Question

The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle (in $degree)$ between $A$ and $B$ is

Answer 1

Given magnitude of the resultant of the two vectors $A$ and $B$ is,
$\frac{B}{2}=\sqrt{A^2+B^2+2AB\cos \theta }\dots \dots \left(i\right)$
And resultant makes $90$ angle with vector $A$
$\tan \theta =\frac{B\sin \theta }{A+B\cos \theta }$

$\therefore \tan 90=\frac{B\sin \theta }{A+B\cos \theta }$

$\Rightarrow A+B\cos \theta =0$

$\Rightarrow \cos \theta =-\frac{A}{B}\dots \dots \left(ii\right)$
Using this value in equation~(i)
$\frac{B^2}{4}=A^2+B^2-2A^2$

$\Rightarrow A=\sqrt{3}\frac{B}{2}$
substitute this in equation~(ii)
$\Rightarrow \cos \theta =-\frac{A}{B}=-\frac{\sqrt{3}}{2}$
$\Rightarrow \theta ={150}^{\circ }$
Final answer:$150$