Question

if the magnitude of two vectors are 3&4 & the magnitude of their scaler product is 6 then find the angle b/w the two vectors

Answer 1

We know that,

Scalar Product = I a I I b I . CosФ

Here Ф refers to the angle between the vectors.

=> A . B = I a I I b I . CosФ

A . B = 6 ; I a I I b I = 3 * 4

Substituting in the formula we get,

6 = 3 * 4 . CosФ

6 = 12 . CosФ

=> Cos Ф = 6 / 12 = 1 / 2

We know that Cos 60° is 1 / 2 . Hence the angle between them would be 60°

Hence Ф = 60°.

Hence solved !

Answer 2

$given \: = |a| = 3 \: , |b| = 4 \: , |a| . |b| = 6$

$cosθ\: = \frac{6}{ |3|. |4| }$

$= > \frac{6}{12} = > \frac{1}{2}$

$cosθ\: = \frac{1}{2}$

$θ = 60°$