitude of a vector a = (a" a2, a3) is defined as

L-_______ V_EC_T_O_R_S_:_B_A_SI_C_N_O_TI_O_N_S _______ --'II 27 a = "'(a~ + a~ + a~). (2.7)

It will also be convenient occasionally to denote the magnitude of a vector a by lal. ~

If a is represented geometrically as OA = (alo a2, a3) it is seen that a is just the length of the line segment ~A. Since this length is invariant under a rotation of the axes (cf. section 1.7, Example 1) it follows that the magnitude of a vector is also invariant. The magnitude of a vector is also sometimes termed its modulus or its norm.

Unit vectors

A vector of unit magnitude is called a unit vector. These are frequently distinguished by a circumflex; thus r = (cos e, sin e, 0) is a unit vector.

If a is any vector, the unit vector whose direction is that of a is denoted by a.

EXERCISES

2.1 Show that equations (2.2) are equivalent to

[Hint. Multiply (2.2) by lik and use the orthonormality conditions (1.29).]

2.2 Relative to axes Oxyz, points P, Q have coordinates (1, 2, 3), (0, 0, 1). Find the components, referred to the axes Oxyz, of: (i) the position vector of P relative to 0; (ii) the position vector of 0 relative to P; (iii) the position vector of P relative to Q.

2.3 Find the magnitudes of the vectors a = (I, 3,4) and b = (2, - 1,0).

2.4 Show that the vectors a = (0, - 3, 3), b = (0, - 5, 5) have the same direction. What is the ratio alb?

2.5 The transformation matrix for a rotation from axes Oxyz to axes Ox'y'z' is

o x x' 0 y' -1 z' 0

y

1 o o

z

o o 1.

Describe (in words or by a diagram) how the positions of the two sets of axes are related.

Relative to the set of axes Oxyz, a vector a has components (2, I, 2). Find the components of a relative to the axes Ox'y'z'.

2.6 Show that a = (cos e, sin e cos qJ, sin e sin qJ) is a unit vector.

28 1 L.I _______ SC_A_L_A_R_AN_D_V_E_C_T_O_R_A_L_G_E_B_R_A ______ ---.J

2.7 Find two unit vectors which are perpendicular to each of the vectors a = (0, 0,1), b = (0,1, 1). Are there any more unit vectors which are per-pendicular to both a and b?

2.3 MULTIPLICATION OF A VECTOR BY A SCALAR

If the components of a vector a in every rectangular cartesian coordinate system are each multiplied by a scalar A, we define the ordered triads of scalars so formed to be components of Aa(or aA). Thus, if a = (ai, a2, a3) when referred to axes Oxyz,

(2.8)

It is easily seen that Aa is a vector. Condition 1 of the definition given in section 2.2 is obviously satisfied, as is condition 2, since the triad (a" a2, a3) is invariant under a translation of the axes. Further, referred to the axes Ox'y'z' (defined in section 2.2),

and hence the components of Aa referred to these axes are (Aa~, Aa~, Aa;). To satisfy 3 we require

Aai = l;j{Aaj), (i = 1,2,3)

and as these are the same as equations (2.2) multiplied by A, the result follows. Note that when A = 0, A a is the null vector defined by equation (2.5).

From equation (2.8) and the definition of the magnitude of a vector

I A a I = ...J(A2a~ + A2a~ + A2ab = IA h/(ai+a~+a~)

= IA la. Thus, when a vector is multiplied by a scalar A, its magnitude is multiplied by I A I.

Multiplication of a non-zero vector by a positive scalar leaves the direction of the vector unchanged, but multiplication by a negative scalar reverses its direc-

~ tion. To see this, let OA = (a" a2, a3) represent a; Aa will then be represented

~ ~,.,. by A OA = (Aa" Aa2, Aa3) = OA, say. SlOce A, A have coordlOates (a" a2, a3), (Aa" Aa2, Aa3) respectively, the direction cosines of OA and OA' will be

ai, a2, a3 and ~, Aa2, Aa3 a a a IAla IAla IAla

where a=...J(ai+a~+~). If A>O, AlIAI=I, whilst if A

L-___ _ M_U_L_T_IP_L_IC_A_T_I_O_N_O_F_ A_V_E_C_T_O_R_B_y_ A_S_C_A_LA_R _ ___ ---'I I 29

z

~~------------------ y

x

Fig. 2.3 The case A < O.

The vector - a

We define

-a=(-I)a. (2.9) ~ ~

If a, - a are represented by OA, OA' respectively, then OA, OA' will be oppositely directed line segments of equal length (Fig. 2.4).

~ ~ Note that OA = - AO since AO and OA', being parallel directed line seg-

ments of equal length, represent the same vector.

A

A'

Fig. 2.4 Oppositely directed line segments OA and OA' of equal lengths.

