Vectors & 3D GeometrymediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Rotating a Parallelogram Side to Perpendicular: Value = √3/2 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Two adjacent sides of a parallelogram PQRSPQRS are given by PQ⃗=j^+k^\vec{PQ}=\hat j+\hat k and PS⃗=i^−j^\vec{PS}=\hat i-\hat j. If the side PSPS is rotated about the point PP by an acute angle α\alpha in the plane of the parallelogram so that it becomes perpendicular to the side PQPQ, then sin⁡2(5α2)−sin⁡2(α2)\sin^2\left(\dfrac{5\alpha}{2}\right)-\sin^2\left(\dfrac{\alpha}{2}\right) is equal to
A12\dfrac12
B32\dfrac{\sqrt3}{2}correct
C34\dfrac{\sqrt3}{4}
D235\dfrac{2\sqrt3}{5}
Solution
Step 1: Let θ\theta be the angle between PQ⃗\vec{PQ} and PS⃗\vec{PS}:
cos⁡θ=PQ⃗⋅PS⃗∣PQ⃗∣∣PS⃗∣=0−1+02⋅2=−12 ⇒ θ=2π3.\cos\theta=\frac{\vec{PQ}\cdot\vec{PS}}{|\vec{PQ}||\vec{PS}|}=\frac{0-1+0}{\sqrt2\cdot\sqrt2}=-\frac12\ \Rightarrow\ \theta=\frac{2\pi}{3}.
 Step 2: To become perpendicular to PQPQ, rotate PSPS through
α=2π3−π2=π6.\alpha=\frac{2\pi}{3}-\frac{\pi}{2}=\frac{\pi}{6}.
 Step 3: Use sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B) with A=5α2, B=α2A=\dfrac{5\alpha}{2},\ B=\dfrac{\alpha}{2}:
sin⁡2(5α2)−sin⁡2(α2)=sin⁡(5α2+α2)sin⁡(5α2−α2)=sin⁡(3α)sin⁡(2α).\sin^2\left(\frac{5\alpha}{2}\right)-\sin^2\left(\frac{\alpha}{2}\right)=\sin\left(\frac{5\alpha}{2}+\frac{\alpha}{2}\right)\sin\left(\frac{5\alpha}{2}-\frac{\alpha}{2}\right)=\sin(3\alpha)\sin(2\alpha).
 Step 4: Substitute α=π6\alpha=\dfrac{\pi}{6}:
sin⁡(π2)sin⁡(π3)=1⋅32=32.\sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{3}\right)=1\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{2}.
 Correct answer: (2)
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