Understanding Engineering Mathematics

Note the special case:

a·a=a^21 +a 22 +a 32 =a^2 ≡|a|^2

=square of magnitude ofa

In the above form of the scalar product it is not difficult to show that the scalar product
has the following properties:

(i) a·bis a scalar

(ii) a·b=b·a

i.e. the scalar product is commutative

(iii) a·(b+c)=a·b+a·c

i.e. the scalar product is distributive over addition

(iv) (ka)·b=k(a·b)=a·(kb)for any scalark

(v) Ifa·b=0anda,bare not zero thenais perpendicular tob.

Problem 11.7

If a=2i−jY2k and b=2iY4j−3k evaluate a·b, b·a,a^2 ,b^2

a·b= 2 ( 2 )+(− 1 )( 4 )+ 2 (− 3 )= 4 − 4 =− 6

b·a= 2 ( 2 )+ 4 (− 1 )+(− 3 )( 2 )=− 6 , agreeing with property (ii)

a^2 =|a|^2 =a·a=( 2 )^2 +(− 1 )^2 + 22 = 9

b^2 = 22 + 42 +(− 3 )^2 = 29

To see the connection of the above ‘component expression’ for the scalar product with the
a·b=abcosθ definition, we appeal to the work we did on direction cosines. You can
omit this if you are prepared to take the result on trust. We have

a·b

ab

=

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

ab

=

(a

1

a

)(b

1

b

)

+

(a

2

a

)(b

2

b

)

=

(a

3

a

)(b

3

b

)

=lalb+mamb+nanb

=cosθ

where(la,ma,na)(lb,mb,nb)are the direction cosines ofaandbandθ is the angle
betweenaandb(327
➤
). Soa·b=abcosθ=a 1 b 1 +a 2 b 2 +a 3 b 3.

Problem 11.8

Find the scalar product of a=.−1, 2, 1/and b=.0, 2, 3/and hence find

the acute angle between these vectors.