EXERCISES

2.8 A point P on a line AB is such that I AP I : I P B I = 3 : 2. Show that ~ ~

2AP =3PB.

2.9 Show that, for any vector a, a=aa

where a is a unit vector in the same direction as a.

30 I LI _______ SC_A_L_A_R_AN_D_V_E_C_T_O_R_A_L_G_E_B_R_A ______ --'

2.10 Show that the four points with position vectors

where rl *- 0, r2 *- 0, lie on a circle.

2.4 ADDITION AND SUBTRACTION OF VECTORS

Addition

The sum of two vectors a = (ai, a2, a3) and b = (bb b2, b3) is defined as (2.10)

The set of all vectors is closed under the operation of addition, i.e. the sum of two vectors a, b is a vector. For, considering the requirements 1-3 of the definition of a vector (section 2.2), it is seen at once that 1 and 2 are satisfied. Also, if (a~, a~, a;), (b~, b~, b;) are the components of a, b referred to the axes Ox'y'z', equations (2.2) give

at = lij aj and bi = lij bj (i = 1, 2, 3).

Adding, we have

ai + bi = lij (aj + bj)

showing that the components of a + b satisfy requirement 3. As

ai+bi=bi+ai (i= 1,2,3),

it follows from (2.10) that

a+b=b+a,

i.e. the operation of addition is commutative. Also, if c = (CI. C2, C3) is a third vector, then since

(ai+b;)+ci=ai+(bi+Ci) (i= 1,2,3)

we have

(a + b) + c = a + (b + c),

i.e. the operation of addition is associative.

(2.11)

(2.12)

It is important to appreciate that rules like the commutative and associative laws of addition cannot be taken for granted by analogy with similar laws for real numbers. A new mathematical system is being established involving entities which are not real numbers and there is no reason to suppose, until this has been proved, that such rules of manipUlation are obeyed.

ADDITION AND SUBTRACTION OF VECTORS II 31 L-____________________________________________________ ~ The triangle law of addition

From the definition of addition of vectors, we may deduce the triangle law of addition (also sometimes called the parallelogram law) as follows :

~ ~ If two vectors are represented geometrically by AB and BC, then their sum

~ is represented by AC (Fig. 2.5). Thus

~ ~ ~ AB+BC=Ae.

~ ~ ~ AC is often called the resultant of AB and Be.

Proof

Construct axes Axyz through the point A and draw the line AD so as to complete the parallelogram ABCD (Fig. 2.6). Let the coordinates of B, D be (b), b2, b3), (d), d2, d3) respectively. Then, as BC is parallel to AD, the coordi-nates of C will be (b l + d), b2 + d2, b3 + d3). Thus, referred to the axes Axyz,

A..c....------:~------lo.C

-7 -7 -7 Fig. 2.5 AB + BC = AC (the triangle law of addition).

z

~~=----------------- y

x

-7 -7 -7 Fig. 2.6 If ABCD is a parallelogram, AB+AD=AC (the parallelogram law).

32 I LI _______ S_C_A_L_A_R_A_N_D_V_E_C_T_O_R_A_LG_EB_R_A ______ _

~ AB = (bt. b2, b3), ~ ~

BC = AD = (dt. d2, d3), ~

AC = (b l + dt. b2 + d2, b3 + d3).

It follows at once from the law of addition of vectors that, as required, ~ ~ ~

AB +BC =AC.

Subtraction

The difference of two vectors a, b is defined as

a-b=a+(-b). (2.13)

Note that for any vector a

a -a=O.

Figure 2.7 shows how the sum and difference of vectors a, b may be repre-sented geometrically.

-b b

Fig. 2.7 The sum and difference of vectors a and b.

EXAMPLE 1

The position vectors of three points, A, B, P, relative to an origin 0, are such that

where A, J1. are non-zero real numbers. Prove that P lies on AB (possibly produced) and that AP : PB = 1J1.1 : I A I (Fig. 2.8).

Solution

By the triangle law, ~ ~ ~

OA=OP+PA

'----___ ~A~D~D_IT~IO~N_A~N~D_SU~B~T~R~A~C~T~IO~N_O~F~V~E~C~T~O~R~S ____ ---,II 33

A

Fig. 2.8

and

p

~ ~ ~ OB=OP+PB.

o

~ ~ ~

B

To satisfy the given relation between OP, OA , OB we require ~ ~ ~ ~ oJ> =A(OP+PA)+fl(OP+PB),

A + fl

and hence ~ ~

0= APA + flPB. ~ ~

Since - PA = AP this may be expressed as

~ fl ~ AP = T PB (A, fl to 0). (2.14)

Equating the magnitudes of each side of (2.14) gives AP= IfllA I PB. Thus AP:PB= Ifll:IAI.

