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374 Vectors<br />

OC<br />

=<br />

<br />

m n<br />

Cor. If m = n = 1, then C will be the mid-point, and<br />

<br />

mb na<br />

OC<br />

a b<br />

=<br />

2<br />

5.7 PRODUCT OF TWO VECTORS<br />

<br />

<br />

The product of two <strong>vector</strong>s results in two different ways, the one is a number and<br />

the other is <strong>vector</strong>. So, there are two types of product of two <strong>vector</strong>s, namely scalar<br />

product and <strong>vector</strong> product. They are written as <br />

a .<br />

<br />

b and a <br />

<br />

b .<br />

5.8 SCALAR, OR DOT PRODUCT<br />

The scalar, or dot product of two <strong>vector</strong>s a and b is defined to be <br />

a<br />

scalar where is the angle between a and b .<br />

Symbolically, <br />

a .<br />

<br />

b = a b cos <br />

<br />

b cos i.e.,<br />

Due to a dot between a and <br />

b this product is also called dot product.<br />

The scalar product is commutative<br />

To Prove.<br />

Proof.<br />

<br />

<br />

<br />

<br />

a . b = b . a<br />

<br />

<br />

b . a = b a<br />

cos ( )<br />

b<br />

B<br />

= <br />

a b cos <br />

0<br />

A<br />

a<br />

= a .<br />

<br />

b Proved.<br />

Geometrical interpretation. The scalar product of two <strong>vector</strong>s is the product of one<br />

<strong>vector</strong> and the length of the projection of the other in the direction of the first.<br />

Let<br />

then<br />

<br />

=<br />

OA<br />

<br />

<br />

<br />

<br />

a and OB b<br />

a . b = (OA) . (OB) cos <br />

ON<br />

= OA . OB . OB<br />

= OA . ON<br />

= (Length of a ) (projection of b along a )<br />

5.9 USEFUL RESULTS<br />

^<br />

^<br />

i . i = (1) (1) cos 0° = 1 Similarly, ^ j .<br />

^<br />

j = 1,<br />

^<br />

^<br />

^<br />

^<br />

k . k = 1<br />

i . j = (1) (1) cos 90° = 0 Similarly, ^ j . k ^<br />

= 0, k . i = 0<br />

Note. If the dot product of two <strong>vector</strong>s is zero then <strong>vector</strong>s are prependicular to each other.<br />

5.10 WORK DONE AS A SCALAR PRODUCT<br />

If a constant force F acting on a particle displaces it from A to B then,<br />

Work done = (component of F along AB). Displacement<br />

= F cos . AB<br />

= <br />

F . AB<br />

Work done = Force . Displacement<br />

0<br />

A<br />

<br />

<br />

^<br />

b<br />

^<br />

a<br />

N<br />

B<br />

F<br />

A<br />

B

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