Two adjacent sides of a parallelogram ABCD ar | vedclass

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

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

(A) Let $\vec{u} = \overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{v} = \overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitudes: $|\vec{u}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Since $AD'$ is in the plane of the parallelogram and perpendicular to $AB$,it must be in the direction of the projection of $AD$ onto the plane perpendicular to $AB$ within the parallelogram plane.The vector $\vec{AD'}$ is the component of $\vec{AD}$ perpendicular to $\vec{AB}$ in the plane. Specifically,$\vec{AD'} = \vec{AD} - 	ext{proj}_{\vec{AB}}(\vec{AD}) = \vec{AD} - \frac{\vec{AD} \cdot \vec{AB}}{|\vec{AB}|^2} \vec{AB}$.Calculate $\vec{AD} \cdot \vec{AB} = (-1)(2) + (2)(10) + (2)(11) = -2 + 20 + 22 = 40$.So,$\vec{AD'} = \vec{AD} - \frac{40}{225} \vec{AB} = \vec{AD} - \frac{8}{45} \vec{AB}$.Alternatively,using the property of rotation,$\cos \alpha = \frac{|\vec{AD} \cdot \vec{AB}|}{|\vec{AD}| |\vec{AB}|} = \frac{40}{3 	imes 15} = \frac{40}{45} = \frac{8}{9}$.

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