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Mathematics · Straight lines

JEE Main 2026 — 2 April, Evening Shift — Question 38

Two adjacent sides of a parallelogram PQRS are given by PQ→=j+k\overrightarrow{\mathrm{PQ}} = \mathbf{j} + \mathbf{k} and PS→=i−j\overrightarrow{\mathrm{PS}} = \mathbf{i} - \mathbf{j}. If the side PS is rotated about the point P by an acute angle α\alpha in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then sin⁡2(5α2)−sin⁡2(α2)\sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right) is equal to :

  1. Option A:

    12\frac{1}{2}

  2. Option B:

    32\frac{\sqrt{3}}{2}

    Correct
  3. Option C:

    34\frac{\sqrt{3}}{4}

  4. Option D:

    235\frac{2\sqrt{3}}{5}

Answer: B

Step-by-step solution

Let angle between PQ→\overrightarrow{\mathrm{PQ}} & PS→\overrightarrow{\mathrm{PS}} is θ\theta

cos⁡θ=PQ→⋅PS→∣PQ→∣∣PS→∣cos⁡θ=0−1+022cos⁡θ=−12θ=2π3α=2π3−π2 So α=π6sin⁡25α2−sin⁡2α2=sin⁡(5α2+α2)sin⁡(5α2−α2)=sin⁡6α2×sin⁡4α2=sin⁡3α×sin⁡2α=sin⁡π2×sin⁡π3=1×32=32\begin{aligned} & \cos \theta=\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{PS}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{PS}}|} \\& \cos \theta=\frac{0-1+0}{\sqrt{2} \sqrt{2}} \\& \cos \theta=\frac{-1}{2} \\& \theta=\frac{2 \pi}{3} \\& \alpha=\frac{2 \pi}{3}-\frac{\pi}{2} \text { So } \alpha=\frac{\pi}{6} \\& \sin ^{2} \frac{5 \alpha}{2}-\sin ^{2} \frac{\alpha}{2}=\sin \left(\frac{5 \alpha}{2}+\frac{\alpha}{2}\right) \sin \left(\frac{5 \alpha}{2}-\frac{\alpha}{2}\right) \\& =\sin \frac{6 \alpha}{2} \times \sin \frac{4 \alpha}{2} \\& =\sin 3 \alpha \times \sin 2 \alpha \\& =\sin \frac{\pi}{2} \times \sin \frac{\pi}{3} \\& =1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Mathematics
Chapter
Straight lines
Topic
Angle between lines, perpendicular distance & distance between parallel lines, foot, image