The angle between the lines whose direction c | vedclass

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(B) Given equations are $\ell+m+n=0$ $(1)$ and $\ell^2+m^2-n^2=0$ $(2)$. From $(1)$,$n = -(\ell+m)$. Substituting this into $(2)$: $\ell^2+m^2-(-\ell-

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$\frac{\pi}{3}$

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(B) Given equations are $\ell+m+n=0$ $(1)$ and $\ell^2+m^2-n^2=0$ $(2)$. From $(1)$,$n = -(\ell+m)$. Substituting this into $(2)$: $\ell^2+m^2-(-\ell-m)^2 = 0$. $\ell^2+m^2-(\ell^2+m^2+2\ell m) = 0$. $-2\ell m = 0$,which implies $\ell=0$ or $m=0$. Case $1$: If $\ell=0$,then $n=-m$. The direction ratios are $(0, m, -m)$,which simplifies to $(0, 1, -1)$. Case $2$: If $m=0$,then $n=-\ell$. The direction ratios are $(\ell, 0, -\ell)$,which simplifies to $(1, 0, -1)$. Let the direction vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$. The angle $	heta$ between them is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. $\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$. $|\vec{a}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$. $|\vec{b}| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$. $\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$.

The angle between the lines whose direction cosines $(\ell, m, n)$ satisfy the equations $\ell+m+n=0$ and $\ell^2+m^2-n^2=0$ is:

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