Let bar{a}=bar{i}-bar{j}+bar{k}, bar{b}=bar{i | vedclass

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(B) Since $\bar{d}$ is perpendicular to both $\bar{a}$ and $\bar{b}$, $\bar{d}$ must be parallel to $\bar{a} 	imes \bar{b}$.$\bar{a} 	imes \bar{b} =

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$70$

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(B) Since $\bar{d}$ is perpendicular to both $\bar{a}$ and $\bar{b}$, $\bar{d}$ must be parallel to $\bar{a} 	imes \bar{b}$.$\bar{a} 	imes \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 1 & -1 & 1 \ 1 & -2 & -2 \end{vmatrix} = \bar{i}(2+2) - \bar{j}(-2-1) + \bar{k}(-2+1) = 4\bar{i} + 3\bar{j} - \bar{k}$.Let $\bar{d} = k(4\bar{i} + 3\bar{j} - \bar{k})$ for some scalar $k$.Given $|\bar{d} 	imes \bar{c}| = 14$. Note that $\bar{d} 	imes \bar{c} = k(4\bar{i} + 3\bar{j} - \bar{k}) 	imes (6\bar{i} + 3\bar{j} - 2\bar{k})$.$(4\bar{i} + 3\bar{j} - \bar{k}) 	imes (6\bar{i} + 3\bar{j} - 2\bar{k}) = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 4 & 3 & -1 \ 6 & 3 & -2 \end{vmatrix} = \bar{i}(-6+3) - \bar{j}(-8+6) + \bar{k}(12-18) = -3\bar{i} + 2\bar{j} - 6\bar{k}$.Magnitude is $\sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$.So, $|\bar{d} 	imes \bar{c}| = |k| 	imes 7 = 14$, which implies $|k| = 2$.Now, $|\bar{d} \cdot \bar{c}| = |k(4\bar{i} + 3\bar{j} - \bar{k}) \cdot (6\bar{i} + 3\bar{j} - 2\bar{k})| = |k| |24 + 9 + 2| = 2 	imes 35 = 70$.

Let $\bar{a}=\bar{i}-\bar{j}+\bar{k}, \bar{b}=\bar{i}-2\bar{j}-2\bar{k}, \bar{c}=6\bar{i}+3\bar{j}-2\bar{k}$ be three vectors. If $\bar{d}$ is a vector perpendicular to both $\bar{a}$ and $\bar{b}$, and $|\bar{d} \times \bar{c}|=14$, then $|\bar{d} \cdot \bar{c}|=$

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