Understanding Engineering Mathematics

a·b=(− 1 )( 0 )+ 2 ( 2 )+ 1 ( 3 )= 7

=abcosθ=

√

(− 1 )^2 + 22 + 12

√

02 + 22 + 32 cosθ

=

√

6

√

13 cosθ

So

cosθ=

7

√

6

√

13

giving, to two decimal places,

θ= 37. 57 °

for the acute angle between the vectorsaandb.
Two non-zero vectorsa,bare perpendicular to each other, ormutually orthogonalif

a·b= 0

This follows directly from cosθ=0forθ=±π/2.
For thei,j,kbasis vectors we easily find: (Reinforcement Exercise 16)

i·i=j·j=k·k= 1

i·j=i·k=j·k= 0

Using these provides a direct derivation of the result:

a·b=a 1 b 1 +a 2 b 2 +a 3 b 3

Exercises on 11.10

1. Using the component form for the scalar product prove the properties (i) – (v).


2. Find all possible scalar products between the vectors a=−i+ 2 j, b=j+ 2 k,
   c=i+ 2 j+ 3 kand determine the angles between each pair of vectors.

Answers

2.a·b=2,a·c=3,b·c= 8
Angle betweenaandbis 66. 42 °, betweenaandcis 68. 99 °,betweenbandcis 17. 02 °
all to two decimal places.

11.11 The vector product of two vectors

Now what about a vector valued ‘product’ of vectors? In this case we will simply define
the vector product in terms of components first and give the connection with thea,b,θ
form later. The mechanical application of the vector product as a moment of a force is
given in the Applications section.
Thevector productofa=a 1 i+a 2 j+a 3 kwithb=b 1 i+b 2 j+b 3 kis denoteda×b
(‘acrossb’) and defined by:

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