~ ~

If A, fl have the same sign, (2.14) shows that AP, PB are in the same direction and so P lies on AB and between A and B. If A, fl have opposite

signs, Xp, Ph are in opposite directions, hence P lies on AB produced beyond B (if Ifll/lA I> 1) or beyond A (if Ifli/l A 1< 1).

EXERCISES

2.11 If a and b are vectors as given below, verify that their sums and differen-ces are as shown.

a (i) (2,2,2)

(ii) (3,0, 0) (iii) (1, - 2, 6)

b (1,0, 1) (5,0,0) (-1,-3,7)

a+b (3,2,3) (8,0,0) (0, - 5, 13)

a-b (1,2,1) (-2,0,0) (2,1, - 1).

34 I LI _______ SC_A_L_A_R_AN_D_V_E_C_T_O_R_A_L_G_E_B_R_A ______ --.J

2.12 Relative to axes Oxyz, points A, B are such that ~ ~

OA = (1,1,1), AB = (0, - 1,3).

What is the position vector of: (i) B relative to 0; (ii) 0 relative to B?

2.13 On a flat horizontal plane an observer walks one mile N followed by one mile E, one mile S and one mile W. Explain vectorially why he finds himself back at the starting point.

2.14 If the angle between vectors a and b is 60, and if a = b = 3, show that

la-bl =3. 2.15 From the property AC ~ AB + BC of a triangle ABC, prove that

la+ bl ~ a+b. For what particular cases is it true that

la+bl =a+b? 2.16 Prove that

la-bl ~ a+b. 2.17 If 0, v are unit vectors with different directions, show that the vector

0+ v bisects internally the angle between 0 and V. Is 4 (0 + v) a unit vector?

2.5 THE UNIT VECTORS i, j, k

Let i, j, k denote unit vectors in the directions of the x-axis, y-axis, z-axis respectively. Then

i = (1, 0, 0), j = (0,1,0), k = (0, 0,1). Using the rules for multiplication of a vector by a scalar and for addition of vectors, a vector a = (a" a2, a3) may be written

It is easily shown that, when the triad i, j, k is given, this representation of a is unique.

The three vectors i, j, k are unit vectors which are mutually perpendicular. Any set of three mutually perpendicular unit vectors is said to be an ortho-normal set. Because any arbitrary vector can be represented as a linear com-bination of i, j, k, these vectors are also said to form an orthonormal basis for the totality of all vectors. Orthonormal bases play an important role in vector analysis.

SCALAR PRODUCTS II 35 ~----------------------------------------------------~

EXERCISES

2.18 A unit vector a in the positive quadrant of the xy-plane makes an angle of 45 with each of the axes Ox and Oy. Show that

a=(i+j)N2.

2.19 The position vectors of points A and B relative to the origin 0 of axes Oxyz are i - j + 2k and 5i + j + 6k respectively. Show that AB = 6.

2.20 Find a, b, c if

(a + b - 2)i + (c - 1) j + (a + c)k = o.

2.6 SCALAR PRODUCTS

The scalar product (or dot product) of two vectors a = (aI' a2, a3) and b = (b], b2, b3) is defined as

a.b = al bl + a2b2 + a3b3.

This operation between two vectors is commutative because

b.a = blal + b2a2 + b3a3 =a.b.

The scalar product of a with itself is

thus a.a is the square of the magnitude of a. If a = (a], a2, a3), b = (b l , b2, b3) and c = (c], C2, C3), then

a.(b + c) = a.b + a.c;

this is the distributive law. It is easily proved, for we have

a. (b + c) = (a], a2, a3) . (b l + c], b2 + C2, b3 + C3)

Scalar invariants

= (a l bl + a l CI + a2 b2 + a2 C2 + a3 b3 + a3 C3) = (al bl + a2 b2 + a3 b3) + (al CI + a2 C2 + a3 C3) = a.b + a.c.

(2.15)

(2.16)

(2.17)

Any scalar which takes the same value in each coordinate system with which it may be associated is called a scalar invariant. Thus the components of a vector a = (aI, a2, a3) are not scalar invariants because they may take different values in different coordinate systems. However the magnitude of a, i.e. a = -J(af + ai + aj), is a scalar invariant.

36 I LI _____________ S_C_A_L_A_R_A_N_D __ V_E_CT __ O_R_A __ LG_E_B_R_A ____________ ~

Since a.a = a2, the scalar product of a with itself is a scalar invariant. This is a special case of the following more general result.

Scalar products are scalar invariants

To prove this, let (a~, a~, a;), (b~, b~, b;) be the components of a, b relative to the axes Ox'y'z'. From equation (2.2)

ai=/ijaj and bi=/ikbk (i=1,2,3)

and so

But, from the orthonormality conditions (1.29)

lij lik = Jjk.

