If the direction cosines l, m, n of two lines | vedclass

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(D) Given the relations $l+m+n=0$ and $lm=0$. From $lm=0$,we have either $l=0$ or $m=0$. Case $1$: If $l=0$,then $m+n=0$,so $n=-m$. The direction cosi

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$\frac{\pi}{3}$

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(D) Given the relations $l+m+n=0$ and $lm=0$. From $lm=0$,we have either $l=0$ or $m=0$. Case $1$: If $l=0$,then $m+n=0$,so $n=-m$. The direction cosines are $(0, m, -m)$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1$,which gives $2m^2=1$,so $m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(0, 1, -1)$. Case $2$: If $m=0$,then $l+n=0$,so $n=-l$. The direction cosines are $(l, 0, -l)$. Similarly,$l^2+0^2+(-l)^2=1$,which gives $2l^2=1$,so $l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction ratios are $(1, 0, -1)$. Let the two lines have direction vectors $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$. The cosine of the angle $	heta$ between them is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. $\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$. $|\vec{a}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$ and $|\vec{b}| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$. $\cos 	heta = \frac{1}{\sqrt{2} 	imes \sqrt{2}} = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$.

If the direction cosines $l, m, n$ of two lines satisfy the relations $l+m+n=0$ and $lm=0$,then the angle between those two lines is

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