Higher Engineering Mathematics, Sixth Edition

Scalar and vector products 277

Let a=a 1 i+a 2 j+a 3 k

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

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

Multiplying out the brackets gives:

a•b=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

However, the unit vectorsi,jandkall have a magnitude
of 1 andi•i=(1)(1) cos 0◦=1,i•j=(1)(1) cos 90◦=0,
i•k=(1)(1) cos 90◦=0 and similarlyj•j=1,j•k= 0
andk•k=1. Thus, only terms containingi•i, j•jor
k•kin the expansion above will not be zero.
Thus, the scalar product

a•b=a 1 b 1 +a 2 b 2 +a 3 b 3 (2)

Bothaandbin equation (1) can be expressed in terms
ofa 1 ,b 1 ,a 2 ,b 2 ,a 3 andb 3.

P c

b

a

A B

O

Figure 26.7

FromthegeometryofFig.26.7,thelengthofdiagonal
OPin terms of side lengthsa,bandccan be obtained
from Pythagoras’ theorem as follows:

OP^2 =OB^2 +BP^2 and

OB^2 =OA^2 +AB^2

Thus, OP^2 =OA^2 +AB^2 +BP^2

=a^2 +b^2 +c^2 ,

in terms of side lengths

Thus, thelengthormodulusormagnitudeornorm of

vectorOPis given by:

OP=

√

(a^2 +b^2 +c^2 ) (3)

Relating this result to the two vectorsa 1 i+a 2 j+a 3 k

andb 1 i+b 2 j+b 3 k,gives:

a=

√

(a 12 +a 22 +a 32 )

and b=

√

(b 12 +b^22 +b^23 ).

That is, from equation (1),

cosθ=

a 1 b 1 +a 2 b 2 +a 3 b 3

√

(a^21 +a^22 +a^23 )

√

(b^21 +b^22 +b^23 )

(4)

Problem 2. Find vectorajoining pointsPandQ

where pointPhas co-ordinates (4,−1, 3) and point

Qhas co-ordinates (2, 5, 0). Also, find|a|,the

magnitude or norm ofa.

Let O be the origin,i.e. its co-ordinates are (0, 0, 0). The

position vector ofPandQare given by:

OP= 4 i−j+ 3 kandOQ= 2 i+ 5 j

By the addition law of vectorsOP+PQ=OQ.

Hence a=PQ=OQ−OP

i.e. a=PQ=( 2 i+ 5 j)−( 4 i−j+ 3 k)

=− 2 i+ 6 j− 3 k

From equation (3), the magnitude or norm ofa,

|a|=

√

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

=

√

[(− 2 )^2 + 62 +(− 3 )^2 ]=

√

49 = 7

Problem 3. Ifp= 2 i+j−kandq=i− 3 j+ 2 k

determine:

(i)p•q (ii)p+q

(iii)|p+q| (iv)|p|+|q|