Higher Engineering Mathematics

242 VECTOR GEOMETRY

the sine of the angle between them. The vector prod-

uct of vectorsoaandobis written asoa×oband is

defined by:

|oa×ob|=oa obsinθ

whereθis the angle between the two vectors.
The direction ofoa×obis perpendicular to bothoa
andob, as shown in Fig. 22.9.

o θ

a

oa×ob b

θ

a

b

ob×oa

(a) (b)

o

Figure 22.9

The direction is obtained by considering that a
right-handed screw is screwed alongoa×obwith
its head at the origin and if the direction ofoa×ob
is correct, the head should rotate fromoatoob,
as shown in Fig. 22.9(a). It follows that the direc-
tion ofob×oais as shown in Fig. 22.9(b). Thus
oa×obis not equal toob×oa. The magnitudes of
oa obsinθare the same but their directions are 180◦
displaced, i.e.

oa×ob=−ob×oa

The vector product of two vectors may be expressed
in terms of the unit vectors. Let two vectors,aand
b, be such that:

a=a 1 i+a 2 j+a 3 kand

b=b 1 i+b 2 j+b 3 k

Then,

a×b=(a 1 i+a 2 j+a 3 k)×(b 1 i+b 2 j+b 3 k)

=a 1 b 1 i×i+a 1 b 2 i×j

+a 1 b 3 i×k+a 2 b 1 j×i+a 2 b 2 j×j

+a 2 b 3 j×k+a 3 b 1 k×i+a 3 b 2 k×j

+a 3 b 3 k×k

But by the definition of a vector product,

i×j=k,j×k=iandk×i=j

Alsoi×i=j×j=k×k=(1)(1) sin 0◦=0.

Remembering thata×b=−b×agives:

a×b=a 1 b 2 k−a 1 b 3 j−a 2 b 1 k+a 2 b 3 i

+a 3 b 1 j−a 3 b 2 i

Grouping thei,jandkterms together, gives:

a×b=(a 2 b 3 −a 3 b 2 )i+(a 3 b 1 −a 1 b 3 )j

+(a 1 b 2 −a 2 b 1 )k

The vector product can be written in determinant

form as:

a×b=

∣

∣

∣

∣

∣

ijk

a 1 a 2 a 3

b 1 b 2 b 3

∣

∣

∣

∣

∣

(5)

The 3×3 determinant

∣

∣

∣

∣

∣

ijk

a 1 a 2 a 3

b 1 b 2 b 3

∣

∣

∣

∣

∣

is evaluated as:

i

∣

∣

∣

∣

a 2 a 3

b 2 b 3

∣

∣

∣

∣−j

∣

∣

∣

∣

a 1 a 3

b 1 b 3

∣

∣

∣

∣+k

∣

∣

∣

∣

a 1 a 2

b 1 b 2

∣

∣

∣

∣

where

∣

∣

∣

∣

a 2 a 3

b 2 b 3

∣

∣

∣

∣=a^2 b^3 −a^3 b^2 ,

∣

∣

∣

∣

a 1 a 3

b 1 b 3

∣

∣

∣

∣=a^1 b^3 −a^3 b^1 and

∣

∣

∣

∣

a 1 a 2

b 1 b 2

∣

∣

∣

∣=a^1 b^2 −a^2 b^1

The magnitude of the vector product of two vectors

can be found by expressing it in scalar product form

and then using the relationship

a•b=a 1 b 1 +a 2 b 2 +a 3 b 3

Squaring both sides of a vector product equation

gives:

(|a×b|)^2 =a^2 b^2 sin^2 θ=a^2 b^2 (1−cos^2 θ)

=a^2 b^2 −a^2 b^2 cos^2 θ (6)

It is stated in Section 22.2 thata•b=abcosθ, hence

a•a=a^2 cosθ.

Butθ= 0 ◦, thusa•a=a^2