If the direction cosines of two lines are suc | vedclass

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(B) Given equations for direction cosines $(l, m, n)$ are:$l+m+n=0$ $\dots(i)$$l^2+m^2-n^2=0$ $\dots(ii)$We also know that for direction cosines,$l^2+

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$\pi / 3$

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(B) Given equations for direction cosines $(l, m, n)$ are:$l+m+n=0$ $\dots(i)$$l^2+m^2-n^2=0$ $\dots(ii)$We also know that for direction cosines,$l^2+m^2+n^2=1$ $\dots(iii)$From $(i)$,$l+m = -n$. Squaring both sides: $l^2+m^2+2lm = n^2$. Substituting $l^2+m^2 = n^2$ from $(ii)$ into this,we get $n^2+2lm = n^2$,which implies $2lm = 0$,so $l=0$ or $m=0$.Case $1$: If $l=0$,then $m+n=0 \implies m=-n$. Substituting into $(iii)$,$0^2+(-n)^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$. Thus,$(l_1, m_1, n_1) = (0, -1/\sqrt{2}, 1/\sqrt{2})$ and $(0, 1/\sqrt{2}, -1/\sqrt{2})$.Case $2$: If $m=0$,then $l+n=0 \implies l=-n$. Substituting into $(iii)$,$(-n)^2+0^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$. Thus,$(l_2, m_2, n_2) = (-1/\sqrt{2}, 0, 1/\sqrt{2})$ and $(1/\sqrt{2}, 0, -1/\sqrt{2})$.The angle $	heta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos 	heta = |l_1l_2 + m_1m_2 + n_1n_2|$.Taking $L_1 = (0, -1/\sqrt{2}, 1/\sqrt{2})$ and $L_2 = (-1/\sqrt{2}, 0, 1/\sqrt{2})$:$\cos 	heta = |(0)(-1/\sqrt{2}) + (-1/\sqrt{2})(0) + (1/\sqrt{2})(1/\sqrt{2})| = |0 + 0 + 1/2| = 1/2$.Therefore,$	heta = \cos^{-1}(1/2) = \pi/3$.

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$,then the angle between them is:

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