If bar{a}=2 hat{i}+3 hat{j}+4 hat{k},bar{b}=h | vedclass

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(C) Since $\bar{d}$ is perpendicular to both $\bar{b}$ and $\bar{c}$,$\bar{d}$ must be parallel to $\bar{b} 	imes \bar{c}$.First,calculate $\bar{b} \

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$720$

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(C) Since $\bar{d}$ is perpendicular to both $\bar{b}$ and $\bar{c}$,$\bar{d}$ must be parallel to $\bar{b} 	imes \bar{c}$.First,calculate $\bar{b} 	imes \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & -2 \ -1 & 4 & 3 \end{vmatrix} = \hat{i}(-6+8) - \hat{j}(3-2) + \hat{k}(4-2) = 2 \hat{i} - \hat{j} + 2 \hat{k}$.Let $\bar{d} = k(2 \hat{i} - \hat{j} + 2 \hat{k})$.Given $\bar{a} \cdot \bar{d} = 18$,we have $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot k(2 \hat{i} - \hat{j} + 2 \hat{k}) = 18$.$k(4 - 3 + 8) = 18 \implies 9k = 18 \implies k = 2$.Thus,$\bar{d} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}$.Now,calculate $\bar{a} 	imes \bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 4 & -2 & 4 \end{vmatrix} = \hat{i}(12+8) - \hat{j}(8-16) + \hat{k}(-4-12) = 20 \hat{i} + 8 \hat{j} - 16 \hat{k}$.Finally,$|\bar{a} 	imes \bar{d}|^2 = 20^2 + 8^2 + (-16)^2 = 400 + 64 + 256 = 720$.

If $\bar{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\bar{b}=\hat{i}-2 \hat{j}-2 \hat{k}$,$\bar{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$ and if $\bar{d}$ is a vector perpendicular to both $\bar{b}$ and $\bar{c}$,and $\bar{a} \cdot \bar{d}=18$,then $|\bar{a} \times \bar{d}|^2=$

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