Find the angle between two lines whose direct | vedclass

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Question

(A) Given the relations: $l + m + n = 0 \implies n = -(l + m)$ Substitute $n$ into the second equation: $l^2 + m^2 - (-(l + m))^2 = 0$ $l^2 + m^2 - (l

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$2\pi / 3$

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(A) Given the relations: $l + m + n = 0 \implies n = -(l + m)$ Substitute $n$ into the second equation: $l^2 + m^2 - (-(l + m))^2 = 0$ $l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$ $-2lm = 0 \implies lm = 0$ This implies either $l = 0$ or $m = 0$. Case $1$: If $l = 0$,then $m + n = 0 \implies n = -m$. The direction ratios are $(0, m, -m)$,which simplifies to $(0, 1, -1)$. Case $2$: If $m = 0$,then $l + n = 0 \implies n = -l$. The direction ratios are $(l, 0, -l)$,which simplifies to $(1, 0, -1)$. Let the angle between the lines be $	heta$. The direction ratios are $\vec{a} = (0, 1, -1)$ and $\vec{b} = (1, 0, -1)$. $\cos 	heta = \frac{|(0)(1) + (1)(0) + (-1)(-1)|}{\sqrt{0^2 + 1^2 + (-1)^2} \sqrt{1^2 + 0^2 + (-1)^2}}$ $\cos 	heta = \frac{|0 + 0 + 1|}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$ Since the angle between lines is usually taken as the acute angle,$\cos 	heta = 1/2 \implies 	heta = \pi / 3$. However,if we consider the direction cosines as vectors,the angle $	heta$ between vectors $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{-1}{\sqrt{2}\sqrt{2}} = -1/2$,which gives $	heta = 2\pi / 3$.

Find the angle between two lines whose direction cosines are given by the relations $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$.

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