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(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $\langle -1, 2, 2 \

Vedclass · Accepted Answer

$\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}$

Vedclass · Answer

(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $\langle -1, 2, 2 angle$ and $\langle 0, 2, 1 angle$,the vector $\vec{n} = \langle a, b, c angle$ is parallel to the cross product of the two given vectors.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 2 \ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-1-0) + \hat{k}(-2-0) = -2\hat{i} + \hat{j} - 2\hat{k}$.Thus,the direction ratios are $\langle -2, 1, -2 angle$ or $\langle 2, -1, 2 angle$.The magnitude of the vector is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} angle$.

The direction cosines of the line,which is perpendicular to the lines with direction ratios $-1, 2, 2$ and $0, 2, 1$,are respectively

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