Find the angle between the lines whose direct | vedclass

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(A) Given equations are $l+m+n=0$ $(1)$ and $l^{2}+m^{2}-n^{2}=0$ $(2)$.From $(1)$,$n = -(l+m)$. Substituting this into $(2)$:$l^{2}+m^{2}-(-(l+m))^{2

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$\frac{\pi}{3}$

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(A) Given equations are $l+m+n=0$ $(1)$ and $l^{2}+m^{2}-n^{2}=0$ $(2)$.From $(1)$,$n = -(l+m)$. Substituting this into $(2)$:$l^{2}+m^{2}-(-(l+m))^{2}=0$$l^{2}+m^{2}-(l^{2}+m^{2}+2lm)=0$$-2lm=0 \Rightarrow lm=0$.Case $1$: $l=0$. From $(1)$,$m+n=0 \Rightarrow m=-n$. Direction ratios are $(0, -n, n)$ or $(0, -1, 1)$.Case $2$: $m=0$. From $(1)$,$l+n=0 \Rightarrow l=-n$. Direction ratios are $(-n, 0, n)$ or $(-1, 0, 1)$.Let $\vec{a} = 0\hat{i} - 1\hat{j} + 1\hat{k}$ and $\vec{b} = -1\hat{i} + 0\hat{j} + 1\hat{k}$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(0)(-1) + (-1)(0) + (1)(1)|}{\sqrt{0^2+(-1)^2+1^2} \sqrt{(-1)^2+0^2+1^2}} = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

Find the angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $l^{2}+m^{2}-n^{2}=0$.

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