If the direction cosines of two lines satisfy | vedclass

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

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$\frac{\pi}{3}$

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

If the direction cosines of two lines satisfy the equations $l+m+n=0$ and $2lm+2ln-mn=0$, then the acute angle between those two lines is

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