Question
If $$\vec A = \vec B - \vec C,$$ then, the angle between $${\vec A}$$ and $${\vec B}$$ is
A.
$${\tan ^{ - 1}}\frac{{{B^2} + {A^2} - {C^2}}}{{2AB}}$$
B.
$${\sin ^{ - 1}}\frac{{{B^2} + {A^2} - {C^2}}}{{2AB}}$$
C.
$${\cos ^{ - 1}}\frac{{{A^2} + {B^2} - {C^2}}}{{2AB}}$$
D.
$${\sec ^{ - 1}}\frac{{{A^2} + {B^2} - {C^2}}}{{2AB}}$$
Answer :
$${\cos ^{ - 1}}\frac{{{A^2} + {B^2} - {C^2}}}{{2AB}}$$
Solution :
Given, $$\vec A = \vec B - \vec C$$
$$\therefore \vec C = \vec B - \vec A$$
If $$\theta $$ is the angle between $${\vec A}$$ and $${\vec B},$$ then
$$\eqalign{
& {C^2} = {B^2} + {A^2} - 2AB\cos \theta \cr
& \therefore \cos \theta = \frac{{{B^2} + {A^2} - {C^2}}}{{2AB}} \cr} $$