ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 10. vector algebra

Two adjacent sides of a parallelogram PQRS are given by \(\vec{PQ} = \hat{j} + \hat{k}\) and \(\vec{PS} = \hat{i} - \hat{j}\). 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 \(\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right)\) is equal to:

  1. A \(\dfrac{1}{2}\)
  2. B \(\dfrac{\sqrt{3}}{2}\)
  3. C \(\dfrac{\sqrt{3}}{4}\)
  4. D \(\dfrac{2\sqrt{3}}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{\sqrt{3}}{2}\)

Step-by-step Solution

Detailed explanation

Let \(\theta\) be the angle between the vectors \(\vec{PQ}\) and \(\vec{PS}\). \(\cos \theta = \dfrac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| |\vec{PS}|}\) \(\cos \theta = \dfrac{(\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j})}{\sqrt{1^2 + 1^2} \sqrt{1^2 + (-1)^2}}\)…
From JEE Mains
Explore more questions on app