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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