Two adjacent sides of a parallelogram ABCD ar | vedclass

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(B) Step $1$: Calculate the magnitudes of $\vec{AB}$ and $\vec{AD}$.$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$

Vedclass · Accepted Answer

$\sqrt{17}/9$

Vedclass · Answer

(B) Step $1$: Calculate the magnitudes of $\vec{AB}$ and $\vec{AD}$.$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$|\vec{AD}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Step $2$: Find the angle $	heta$ between $\vec{AB}$ and $\vec{AD}$.$\cos 	heta = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{(2)(-1) + (10)(2) + (11)(2)}{15 	imes 3} = \frac{-2 + 20 + 22}{45} = \frac{40}{45} = \frac{8}{9}$.Step $3$: Since $\vec{AD'}$ is in the same plane and perpendicular to $\vec{AB}$, the angle between $\vec{AD'}$ and $\vec{AB}$ is $90^\circ$. The angle between $\vec{AD}$ and $\vec{AD'}$ is $\alpha$. Thus, $	heta = 90^\circ - \alpha$ (or $\alpha = 90^\circ - 	heta$).Step $4$: Calculate $\cos \alpha = \cos(90^\circ - 	heta) = \sin 	heta$.Since $\cos 	heta = 8/9$, $\sin 	heta = \sqrt{1 - (8/9)^2} = \sqrt{1 - 64/81} = \sqrt{17/81} = \sqrt{17}/9$.

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