The acute angle between the lines whose direc | vedclass

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(C) Given equations are:$l+m+n=0$  ... $(i)$$2lm+2ln-mn=0$  ... (ii)From $(i)$,$m+n = -l$. Substituting this into (ii):$2l(m+n) - mn = 0$$2l(-l) - mn

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$\frac{\pi}{3}$

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(C) Given equations are:$l+m+n=0$  ... $(i)$$2lm+2ln-mn=0$  ... (ii)From $(i)$,$m+n = -l$. Substituting this into (ii):$2l(m+n) - mn = 0$$2l(-l) - mn = 0 \Rightarrow mn = -2l^2$  ... (iii)We know $l^2+m^2+n^2 = 1$. Also,$(m+n)^2 = m^2+n^2+2mn = (-l)^2 = l^2$.So,$m^2+n^2 = l^2 - 2mn = l^2 - 2(-2l^2) = 5l^2$.Substituting into $l^2+m^2+n^2 = 1$:$l^2 + 5l^2 = 1 \Rightarrow 6l^2 = 1 \Rightarrow l^2 = \frac{1}{6}$.Let the direction cosines of the two lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.From $mn = -2l^2$ and $m+n = -l$,$m$ and $n$ are roots of the quadratic $t^2 + lt - 2l^2 = 0$.$(t+2l)(t-l) = 0 \Rightarrow t = -2l, l$.Thus,the direction ratios are proportional to $(l, -2l, l)$ and $(l, l, -2l)$.Normalizing these,the direction cosines are proportional to $(1, -2, 1)$ and $(1, 1, -2)$.Let $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1^2+(-2)^2+1^2} \sqrt{1^2+1^2+(-2)^2}} = \frac{|1-2-2|}{\sqrt{6}\sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

The acute angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $2lm+2ln-mn=0$ is

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