Hence

ai bi = Jjk ajbk = ak bko

i.e. a~ b~ + a~ b~ + a; b; = a1bl + a2b2 + a3b3. This shows that the scalar product of a and b is invariant under a rotation of the axes. Since the scalar product is also obviously invariant under a translation of the axes (for the components of a, b are themselves invariant under such a translation) it now follows that it is a scalar invariant.

Geometrical representation

Let two non-zero vectors a = (al> a2, a3), b = (bl> b2, b3) be represented by -+ -+ OA, OB and let the direction cosines relative to the axes Oxyz of the directed line segments OA and OB be b2, b3), we have by equations (1.4)

ai bi

SCALAR PRODUCTS II 37 L-__________________________________________________ ~

Note that since cos (2n- e) = cos e, no ambiguity arises if 2n- e is taken as the angle between OA, OB.

277 -()

Fig. 2.9

Two non-zero vectors a and b are at right angles to each other if and only if a. b = 0

Proof

If a and b are at right angles, e = ~ n(or ~ n) so, by equation (2.18), a . b = O. Also, if a . b = 0 and a:;t 0, b:;t 0 then cos e= 0; hence e = ~ n (or ~ n) showing that a and b are at right angles.

Scalar products of pairs of i, j, k

The unit vectors i, j, k, introduced in the previous section, are such that the scalar product of anyone with itself is unity and the scalar product of anyone with any other is zero, for these vectors are each of unit magnitude and are mutually perpendicular. Thus

i. i = j . j = k . k = 1,) and

i . j = j . k = k . i = O. (2.19)

Using these relations and the distributive law, the scalar product of two vectors

a = ali + ad + a3k and b = bli + bd + b3k

may be evaluated as follows:

a . b = (ali + ad + a3k) . (bli + bd + b3k) =albli. i+alb2i.j +a l b3i. k

+ a2 bd . i + a2 bd . j + a2 bd . k + a3 blk . i + a3 b2k . j + a3 b3k . k

= al bl + a2b2 + a3 b3'

Of course, the definition (2.15) gives this at once.

38 IIL _______ S_C_A_L_A_R_A_N_D_V_E_C_T_O_R_A_L_G_E_B_R_A ______ -----.J Resolution of vectors

The resolute of a vector a in the direction of (or along) a unit vector n is defined as

an=a.n. (2.20)

If e is the angle between a and n, resolving a in the direction of n gives an = a.n = alnl cos e= a cos e.

Figure 2.10 shows the geometrical interpretation of this. The resolute of a in the direction of n is the sensed projection ON of a on n (produced if necessary).

A

a

"'--'--''r-.... - - - - - - --.J o a N

Fig. 2.10 The resolute of a in the direction n is the sensed projection ON of a on n . The resolute of a along any non-zero vector b is defined as a . b where

is the unit vector in the same direction as b. If

then the resolutes of a in the directions i, j, k are respectively

Thus the resolutes of a along the X-, y-, z-axes are identical to the correspond-ing components of a. It should be noted however that this result is not true if a is referred to a non-orthogonal coordinate system (see Exercise 2.45, section 2.8).

Using the above results, we can write down at once the components, relative to given orthogonal axes, of a given vector of given magnitude and direction. For example, suppose that a is of magnitude 2 and is in the direction making an angle of 60" with Ox, 1200 with Oy and 135' with Oz (cf Exercise 1.9). Then the components of a are

L-________________ S_C_A_LA_R __ P_R_O_D_U_C_T_S ________________ ~I I 39 al = 2 cos 60 = 1,

a2 = 2 cos 120 = - 1,

a3 = 2 cos 135 = - ...j2.

The vector is then said to have been resolved into components.

EXERCISES

2.21 If a and b are vectors as given in the table below, verify that the scalar products are as shown. Which of the three pairs of vectors is perpendicular?

a (i) (1, - 1, 0)

(ii) (4, 1, - 3) (iii) (3,1,4)

b (3,4,5) (-1,3,-7) (2,-2,-1)

a.b -1 20

O.

2.22 Using the formula a. b = ab cos e, find the angle between the vectors a=(O,-I,1), b=(3,4,5).

2.23 A set of rectangular cartesian axes is so arranged that the x-axis points east, the y-axis points north and the z-axis points vertically upwards. Evaluate the scalar products of the vectors a and b in the following cases.

(i) a is of magnitude 3 and points SE, b is of magnitude 2 and points E; (ii) a is of unit magnitude and points NE, b is of magnitude 2 and points

vertically upwards; (iii) a is of unit magnitude and points NE, b is of magnitude 2 and points

W.

2.24 If

a=i-j, b=-j+2k,

show that

(a + b) . (a - 2b) = - 9.

