Two adjacent sides of a parallelogram ABCD ar | vedclass

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Question

(C) Let $	heta$ be the angle between $\overline{AB}$ and $\overline{AD}$.$\cos 	heta = \frac{\overline{AB} \cdot \overline{AD}}{|\overline{AB}||\ove

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $	heta$ be the angle between $\overline{AB}$ and $\overline{AD}$.$\cos 	heta = \frac{\overline{AB} \cdot \overline{AD}}{|\overline{AB}||\overline{AD}|} = \frac{(2 \hat{i}+10 \hat{j}+11 \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{4+100+121} \sqrt{1+4+4}} = \frac{-2+20+22}{\sqrt{225} \sqrt{9}} = \frac{40}{15 	imes 3} = \frac{40}{45} = \frac{8}{9}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - (\frac{8}{9})^2} = \sqrt{\frac{81-64}{81}} = \frac{\sqrt{17}}{9}$.Let $\alpha$ be the angle of rotation of $AD$ such that $AD^{\prime} \perp AB$. The angle between $AB$ and $AD^{\prime}$ is $90^{\circ}$.Thus,$\alpha + 	heta = 90^{\circ}$,which implies $\alpha = 90^{\circ} - 	heta$.Therefore,$\cos \alpha = \cos(90^{\circ} - 	heta) = \sin 	heta$.Substituting the value of $\sin 	heta$,we get $\cos \alpha = \frac{\sqrt{17}}{9}$.

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