JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Vector Algebra

Two adjacent sides of a parallelogram PQRS are given by

PQ = \hat{j} + \hat{k}

and

PS = \hat{i} - \hat{j}

. If the side PS is rotated about the point P by an acute angle

α

in the plane of the parallelogram so that it becomes perpendicular to the side PQ , then

sin^{2} (\frac{5 α}{2}) - sin^{2} (\frac{α}{2})

is equal to :

Answer: (B)

Solution

The angle

θ

between

P Q

and

P S

satisfies

cos θ = \frac{P Q \cdot P S}{∣ P Q ∣ ∣ P S ∣} = \frac{- 1}{2 \cdot 2} = - \frac{1}{2}

, so

θ = \frac{2 π}{3}

. Rotating

P S

until it is perpendicular to

P Q

reduces the angle to

\frac{π}{2}

, so the rotation is

α = \frac{2 π}{3} - \frac{π}{2} = \frac{π}{6}

. Using

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)

sin^{2} \frac{5 α}{2} - sin^{2} \frac{α}{2} = sin 3 α sin 2 α = sin \frac{π}{2} sin \frac{π}{3} = \frac{3}{2} .

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option