Question

Show that dot product of two equal vectors is equal to square of magnitude of either of them.

Answer 1

Answer:

Dot product of two vectors a and b is given by:

a.b=|a||b|cosФ

In this case as both the vectors are equal the magnitude of a=b and the angle between them is 0

a.b=|a||a|cos0

cos0=1

therefore a.b=a.a=|a|²

Hence proved.

Answer 2

To show :

Dot product of two equal vectors is equal to square of magnitude of either of them

Solution:

Let two equal vectors  x and y

By definition of equal vectors

|x|=|y|

$\theta=0^{\circ}$

Because direction of two vectors are same when vectors are parallel.

Now, we know that

$A.B=|A||B|cos\theta$

Substitute the values

$x\cdot y=|x||y|cos0=|x||x|=|x|^2$

or

$x\cdot y=|x||y|cos0=|y||y|=|y|^2$

Hence, proved.