2.25 Show that the vectors i + j + k, ;,.2 i - 2A.j + k are perpendicular if and only if A. = 1.

2.26 Find the component of i in the direction of the vector i + j + 2k.

2.27 Resolve the vector 3i + 4j in the directions of the vectors 4i - 3j, 4i + 3j, and k.

2.28 If a and b are such that a = b, use Fig. 2.11 to interpret geometrically the relation

40 I I~ ____________ S_C_A_L_A_R_A_N_D __ V_E_C_T_O_R_A_L_G_E_B_R_A ____________ ~

I I

I

Fig. 2.11

I I

I I

I I

I I

-b b

, I

I I

I I

I I , , , , ,

2.29 Prove vectorially that the perpendiculars onto the sides of a triangle from the opposite vertices are concurrent. [Hint. Draw the perpendiculars from the vertices A, B of a triangle ABC and let them intersect at O. Let the position vectors of A, B, C relative to 0 be a, b, c. Show that

a. (b - c) = b . (c - a) = 0. Deduce that c . (a - b) = 0, and give the interpretation.]

2.7 VECTOR PRODUCTS

The vector product of a vector a = (aJ, a2, a3) with a vector b = (bJ, b2, b3) is defined as

a x b = (aZb3 - a3bZ, a3bl - a lb3, albz - a2bl)'

Alternatively the vector product of

is defined to be

a = ali + azj + a3k and b = bli + bzj + b3k

j k a x b = al az a3

b l bz b3

(2.21)

(2.22)

The notation used in (2.21) anticipates that the vector product of two vectors is a vector and this result will now be established.

It should be noted first that a x b clearly satisfies conditions I and 2 of the definition of a vector, given in section 2.2, and so it only remains to be shown that 3 is satisfied.

For convenience put

a x b = (a2b3 - a3b2, a3bl - a lb3, a lb2 - a2bl) = (cJ, Cz, C3)'

~ _________________ V_EC_T_O __ R_P_R_O_D_U_C_T_S ________________ ~I I 41 Referred to the axes Ox'y'z' (defined in section 2.2),

and so, relative to these axes,

a x b = (a; b; - a; b;, a; b~ - a~ K a~ b; - a; b~) = (c~, c~, c;).

Condition 3 will be satisfied if it can be proved that

c~ = III C, + lI2c2 + lI3c3, c; = 12'c, + 122 c2+ 123 c3, c; = 13, C, + 132c2 + 133c3.

(2.23)

U sing the transformation law (2.1) for the vectors a, b the quantity c~ = a; b; - a; b; becomes

c~ = (12' a, + 122a2 + 123a3)(l3,b, + 132 b2 + 133b3) - (13' a, + 132 a2 + 133a3)(l2, b, + 122b2 + 123b3)

= (122 133 -123 132)(a2b3 - a3 b2) + (123 13, -12,133)(a3b, - a, b3) + (12,132 -12213,)(a, b2 - a2 b,).

Substituting the results of Exercise 1.24, section 1.5, and using the definitions of c" C2, C3, this reduces to

which is the first of relations (2.23). The other two relations may be verified in a similiar way, and the proof that a x b is a vector is then complete.

Interchanging a and b in (2.21) gives

b x a = (b2a3 - b3a2, b3a, - b,a3, b,a2 - b2 a,);

thus

bXa=-axb (2.24)

showing that the operation of vector multiplication is non-commutative. It is essential therefore to preserve the order of the vectors in a vector product.

If a, b, c are any three vectors it is easily verified that

a x (b + c) = a x b + a x c;

thus the vector product obeys the distributive law.

Vector products of pairs of i, j, k

Since

i=(l,O,O), j=(O, 1,0), k=(O,O, 1)

(2.25)

42 II SCALAR AND VECTOR ALGEBRA

we have

i j k ixi= I 0 0 =0,

I 0 0

i j k ixj= I 0 0 =k

0 I 0

and four similar relations obtained by cyclic permutation of i, j, k. In all

i x i = j x j = k x k = 0, } ixj=k, jxk=i, kxi=j.

(2.26)

These identities should be compared with the corresponding identities (2.19) involving scalar products of pairs of the vectors i, j, k. Note that because of (2.24), interchanging the two vectors on the left-hand side of anyone of the second set of identities (2.26) changes the sign in that identity; for example jxi=-k.

Equations (2.26) may be used in conjunction with the distributive law (2.25) to evaluate vector products. For example if

then

u = i + 3j + k, v = 2i - j + 2k

u x v = (i + 3j + k) x (2i - j + 2k)

= (2i x i - i x j + 2i x k) + (6j xi - 3j x j + 6j x k)

+ (2k x i - k x j + 2k x k)

= (- k - 2j) + (- 6k + 6i) + (2j + i)

= 7i-7k.

