Q 11-03-075JEE MainJEE Main 2024 (9 Apr, Shift 2)Medium
The resultant of two vectors $\vec A$ and $\vec B$ is perpendicular to $\vec A$ and its magnitude is half that of $\vec B$. The angle between vectors $\vec A$ and $\vec B$ is ______ $^\circ$.
Numerical value type. Enter your answer.
Answer: 150
$\vec R = \vec A + \vec B$ with $\vec R\perp\vec A$, so the three vectors form a right triangle with hypotenuse $B$:
$$A^2 + R^2 = B^2 \;\Rightarrow\; A^2 = B^2 - \frac{B^2}{4} \;\Rightarrow\; A = \frac{\sqrt3}{2}B$$
From $\vec R\cdot\vec A = 0$: $A^2 + AB\cos\theta = 0 \Rightarrow \cos\theta = -\dfrac AB = -\dfrac{\sqrt3}{2}$, so $\theta = 150^\circ$.
Solution by Sreeraj P, M.Sc Physics