Suppose L_1 and L_2 are two lines having the | vedclass

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(B) Let the direction cosines of the line perpendicular to both $L_1$ and $L_2$ be $(l, m, n)$.Since the line is perpendicular to $L_1$ with direction

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$\frac{5}{3}$

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(B) Let the direction cosines of the line perpendicular to both $L_1$ and $L_2$ be $(l, m, n)$.Since the line is perpendicular to $L_1$ with direction ratios $(1, -2, -2)$, we have:$l - 2m - 2n = 0$  $(i)$Since the line is perpendicular to $L_2$ with direction ratios $(0, 2, 1)$, we have:$0l + 2m + n = 0 \Rightarrow n = -2m$  $(ii)$Substituting $(ii)$ into $(i)$:$l - 2m - 2(-2m) = 0$$l - 2m + 4m = 0$$l + 2m = 0 \Rightarrow l = -2m$We know that for direction cosines, $l^2 + m^2 + n^2 = 1$.Substituting $l = -2m$ and $n = -2m$:$(-2m)^2 + m^2 + (-2m)^2 = 1$$4m^2 + m^2 + 4m^2 = 1$$9m^2 = 1 \Rightarrow m^2 = \frac{1}{9} \Rightarrow |m| = \frac{1}{3}$Since $l = -2m$, $|l| = |-2m| = 2|m| = 2(\frac{1}{3}) = \frac{2}{3}$.Since $n = -2m$, $|n| = |-2m| = 2|m| = 2(\frac{1}{3}) = \frac{2}{3}$.Therefore, $|l| + |m| + |n| = \frac{2}{3} + \frac{1}{3} + \frac{2}{3} = \frac{5}{3}$.

Suppose $L_1$ and $L_2$ are two lines having the direction ratios $1, -2, -2$ and $0, 2, 1$ respectively. If the direction cosines of a line perpendicular to both $L_1$ and $L_2$ are $l, m, n$, then $|l| + |m| + |n| =$

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