Let vec{a}=2 hat{i}-hat{j}+2 hat{k} and vec{b | vedclass

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(D) Given $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}+2 \hat{j}-\hat{k}$.Since $\vec{v}$ is in th

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$1494$

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(D) Given $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}+2 \hat{j}-\hat{k}$.Since $\vec{v}$ is in the plane of $\vec{a}$ and $\vec{b}$,$\vec{v} = x\vec{a} + y\vec{b}$.Since $\vec{v} \perp \vec{c}$,$\vec{v} \cdot \vec{c} = 0$. Also,$\vec{v}$ is perpendicular to the normal of the plane,which is $\vec{n} = \vec{a} 	imes \vec{b}$.Thus,$\vec{v}$ is parallel to $\vec{c} 	imes (\vec{a} 	imes \vec{b})$.Using the vector triple product formula,$\vec{v} = \lambda [(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}]$.Calculate dot products: $\vec{c} \cdot \vec{b} = (3)(1) + (2)(2) + (-1)(-1) = 3+4+1 = 8$.$\vec{c} \cdot \vec{a} = (3)(2) + (2)(-1) + (-1)(2) = 6-2-2 = 2$.So,$\vec{v} = \lambda [8(2 \hat{i}-\hat{j}+2 \hat{k}) - 2(\hat{i}+2 \hat{j}-\hat{k})] = \lambda [16 \hat{i}-8 \hat{j}+16 \hat{k} - 2 \hat{i}-4 \hat{j}+2 \hat{k}] = \lambda [14 \hat{i}-12 \hat{j}+18 \hat{k}]$.The projection of $\vec{v}$ on $\vec{a}$ is $\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|} = 19$.$|\vec{a}| = \sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3$.$\vec{v} \cdot \vec{a} = \lambda [14(2) - 12(-1) + 18(2)] = \lambda [28+12+36] = 76\lambda$.So,$\frac{76\lambda}{3} = 19 \Rightarrow 76\lambda = 57 \Rightarrow \lambda = \frac{57}{76} = \frac{3}{4}$.Thus,$\vec{v} = \frac{3}{4} [14 \hat{i}-12 \hat{j}+18 \hat{k}] = \frac{3}{2} [7 \hat{i}-6 \hat{j}+9 \hat{k}]$.$|2\vec{v}|^2 = 4|\vec{v}|^2 = 4 	imes \left(\frac{3}{4}ight)^2 	imes (14^2 + (-12)^2 + 18^2) = 4 	imes \frac{9}{16} 	imes (196 + 144 + 324) = \frac{9}{4} 	imes 664 = 9 	imes 166 = 1494$.

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