It is however much more convenient to use (2.22). Thus

i j uxv= 1 3

2 -1

k 1 =7i-7k. 2

Geometrical interpretation of the vector product

Let a, b be given vectors and choose rectangular axes Oxyz such that a, b are parallel to the xy-plane and Ox is in the same direction as a. Let (l be the angle between a and b, measured in the sense turning from Ox into the positive quadrant of the xy-plane (Fig. 2.12). Then

a=a,i, b=b,i+bd

~ ________________ V __ EC_T_O_R __ P_R_O_D_U_C_T_S ________________ ~I I 43 z

~~------------------ y

x

Fig. 2.12 The vector product a x b is of magnitude ab sin a and in the direction of k.

giving

a x b =a,b2 k.

Also, for this special configuration,

a, = a, b2 = b sin

44 I LI _______ sc_A_L_A_R_AN_D_v_E_c_T_o_R_A_L_G_E_B_R_A ______ --.J

between a and b is defined suitably, the geometrical interpretation may be expressed more conveniently, as follows.

Let () be the angle between a and b, measured in the sense turning from a to b and chosen so that 0 :s::; () :s::; n. Then the vector product a x b is the vector ab sin () C, where c is a unit vector at right angles to both a and b, and such that, to an observer looking in the direction of ~, the sense in which () is measured is clockwise. Fig. 2.13 makes the situation clear. Since 0 :s::; () :s::; n, the magnitude of a x b is ab sin () and its direction is that of C.

Two non-zero vectors a, b are parallel or anti-parallel if and only if a x b = 0

Proof

If a and b are parallel, () = 0, and if a and b are anti-parallel, () = n. In either case a x b = 0 since sin () = O. Also, if ax b = 0 and if a * 0, b * 0, then sin () = 0; hence () = 0 or n, showing that a and b are either parallel or anti-parallel.

EXAMPLE 2

Show that the area of a parallelogram with adjacent sides a and b is 1 a x b I.

Solution

Denote by () the smaller angle between a and b (Fig. 2.14). Drop a perpen-dicular from the end of b onto a. This perpendicular will be of length b sin (). Thus the area of the parallelogram is base x height = ab sin () which is laxbl

Fig. 2.14

EXAMPLE 3

I I

:bsin8 I I

Find the most general form for the vector r satisfying the equation

rx(1, 1, 1)=(2,-4,2).

~ ________________ V_E_C_T_O_R_P_R_O_D_U_C_T_S ________________ ~I I 45 Solution

Let r = (a, b, c). Substituting this into the equation given, we have

(a, b, c) x (1,1,1) = (2, - 4, 2).

Thus

(b - c, c - a, a - b) = (2, - 4, 2).

Using the definition of equality of vectors, this gives

b-c=2, c-a=-4, a-b=2.

These equations are not independent but they are consistent, for adding the first two gives

b-a=-2

which is the third equation. Let

a = A.. Then it follows at once that

b = A. - 2, c = A. - 4. Hence the general solution of the equation given may be represented in the form

r = (A., A. - 2, A. - 4) where A. is arbitrary.

Remarks

If rX3=b

the geometrical interpretation of vector products shows that rand 3 must both be perpendicular to b. Hence, if 3 and b are given, the equation can have no solution for runless 3 is perpendicular to b.

It can be shown (see Exercise 2.54 at the end of this chapter) that the general solution for r is

r = A.3 + (3 x b)/a2

The reader may verify these remarks by reference to the example solved above.

EXAMPLE 4

Show vectorically that the bisectors of the angles of a triangle are concurrent.

46 IIL _______ S_C_A_L_A_R_A_N_D_V_E_C_T_O_R_A_L_G_E_B_R_A ______ -----.J

c

A B

Fig. 2.15

Solution

Let the bisectors of the angles A and B of a triangle ABC intersect at O. Let the position vectors of A, B, C relative to 0 be a, b, c, respectively. Then (see Fig. 2.15)

~ ~ ~ AC = c - a, CB = b - c, BA = a-b.

Now if U and v are unit vectors, the internal bisector of the angle between them is parallel to the vector u+v (see Exercise 2.17, section 2.4). Unit

. h d' . f ~ ~ vectors III t e lfectlOns 0 CA and BA are

Thus a is parallel to

Similarly b is parallel to

a-c ---

la-cl and

a-b la-bl

a-c a-b ---+-,-------;-la-cl la-bl

b-a b-c -,-------:-+-,-------;-

Ib-al Ib-cl

These conditions may be expressed as

ax ---+ -0 [ a-c a-b] la-cl la-bl-

and

b X [ b-a + b-c ]=0 Ib-al Ib-cl .

