If the relation between the direction ratios | vedclass

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(B) Given equations are $l+m+n=0 \Rightarrow l = -(m+n)$.Substitute $l$ into the second equation: $2(-(m+n))m + 2mn - (-(m+n))n = 0$.$-2m^2 - 2mn + 2m

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$\frac{2\pi}{3}$

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(B) Given equations are $l+m+n=0 \Rightarrow l = -(m+n)$.Substitute $l$ into the second equation: $2(-(m+n))m + 2mn - (-(m+n))n = 0$.$-2m^2 - 2mn + 2mn + n^2 + mn = 0$.$-2m^2 + mn + n^2 = 0 \Rightarrow 2m^2 - mn - n^2 = 0$.Dividing by $n^2$ (assuming $n 
eq 0$): $2(\frac{m}{n})^2 - (\frac{m}{n}) - 1 = 0$.Let $x = \frac{m}{n}$,then $2x^2 - x - 1 = 0 \Rightarrow (2x+1)(x-1) = 0$.So,$\frac{m}{n} = 1$ or $\frac{m}{n} = -\frac{1}{2}$.Case $1$: If $m=n$,then $l = -(n+n) = -2n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.Case $2$: If $m = -\frac{1}{2}n$,then $l = -(-\frac{1}{2}n + n) = -\frac{1}{2}n$. Direction ratios are $(-\frac{1}{2}n, -\frac{1}{2}n, n)$,i.e.,$(-1, -1, 2)$.Let $\vec{a} = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} - \hat{j} + 2\hat{k}$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|(-2)(-1) + (1)(-1) + (1)(2)|}{\sqrt{4+1+1}\sqrt{1+1+4}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Thus,$	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$. However,the angle between lines is often defined as the obtuse angle if specified by the context of the given options,or the acute angle. Given the options,$\frac{2\pi}{3}$ is the supplementary angle.

If the relation between the direction ratios of two lines in $\mathbb{R}^3$ are given by $l+m+n=0$ and $2lm+2mn-ln=0$,then the angle between the lines is ($l, m, n$ have their usual meaning).

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