Maths Q16 – JEE-MAIN 2026

Two adjacent sides of a parallelogram PQRS are given by

\vec{P Q} = \hat{j} + \hat{k}

and

\vec{P S} = \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 :

Clue 1 of 3: Vector Geometry

Understand how to find the angle between two vectors using their dot product.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate the initial angle between

\vec{P Q}

and

\vec{P S}

.✦ Active

Let $\vec{a} = \vec{P Q} = \hat{j} + \hat{k}$ and $\vec{b} = \vec{P S} = \hat{i} - \hat{j}$ . The dot product is:

\vec{a} \cdot \vec{b} = (0) (1) + (1) (- 1) + (1) (0) = - 1

The magnitudes are $| \vec{a} | = \sqrt{0^{2} + 1^{2} + 1^{2}} = \sqrt{2}$ and $| \vec{b} | = \sqrt{1^{2} + (- 1)^{2} + 0^{2}} = \sqrt{2}$ . The cosine of the angle $θ$ between them is:

\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |} = \frac{- 1}{\sqrt{2} \cdot \sqrt{2}} = - \frac{1}{2}

Thus, $θ = 120^{\circ}$ .

Step 2: Determine the rotation angle

α

.○ Expand

The initial angle between $\vec{P Q}$ and $\vec{P S}$ is $120^{\circ}$ . After rotation, the new vector $\vec{P S^{'}}$ is perpendicular to $\vec{P Q}$ , meaning the angle between them is $90^{\circ}$ . Since $α$ is an acute angle of rotation, it is the difference between the initial angle and $90^{\circ}$ :

α = | 120^{\circ} - 90^{\circ} | = 30^{\circ}

💡 Teacher's Secret Hint

Ensure the rotation angle $α$ is acute as stated in the problem.

Step 3: Evaluate the trigonometric expression.○ Expand

Substitute $α = 30^{\circ}$ into the expression $\sin^{2} (\frac{5 α}{2}) - \sin^{2} (\frac{α}{2})$ :

\sin^{2} (\frac{5 \cdot 30^{\circ}}{2}) - \sin^{2} (\frac{30^{\circ}}{2}) = \sin^{2} (75^{\circ}) - \sin^{2} (15^{\circ})

Using the trigonometric identity $\sin^{2} A - \sin^{2} B = \sin (A + B) \sin (A - B)$ :

\sin^{2} (75^{\circ}) - \sin^{2} (15^{\circ}) = \sin (75^{\circ} + 15^{\circ}) \sin (75^{\circ} - 15^{\circ}) = \sin (90^{\circ}) \sin (60^{\circ})

Since $\sin (90^{\circ}) = 1$ and $\sin (60^{\circ}) = \frac{\sqrt{3}}{2}$ , the value is:

1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}

💡 Teacher's Secret Hint

Remember common trigonometric identities to simplify calculations.

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Vector Geometry

🧠 Full Solution Path