If the direction cosines (l, m, n) of two lin | vedclass

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(A) Given the relations for the direction cosines $(l, m, n)$ of the two lines: $l+m+n=0$ and $lm=0$ From $lm=0$, we have two cases: $l=0$ or $m=0$. C

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$\frac{\pi}{3}$

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(A) Given the relations for the direction cosines $(l, m, n)$ of the two lines: $l+m+n=0$ and $lm=0$ From $lm=0$, we have two cases: $l=0$ or $m=0$. Case $1$: If $l=0$, then $0+m+n=0 \Rightarrow n=-m$. The direction ratios are $(0, m, -m)$, which simplifies to $(0, 1, -1)$. Case $2$: If $m=0$, then $l+0+n=0 \Rightarrow n=-l$. The direction ratios are $(l, 0, -l)$, which simplifies to $(1, 0, -1)$. Let the direction vectors be $\vec{a} = (0, 1, -1)$ and $\vec{b} = (1, 0, -1)$. The cosine of the angle $	heta$ between the two lines is given by: $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(0)(1) + (1)(0) + (-1)(-1)|}{\sqrt{0^2+1^2+(-1)^2} \cdot \sqrt{1^2+0^2+(-1)^2}}$ $\cos 	heta = \frac{|0 + 0 + 1|}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$ Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If the direction cosines $(l, m, n)$ of two lines are connected by the relations $l+m+n=0$ and $lm=0$, then the angle between those lines is

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