L-_________________ V_ECT __ O __ R_P_R_O_D_U_C_T_S ________________ ~I I 47 Since a x a = 0 and b x b = 0 these simplify to

ax ---+ =0 [ e b] la-el la-bl

(2.28)

and

(2.29)

The condition that CO should be the bisector of the angle at C can now be written down (by cyclic symmetry) as

ex + =0. [ b a] le-bl le-al

(2.30)

The result (2.30) follows by adding (2.28) and (2.29), and observing that:

(i) la-el = le-al, with similar results for la-bl and Ib-el; and (ii) a x b + b x a = 0, a x e = - e x a, b x e = - e x b.

EXERCISES

2.30 Show that, if vectors a and b are as given in the first two columns of the table below, a x b is as given in the third column.

a (i) (3,7,2)

(ii) (1, - 3, 0) (iii) (8, 8, - 1)

b (1,3,1) (- 2,5,0) (5,5,2)

axb (1, - 1,2) (0,0,-1) (21, - 21, 0).

2.31 Axes Oxyz are positioned so that Ox points east, Oy points north and Oz points vertically upwards. Find the vector product a x b of vectors a and b in the following cases:

(i) a is of unit magnitude and points E, b is of magnitude 2 and points 30 N ofE;

(ii) a is of unit magnitude and points E, b is of magnitude 2 and points SW;

(iii) a is of unit magnitude and points vertically upwards, b is of unit magnitude and points NE.

2.32 By taking components, prove the distributive law for vector products, namely

a x (b + e) = a x b + a x e.

2.33 Prove that for any scalar A,

a x (A b) = (Aa) x b = A(a x b).

48 I ~I _____________ S_C_A_L_A_R_A_N_D __ V_E_C_T_O_R_A __ LG_E_B_R_A ____________ ~ 2.34 If

axb=a-b,

prove that

a=b.

2.35 Find the most general form for the vector u satisfying the equation

u x (2,1, -1) = (1, 0, 0) x (2,1, -1).

2.36 Find a, b if

(ai + bj + k) x (2i + 2j + 3k) = i - j.

2.37 By constructing an example, show that in general the associative law for vector products does not hold; that is, there exist a, b, c such that

a x (b x c):;t: (a x b) xc.

2.38 Prove vectorially that the medians of a triangle are concurrent. [Hint. Let E, F, G be the mid-points of the sides BC, CA, AB of a triangle ABC. Let AE, BF meet at 0 and denote the position vectors of A, B, C relative to o by a, b, c. Find the position vectors of E and F, and deduce that

a x (b + c) = 0 and b x (c + a) = O.

Hence show that c x (a + b) = 0 and deduce the result required.]

2.8 THE TRIPLE SCALAR PRODUCT

The scalar a . (b x c) is called a triple scalar product. If

a = (ah a2, a3), b = (bh b2, b3), c = (CI, C2, C3)

we have

Thus

It is easily verified that

i j k a . (b x c) = a. b l b2 b3

CI C2 C3

al a2 a3 a . (b x c) = bl b2 b3

CI C2 C3

a. (b x c) = (a x b) . c,

(2.31)

(2.32)

that is, the 'dot' and 'cross' may be interchanged in a triple scalar product.

L--_______ T_H_E_T_R_IP_L_E_S_C_A_L_A_R_PR_O_D_U_C_T ___ ___ ----'II 49

Geometrical interpretation

Consider the parallelepiped with adjacent edges representing vectors a, b, c, as shown in Fig. 2.16, with band c horizontal. The volume V of the paralle-lepiped is 'area of base x height'. That is

V = 1 (be sin O)(a cos 1fJ) 1 ,

where 0 is the angle between band c and IfJ is the angle between a and the upward vertical. But

b x c = be sin 0 k,

where k is a unit vector vertically upwards. Also

a . k =acoslfJ.

It follows that

V = 1 a . (b x c) I.

Condition for coplanar vectors

Three non-zero vectors a, b, c are coplanar if and only if

a. (b x c) = o.

Proof

(2.33)

As the three vectors have non-vanishing magnitudes, the volume V of a parallelepiped with adjacent edges a, b, c is zero if and only if the vectors are coplanar. Thus (2.33) shows that a, b, c are coplanar if and only if a. (b x c) =0.

Note

The triple scalar product a . (b x c) vanishes if any two of the three vectors are parallel or anti-parallel, but the converse of this statement is not true.

b x c

b

Fig. 2.16 The volume of a parallelepiped is a . (b x c).

50 I LI _____________ S_C_A_L_A_R_A_N_D __ V_E_C_T_O_R_A __ LG_E_B_R_A ____________ ~

EXAMPLE 5

If the non-zero vectors a, b, c are not coplanar show that any other vector A may be expressed uniquely in the form

A = Aa +,ub + vc, where A, ,u, v are scalars.

