JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
Two adjacent sides of a parallelogram $PQRS$ are given by $\overrightarrow{PQ}=\hat{j}+\hat{k}$ and $\overrightarrow{PS}=\hat{i}-\hat{j}$. If the side $PS$ is rotated about the point $P$ by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side $PQ$, then $\sin^{2}\left(\frac{5\alpha}{2}\right)-\sin^{2}\left(\frac{\alpha}{2}\right)$ is equal to :
Vector Algebra
Solution
$\cos\theta=\dfrac{\overrightarrow{PS}\cdot\overrightarrow{PQ}}{|PS||PQ|}=\dfrac{-1}{\sqrt2\sqrt2}=-\dfrac12$, so $\theta=120^\circ$. Rotating $PS$ by an acute angle to become perpendicular to $PQ$ gives $\alpha=120^\circ-90^\circ=30^\circ$. $\sin^2\frac{5\alpha}{2}-\sin^2\frac{\alpha}{2}=\sin(3\alpha)\sin(2\alpha)=\sin90^\circ\sin60^\circ=\dfrac{\sqrt3}{2}$. Option (B).