If the direction cosines of two lines are giv | vedclass

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

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$\frac{\pi}{3}$

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(B) Given equations are $l+m+n=0$ and $mn-2lm-2nl=0$. From the first equation,$l = -(m+n)$. Substituting this into the second equation: $mn - 2(-(m+n))m - 2(-(m+n))n = 0$ $mn + 2m^2 + 2mn + 2mn + 2n^2 = 0$ $2m^2 + 5mn + 2n^2 = 0$ $(2m+n)(m+2n) = 0$. Case $1$: $n = -2m$. Substituting into $l+m+n=0$,we get $l+m-2m=0 \Rightarrow l=m$. So,direction ratios are $(1, 1, -2)$. Case $2$: $m = -2n$. Substituting into $l+m+n=0$,we get $l-2n+n=0 \Rightarrow l=n$. So,direction ratios are $(1, -2, 1)$. Let the direction ratios be $\vec{a} = (1, 1, -2)$ and $\vec{b} = (1, -2, 1)$. The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. $\cos 	heta = \frac{|(1)(1) + (1)(-2) + (-2)(1)|}{\sqrt{1^2+1^2+(-2)^2} \sqrt{1^2+(-2)^2+1^2}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$. Thus,$	heta = \frac{\pi}{3}$.

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

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