The angle between the lines whose direction c | vedclass

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(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.From the first equation, $l = -(m+n)$.Substituting this into the second equation: $(-m-n)^2 + m^

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$\frac{\pi}{3}$

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(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.From the first equation, $l = -(m+n)$.Substituting this into the second equation: $(-m-n)^2 + m^2 - n^2 = 0$.$m^2 + 2mn + n^2 + m^2 - n^2 = 0$.$2m^2 + 2mn = 0 \Rightarrow 2m(m+n) = 0$.This gives two cases:Case $1$: $m=0$. Then $l = -n$. The direction ratios are $(-1, 0, 1)$.Case $2$: $m = -n$. Then $l = 0$. The direction ratios are $(0, -1, 1)$.Let the two lines be represented by vectors $\vec{a} = -\hat{i} + \hat{k}$ and $\vec{b} = -\hat{j} + \hat{k}$.The angle $	heta$ between them is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = (-1)(0) + (0)(-1) + (1)(1) = 1$.$|\vec{a}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.$|\vec{b}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.$\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.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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