Q 11-03-017NEETJEE MainMedium
The resultant of two forces $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$, and its magnitude is half the magnitude of $\vec{B}$. The angle between $\vec{A}$ and $\vec{B}$ is
Answer: (B) $150°$
Take $\vec{A}$ along $x$ and let $\theta$ be the angle between the forces. The resultant has components $(A + B\cos\theta,\ B\sin\theta)$.
Perpendicular to $\vec{A}$: $A + B\cos\theta = 0$, so $\cos\theta < 0$.
Magnitude: $R = B\sin\theta = \dfrac{B}{2}$, so $\sin\theta = \dfrac{1}{2}$.
With $\cos\theta$ negative, $\theta = 150°$.
Solution by Sreeraj P, M.Sc Physics