Let bar{a}=4 bar{i}+5 bar{j}-bar{k},bar{b}=ba | vedclass

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

Vedclass · Accepted Answer

$21 \bar{i}-21 \bar{j}-21 \bar{k}$

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(C) Since $\bar{\alpha}$ is perpendicular to both $\bar{a}$ and $\bar{b}$,$\bar{\alpha}$ must be parallel to $\bar{a} 	imes \bar{b}$.$\bar{a} 	imes \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 4 & 5 & -1 \ 1 & -4 & 5 \end{vmatrix} = \bar{i}(25-4) - \bar{j}(20+1) + \bar{k}(-16-5) = 21 \bar{i} - 21 \bar{j} - 21 \bar{k}$.Let $\bar{\alpha} = k(21 \bar{i} - 21 \bar{j} - 21 \bar{k}) = 21k(\bar{i} - \bar{j} - \bar{k})$.Given $\bar{\alpha} \cdot \bar{c} = 63$,we have $21k(\bar{i} - \bar{j} - \bar{k}) \cdot (3 \bar{i} + \bar{j} - \bar{k}) = 63$.$21k(3 - 1 + 1) = 63 \implies 21k(3) = 63 \implies 63k = 63 \implies k = 1$.Therefore,$\bar{\alpha} = 21 \bar{i} - 21 \bar{j} - 21 \bar{k}$.

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$

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