Question

The resultant of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is perpendicular to the vector $\overrightarrow{A}$ and its magnitude is equal to half of the magnitude of $\overrightarrow{B}$ figure. The angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ :-

Answer 1

$\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{a^2+b^2+2ab\cos \theta }\\ OR\\ \cos \theta =\frac{\left|\overrightarrow{R}\right|}{\left|\overrightarrow{b}\right|}=\frac{1}{2}\end{array}$

$\theta ={60}^0$

Angle b/w $\overrightarrow{A}\&\overrightarrow{B}$ is ${150}^0$.