Higher Engineering Mathematics, Sixth Edition

280 Higher Engineering Mathematics

10. Find the angle between the velocity vectors
    υ 1 = 5 i+ 2 j+ 7 kandυ 2 = 4 i+j−k.
    [66.40◦]


11. Calculate the work done by a force
    F=(− 5 i+j+ 7 k)N when its point of appli-
    cation moves from point(− 2 i− 6 j+k)mto
    the point(i−j+ 10 k)m. [53Nm]

26.3 Vector products

A second product of two vectors is called thevectoror

cross productand is defined in terms of its modulus

and the magnitudes of the two vectors and the sine of

the angle between them. The vector product of vectors

oaandobis written asoa×oband is defined by:

|oa×ob|=oa obsinθ

whereθis the angle between the two vectors.

The direction ofoa×obis perpendicular to bothoaand

ob, as shown in Fig. 26.9.

(a) (b)

o

b

a

oaob



b

a

oboa

o 

Figure 26.9

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×obis cor-

rect, the head should rotate fromoatoob,asshown

in Fig. 26.9(a). It follows that the direction ofob×oa

is as shown in Fig. 26.9(b). Thusoa×obis not equal to

ob×oa. The magnitudes ofoa 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,aandb,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