Q. 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 vector $\overrightarrow{B}.$ Then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is
A) 30${^\circ}$
B) 45${^\circ}$
C) 150${^\circ}$
D) 120${^\circ}$
Explanation
(c) tan90${^\circ} = \frac{Bsin\theta}{A + Bcos\theta}$ or $A + Bcos\theta = 0$
or $cos\theta = - A/B$ ….(i)
$R = \frac{B}{2} = \left\lbrack A^{2} + B^{2} + 2ABcos\theta \right\rbrack^{1/2}$
or $\frac{B^{2}}{4} = A^{2} + B^{2} + 2AB( - A/B) = B^{2} - A^{2}$
or $\frac{A^{2}}{B^{2}} = \frac{3}{4}$ or $\frac{A}{B} = \frac{\sqrt{3}}{2}$
From (i), $cos\theta = - \frac{\sqrt{3}}{2}\mathbf{=}\cos 150{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.