Two adjacent sides of a parallelogram PQRS ar | vedclass

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(B) Let $\vec{u} = \vec{PQ} = (1, 0, 1)$ and $\vec{v} = \vec{PS} = (1, -1, 0)$.The angle $	heta$ between $\vec{PQ}$ and $\vec{PS}$ is given by $\cos

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) Let $\vec{u} = \vec{PQ} = (1, 0, 1)$ and $\vec{v} = \vec{PS} = (1, -1, 0)$.The angle $	heta$ between $\vec{PQ}$ and $\vec{PS}$ is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{(1)(1) + (0)(-1) + (1)(0)}{\sqrt{1^2+0^2+1^2} \sqrt{1^2+(-1)^2+0^2}} = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Thus, $	heta = 60^\circ$.The side $PS$ is rotated by an angle $\alpha$ to become perpendicular to $PQ$. Since the initial angle is $60^\circ$, rotating it to make the angle $90^\circ$ implies $\alpha = |90^\circ - 60^\circ| = 30^\circ$.Now, we calculate $\sin^2(\frac{5\alpha}{2}) - \sin^2(\frac{\alpha}{2})$ for $\alpha = 30^\circ$:$\sin^2(\frac{5 	imes 30^\circ}{2}) - \sin^2(\frac{30^\circ}{2}) = \sin^2(75^\circ) - \sin^2(15^\circ)$.Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:$\sin(75^\circ + 15^\circ) \sin(75^\circ - 15^\circ) = \sin(90^\circ) \sin(60^\circ) = 1 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

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