Question

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

• A. \( \frac{1}{2} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{\sqrt{3}}{4} \)
• D. \( \frac{2\sqrt{3}}{5} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first need to find the initial angle between the adjacent sides \(\vec{PQ}\) and \(\vec{PS}\) using the dot product.
The problem states that side PS is rotated by an acute angle \(\alpha\) until it becomes completely perpendicular to PQ.
This implies the new angle between the vectors is \(90^\circ\).
Step 2: Key Formula or Approach:
The angle \(\theta\) between two vectors \(\vec{u}\) and \(\vec{v}\) is given by \(\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}\).
The trigonometric identity \(\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B)\) will be used to simplify the final expression.
Step 3: Detailed Explanation:
Let's calculate the initial angle \(\theta\) between \(\vec{PQ}\) and \(\vec{PS}\).
\[ \vec{PQ} = 0\hat{i} + 1\hat{j} + 1\hat{k} \implies |\vec{PQ}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \] \[ \vec{PS} = 1\hat{i} - 1\hat{j} + 0\hat{k} \implies |\vec{PS}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2} \] \[ \cos\theta = \frac{(0)(1) + (1)(-1) + (1)(0)}{\sqrt{2}\sqrt{2}} = \frac{-1}{2} \] Since \(\cos\theta = -1/2\), the initial angle is \(\theta = 120^\circ\).
The vector is rotated by an acute angle \(\alpha\) to make the new angle \(90^\circ\).
The new angle is \(\theta \pm \alpha = 90^\circ\).
So, \(120^\circ \pm \alpha = 90^\circ\).
Since \(\alpha\) must be an acute angle (\(\alpha<90^\circ\)), we take \(\alpha = 120^\circ - 90^\circ = 30^\circ\).
Now we evaluate the required trigonometric expression.
\[ E = \sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right) \] Substitute \(\alpha = 30^\circ\).
\[ E = \sin^2(75^\circ) - \sin^2(15^\circ) \] Apply the identity \(\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B)\) with \(A = 75^\circ\) and \(B = 15^\circ\).
\[ E = \sin(75^\circ - 15^\circ)\sin(75^\circ + 15^\circ) \] \[ E = \sin(60^\circ)\sin(90^\circ) \] We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\) and \(\sin(90^\circ) = 1\).
\[ E = \left(\frac{\sqrt{3}}{2}\right)(1) = \frac{\sqrt{3}}{2} \] Step 4: Final Answer:
The value of the expression is \(\frac{\sqrt{3}}{2}\).