If the direction cosines of two lines are giv | vedclass

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(D) Given the equations for direction cosines $l, m, n$:$l+m+n=0 \implies n = -(l+m)$Substitute $n$ into $l^2-5m^2+n^2=0$:$l^2-5m^2+(-l-m)^2=0$$l^2-5m

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$\frac{\pi}{3}$

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(D) Given the equations for direction cosines $l, m, n$:$l+m+n=0 \implies n = -(l+m)$Substitute $n$ into $l^2-5m^2+n^2=0$:$l^2-5m^2+(-l-m)^2=0$$l^2-5m^2+l^2+2lm+m^2=0$$2l^2+2lm-4m^2=0$$l^2+lm-2m^2=0$$(l+2m)(l-m)=0$Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$, i.e.,$(1, 1, -2)$.Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$, i.e.,$(-2, 1, 1)$.Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.The cosine of the angle $	heta$ is given by $\cos 	heta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.$\cos 	heta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$$\cos 	heta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.Therefore, $	heta = \frac{\pi}{3}$.

If the direction cosines of two lines are given by $l+m+n=0$ and $l^2-5m^2+n^2=0$, then the angle between them is

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