JEE Physics 2024 Question with Solution

Given: The resultant vector is $\vec{R}=\vec{A}+\vec{B}$, $\vec{R} \perp \vec{A}$, and $|\vec{R}|=\frac{|\vec{B}|}{2}$.

Find: The angle between $\vec{A}$ and $\vec{B}$.

From the condition that $\vec{R}$ is perpendicular to $\vec{A}$, the resultant has no component along $\vec{A}$. Therefore, the component of $\vec{B}$ along $\vec{A}$ must cancel $\vec{A}$ itself.

So,

This gives an acute angle of

But since the resultant is perpendicular to $\vec{A}$, the required angle between $\vec{A}$ and $\vec{B}$ is obtuse, hence

Therefore, the angle between the vectors is $150^\circ$. The correct option is C.

One extracted approach on the page computes $120^\circ$ using dot-product algebra, but another approach and the page's declared correct answer state $150^\circ$. Following the solution authority and its final marked answer, the accepted answer is $150^\circ$ corresponding to option C.