Calculate Trigonometric Expression After Rotating Side of Parallelogram

To find the value of the given trigonometric expression, we first determine the initial angle between the vectors $\vec{PQ}$ and $\vec{PS}$.

Step 1: Calculate the initial angle between $\vec{PQ}$ and $\vec{PS}$

The given vectors representing the adjacent sides of the parallelogram $PQRS$ are: $\vec{PQ} = \hat{j} + \hat{k}$ $\vec{PS} = \hat{i} - \hat{j}$

The magnitudes of these vectors are: $|\vec{PQ}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$ $|\vec{PS}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$

The dot product of $\vec{PQ}$ and $\vec{PS}$ is: $\vec{PQ} \cdot \vec{PS} = (0)(1) + (1)(-1) + (1)(0) = -1$

Let $\theta$ be the angle between $\vec{PQ}$ and $\vec{PS}$. Using the dot product formula: $\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| |\vec{PS}|} = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = -\frac{1}{2}$

Since $\cos\theta = -\frac{1}{2}$, the angle between the two sides is: $\theta = 120^\circ \quad \text{or} \quad \frac{2\pi}{3}\text{ radians}$

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Step 2: Determine the acute angle of rotation $\alpha$

The side $PS$ is rotated about the point $P$ by an acute angle $\alpha$ in the plane of the parallelogram to make it perpendicular to $PQ$.

Since the angle between $PQ$ and the rotated side $PS'$ must be $90^\circ$, the required angle of rotation $\alpha$ is: $\alpha = |120^\circ - 90^\circ| = 30^\circ$

Since $30^\circ < 90^\circ$, $\alpha = 30^\circ$ is indeed an acute angle.

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Step 3: Evaluate the trigonometric expression

We need to calculate: $E = \sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right)$

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

Setting $A = \frac{5\alpha}{2}$ and $B = \frac{\alpha}{2}$: $A + B = \frac{5\alpha}{2} + \frac{\alpha}{2} = 3\alpha$ $A - B = \frac{5\alpha}{2} - \frac{\alpha}{2} = 2\alpha$

Substituting $\alpha = 30^\circ$: $A + B = 3(30^\circ) = 90^\circ$ $A - B = 2(30^\circ) = 60^\circ$

Therefore, the expression simplifies to: $E = \sin(90^\circ) \cdot \sin(60^\circ)$ $E = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

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Conclusion

The correct option is B.