The direction cosines of the line which is pe | vedclass

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(D) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\l

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

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(D) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\langle 1, -2, -2 \rangle$ and $\langle 0, 2, 1 \rangle$,we have: $1(a) - 2(b) - 2(c) = 0$  (Equation $1$) $0(a) + 2(b) + 1(c) = 0$  (Equation $2$) From Equation $2$,we get $c = -2b$. Substituting $c = -2b$ into Equation $1$: $a - 2b - 2(-2b) = 0$ $a - 2b + 4b = 0$ $a + 2b = 0 \implies a = -2b$. Let $b = -1$,then $a = 2$ and $c = 2$. The direction ratios are $\langle 2, -1, 2 \rangle$. The magnitude is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$.

The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to $\langle 1, -2, -2 \rangle$ and $\langle 0, 2, 1 \rangle$ is given by

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