The angle between the lines whose direction c | vedclass

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(C) Given,$l+m+n=0 \implies l = -m-n$ and $l^2+m^2-n^2=0$.Substituting $l = -m-n$ into the second equation:$(-m-n)^2 + m^2 - n^2 = 0$$m^2 + 2mn + n^2

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$\frac{\pi}{3}$

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(C) Given,$l+m+n=0 \implies l = -m-n$ and $l^2+m^2-n^2=0$.Substituting $l = -m-n$ into the second equation:$(-m-n)^2 + m^2 - n^2 = 0$$m^2 + 2mn + n^2 + m^2 - n^2 = 0$$2m^2 + 2mn = 0$$2m(m+n) = 0$.This gives two cases:Case $1$: If $m=0$,then $l = -n$. The direction ratios are $(-n, 0, n)$,which simplifies to $(-1, 0, 1)$. Let $\vec{v_1} = (-1, 0, 1)$.Case $2$: If $m+n=0$,then $m = -n$. Substituting into $l = -m-n$,we get $l = -(-n)-n = 0$. The direction ratios are $(0, -n, n)$,which simplifies to $(0, -1, 1)$. Let $\vec{v_2} = (0, -1, 1)$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (-1)(0) + (0)(-1) + (1)(1) = 1$.$|\vec{v_1}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.$|\vec{v_2}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.$\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

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

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