The angle between the lines whose direction c | vedclass

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(C) We are given the equations for direction cosines $l, m, n$ as $l + m + n = 0$ and $l^2 = m^2 + n^2$.From the first equation,$n = -(l + m)$.Substit

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(C) We are given the equations for direction cosines $l, m, n$ as $l + m + n = 0$ and $l^2 = m^2 + n^2$.From the first equation,$n = -(l + m)$.Substituting this into the second equation: $l^2 = m^2 + (-(l + m))^2$.$l^2 = m^2 + l^2 + m^2 + 2lm$.$0 = 2m^2 + 2lm$.$2m(m + l) = 0$.This gives two cases: $m = 0$ or $m = -l$.Case $1$: If $m = 0$,then $l + 0 + n = 0 \Rightarrow n = -l$. The direction ratios are $(l, 0, -l)$,which is proportional to $(1, 0, -1)$. Let $\vec{a} = \hat{i} - \hat{k}$.Case $2$: If $m = -l$,then $l + (-l) + n = 0 \Rightarrow n = 0$. The direction ratios are $(l, -l, 0)$,which is proportional to $(1, -1, 0)$. Let $\vec{b} = \hat{i} - \hat{j}$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = (1)(1) + (0)(-1) + (-1)(0) = 1$.$|\vec{a}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$.$\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 = m^2 + n^2$ is

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