Two adjacent sides of a parallelogram PQRS are given by PQ = j + k...

Let $\theta$ be the angle between the vectors $\vec{PQ}$ and $\vec{PS}$.

$\cos \theta = \dfrac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| |\vec{PS}|}$

$\cos \theta = \dfrac{(\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j})}{\sqrt{1^2 + 1^2} \sqrt{1^2 + (-1)^2}}$

$\cos \theta = \dfrac{-1}{\sqrt{2} \times \sqrt{2}} = -\dfrac{1}{2}$

$\theta = 120^{\circ}$

The side PS is rotated by an acute angle $\alpha$ in the plane of the parallelogram to become perpendicular to PQ. The new angle between the sides is $90^{\circ}$.

$\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$

The given expression is $\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right)$.

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

$\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right) = \sin\left(\dfrac{5\alpha}{2} + \dfrac{\alpha}{2}\right) \sin\left(\dfrac{5\alpha}{2} - \dfrac{\alpha}{2}\right)$

$= \sin(3\alpha) \sin(2\alpha)$

Substituting $\alpha = 30^{\circ}$:

$= \sin(90^{\circ}) \sin(60^{\circ})$

$= 1 \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}$

Answer: $\dfrac{\sqrt{3}}{2}$