Two adjacent sides of a parallelogram ABCD ar | vedclass

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(B) Let $\vec{u} = \overline{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{v} = \overline{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate th

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$\frac{\sqrt{17}}{9}$

Vedclass · Answer

(B) Let $\vec{u} = \overline{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{v} = \overline{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitudes: $|\vec{u}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.The angle $	heta$ between $\vec{u}$ and $\vec{v}$ is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \frac{(2)(-1) + (10)(2) + (11)(2)}{15 	imes 3} = \frac{-2 + 20 + 22}{45} = \frac{40}{45} = \frac{8}{9}$.Since $\cos 	heta = \frac{8}{9}$,we have $\sin 	heta = \sqrt{1 - (8/9)^2} = \sqrt{17}/9$.When $\vec{v}$ is rotated by $\alpha$ to become $\vec{v}'$ such that $\vec{v}' \perp \vec{u}$,the new angle between $\vec{v}'$ and $\vec{u}$ is $90^\circ$.The rotation $\alpha$ is the difference between the original angle $	heta$ and $90^\circ$ (or vice versa). Since $\alpha$ is acute,$\alpha = |	heta - 90^\circ|$.Thus,$\cos \alpha = \cos |	heta - 90^\circ| = \sin 	heta = \frac{\sqrt{17}}{9}$.

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