Solution

Let A = (At. Az, A3), a = (at. az, a3), b = (bt. bz, b3), c = (ct. CZ, C3). Then

if and only if

A = Aa +,ub + vc,

Al = Aal + ,ubi + VCt. Az = Aaz + ,ubz + vCZ, A3 = Aa3 + ,ub3 + VC3.

These simultaneous equations for A,,u, v have a unique solution if and only if

i.e.

al az a3 bl bz b3 ;t: 0, CI Cz C3

a. (b x c) ;t:O.

This condition is satisfied because a, b, c are not coplanar and are non-zero, and so the required result follows.

EXERCISES

2.39 By writing out components, or otherwise, prove that

a. (b x c) = (a x b) . c.

2.40 If (x, y, z) is any point on the plane through the points (Xl> Yt. Zl), (xz, Yz, zz) and the origin, show that

2.41 Show that for all scalars A,

X Y Z XI YI ZI =0. Xz Yz Zz

(a + A b) . (b x c) = a . (b x c).

~ ____________ T_H_E_T_R_I_PL_E_V_E_C_T_O_R __ P_RO __ D_U_C_T ____________ ~I I 51 2.42 Show that

2.43 If

show that

(a + b + c) . (b x c) = a . (b x c).

Al = Ala + Illb + VIC,

A2 = A2a + 1l2b + V2C,

A3 = A3a + 1l3b + V3C,

Al III VI Al . (A2 x A3) = A2 112 V2 a. (b x c).

A3 113 V3

2.44 Show that, given any four non-zero vectors a, b, c, d, there exist scalars p, q, r, s, not all zero, such that

pa +qb + rc +sd =0.

[Hint. Consider the cases (i) when three of the vectors are not coplanar, (ii) when all four vectors are coplanar.]

2.45 Let OX, OY, OZ be a system of oblique rectilinear axes (i.e. axes such that OX, OY, OZ are straight lines which are not mutually perpendicular and not coplanar) and let I, J, K denote unit vectors in the three coordi-nate directions. If a vector A is expressed in the form

A=AII+A2J +A3K,

then A 10 A2, A3 are called the components of A. Show that these com-ponents are not identical with the resolutes of A along OX, OY, OZ.

2.9 THE TRIPLE VECTOR PRODUCT

Vectors such as (a x b) x c and a x (b x c) are called triple vector products. The following identities (proved below) are often needed:

Proof

(a x b) x c = (a . c) b - (b . c) a;

a x (b x c) = (a . c) b - (a. b) c.

(2.34)

(2.35)

Choose axes Oxyz with the x-axis in the same direction as a and such that b is parallel to the xy-plane (as in Fig. 2.12, section 2.7). Then

a = (at. 0, 0), b = (bl, b2, 0), c = (Clo C2, C3).

52 I LI _____________ S_C_A_L_A_R_A_N_D __ V_E_CT __ O_R_A_L_G_E_B_R_A ____________ ~

Thus

giving

Also

(a. c) b - (b. c) a = a(c(b - (b(c( + b2C2) a = (- a(b2c2, a(b2ch 0).

Comparing (2.36) and (2.37), the identity (2.34) is established.

(2.36)

(2.37)

The second identity (2.35) is easily proved likewise, or by making use of (2.34).

Remark

To remember (2.34) and (2.35) observe that each of the vectors inside the brackets on the left appears once outside the brackets on the right and the 'middle' vector b appears first. Each term contains a, b, c once only.

EXERCISES

2.46 Adjacent sides of a triangle represent vectors a and b. Show that the area of the triangle is t 1 a x b I.

2.47 Establish formula (2.35) by making use of (2.34). [Hint. Use the result that for any two vectors A and B, A x B = - B x A.]

2.48 Prove that, if a, b, c are non-zero and

(a x b) x c = a x (b x c), then either (i) b is perpendicular to both a and c, or (ii) a and c are parallel or anti-parallel. [Hint. Expand using (2.34) and (2.35).]

2.10 PRODUCTS OF FOUR VECTORS

It is sometimes necessary to manipulate products of four vectors. This often involves the formulae (2.34) and (2.35) together with the knowledge that the dot and cross in the triple scalar product are always interchangeable. Thus for example

(a x b) . (c x d) = a . [b x (c x d)]. (2.38)

Expanding the triple vector product gives

~ _________________ B_O_U_ND __ VE __ C_T_O_R_S ________________ ~I I 53 b x (c x d) = (b. d) c - (b. c) d.

Substituting into equation (2.38) we obtain

(a x b) . (c x d) = (b. d) (a . c) - (b. c) (a . d)

so that

(axb).(cXd)=I::~ :::1. (2.39) Other exercises on products of four vectors are given below.

EXERCISES

2.49 Show that