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 between A and B is

• A. 120°
• B. 150°
• C. 135°
• D. None of these

Answer 1

The correct option is B 150°

$\frac{B}{2}$ = $\sqrt{A^2+B^2+2ABcos\theta }$ ..(i)
$\therefore$ tan ${90}^{\circ }$ = $\frac{Bsin\theta }{A+Bcos\theta }$ $\Rightarrow$ A + Bcos$\theta$ = 0
$\therefore$ cos $\theta$ = - $\frac{A}{B}$
Hence, from (i) $\frac{B^2}{4}$ = $A^2+B^2-2A^2$ $\Rightarrow$ A = $\sqrt{3}\frac{B}{2}$
$\Rightarrow$ cos $\theta$ = - $\frac{A}{B}$ = - $\frac{\sqrt{3}}{2}$ $\therefore$ $\theta$ = ${150}^{\circ }$