Question

Let \( \overrightarrow{PS} = \hat{i} + \hat{j} \) and \( \overrightarrow{PQ} = -\hat{j} + \hat{k} \). If \( \overrightarrow{PS} \) must be rotated by an angle \( \alpha \) such that \( \overrightarrow{PS} \) is perpendicular to \( \overrightarrow{PQ} \), then \( \left( \sin^2 \frac{5\alpha}{2} - \sin^2 \frac{\alpha}{2} \right) \) equals

• A. \( \frac{1}{2} \)
• B. \( 1 \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( 0 \)

Hint:

For angle-based vector questions, first find the original angle using the dot product formula. Then compare it with the required angle and substitute the rotation angle into the given trigonometric expression.

Answer 1

The Correct Option is C

Step 1: Write the given vectors and find their angle.
We are given:
\[ \overrightarrow{PS}=\hat{i}+\hat{j} \] and
\[ \overrightarrow{PQ}=-\hat{j}+\hat{k} \] Now, let the angle between \( \overrightarrow{PS} \) and \( \overrightarrow{PQ} \) be \( \theta \). Then:
\[ \cos \theta=\frac{\overrightarrow{PS}\cdot \overrightarrow{PQ}}{|\overrightarrow{PS}|\,|\overrightarrow{PQ}|} \] First, calculate the dot product:
\[ (\hat{i}+\hat{j})\cdot(-\hat{j}+\hat{k})=0-1+0=-1 \] Now, calculate the magnitudes:
\[ |\overrightarrow{PS}|=\sqrt{1^2+1^2}=\sqrt{2} \] \[ |\overrightarrow{PQ}|=\sqrt{(-1)^2+1^2}=\sqrt{2} \] So:
\[ \cos \theta=\frac{-1}{\sqrt{2}\cdot \sqrt{2}}=\frac{-1}{2} \] Hence:
\[ \theta=120^\circ \]
Step 2: Find the angle of rotation \( \alpha \).
Initially, the angle between \( \overrightarrow{PS} \) and \( \overrightarrow{PQ} \) is \(120^\circ\). We need to rotate \( \overrightarrow{PS} \) so that it becomes perpendicular to \( \overrightarrow{PQ} \), that is, the new angle should be \(90^\circ\).
Therefore, the required minimum rotation is:
\[ \alpha=120^\circ-90^\circ=30^\circ \]
Step 3: Evaluate the required expression.
We need to find:
\[ \sin^2\frac{5\alpha}{2}-\sin^2\frac{\alpha}{2} \] Substitute \( \alpha=30^\circ \):
\[ \sin^2\frac{5\cdot 30^\circ}{2}-\sin^2\frac{30^\circ}{2} \] \[ =\sin^2 75^\circ-\sin^2 15^\circ \] Now use the identity:
\[ \sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B) \] So:
\[ \sin^2 75^\circ-\sin^2 15^\circ=\sin(90^\circ)\sin(60^\circ) \] \[ =1\cdot \frac{\sqrt{3}}{2} \] \[ =\frac{\sqrt{3}}{2} \]
Step 4: Conclusion.
Thus, the value of \( \left( \sin^2 \frac{5\alpha}{2}-\sin^2 \frac{\alpha}{2} \right) \) is \( \frac{\sqrt{3}}{2} \).
Final Answer: \( \frac{\sqrt{3}}{2} \)