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$ $\dots(ii)$Also, we know that $l^2+m^2+n^2=1$ $\dots(iii

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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$ $\dots(ii)$Also, we know that $l^2+m^2+n^2=1$ $\dots(iii)$From $(i)$, $l+m = -n$. Squaring both sides, $l^2+m^2+2lm = n^2$, so $l^2+m^2 = n^2-2lm$.Substituting this into $(ii)$, we get $(n^2-2lm) - n^2 = 0$, which implies $-2lm = 0$, so $l=0$ or $m=0$.Case $1$: If $l=0$, then from $(i)$, $m+n=0 \implies m=-n$. Substituting into $(iii)$, $0^2+(-n)^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$.Thus, the direction cosines are $(0, 1/\sqrt{2}, -1/\sqrt{2})$ and $(0, -1/\sqrt{2}, 1/\sqrt{2})$.Case $2$: If $m=0$, then from $(i)$, $l+n=0 \implies l=-n$. Substituting into $(iii)$, $(-n)^2+0^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$.Thus, the direction cosines are $(1/\sqrt{2}, 0, -1/\sqrt{2})$ and $(-1/\sqrt{2}, 0, 1/\sqrt{2})$.Let the two lines have direction cosines $(l_1, m_1, n_1) = (0, 1/\sqrt{2}, -1/\sqrt{2})$ and $(l_2, m_2, n_2) = (1/\sqrt{2}, 0, -1/\sqrt{2})$.The cosine of the angle $	heta$ between them is given by $\cos 	heta = |l_1l_2 + m_1m_2 + n_1n_2| = |0(1/\sqrt{2}) + (1/\sqrt{2})(0) + (-1/\sqrt{2})(-1/\sqrt{2})| = |1/2| = 1/2$.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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