The angle between the lines whose direction c | vedclass

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(C) Given,$l^2+m^2-n^2=0$ $(i)$ and $l+m+n=0$ $(ii)$.From $(ii)$,$n=-(l+m)$. Substituting this into $(i)$:$l^2+m^2=(-(l+m))^2 = l^2+m^2+2lm$.This impl

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$\frac{\pi}{3}$

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(C) Given,$l^2+m^2-n^2=0$ $(i)$ and $l+m+n=0$ $(ii)$.From $(ii)$,$n=-(l+m)$. Substituting this into $(i)$:$l^2+m^2=(-(l+m))^2 = l^2+m^2+2lm$.This implies $2lm=0$,so $l=0$ or $m=0$.Case $1$: If $l=0$,then $n=-m$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1 \Rightarrow 2m^2=1 \Rightarrow m=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.Case $2$: If $m=0$,then $n=-l$. Since $l^2+m^2+n^2=1$,we have $l^2+0^2+(-l)^2=1 \Rightarrow 2l^2=1 \Rightarrow l=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.Let the direction vectors of the two lines be $\vec{a} = (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.The angle $	heta$ between the lines is given by $\cos 	heta = |l_1l_2 + m_1m_2 + n_1n_2|$.$\cos 	heta = |(0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

The angle between the lines whose direction cosines are given by the equations $l^2+m^2-n^2=0$ and $l+m+n=0$ is

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