The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$, $l^2+m^2-n^2=0$ is

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Aicharya · Accepted Answer

Given, $l+m+n=0, \quad \Rightarrow \quad l=-m-n$ and $l^2+m^2-n^2=0$ $\begin{aligned} & 	herefore \quad(-m-n)^2+m^2-n^2=0 \ & \Rightarrow \quad 2 m^2+2 m n=0 \ & \Rightarrow \quad 2 m(m+n)=0 \ & \Rightarrow \quad m=0 	ext { or } m+n=0 \ & \end{aligned}$ If $m=0$, then $l=-n$ $	herefore \quad \frac{l_1}{-1}=\frac{m_1}{0}=\frac{n}{1}$ and if $m+n=0 \Rightarrow m=-n$, then $l=0$ $\begin{array}{ll} 	herefore & \frac{l_2}{0}=\frac{m_2}{-1}=\frac{n_2}{1} \ 	ext { ie, } & \left(l_1, m_1, n_1ight)=(-1,0,1) \ 	ext { and } & \left(l_2, m_2, n_2ight)=(0,-1,1) \ 	herefore & \cos 	heta=\frac{0+0+1}{\sqrt{1+0+1} \sqrt{0+1+1}}=\frac{1}{2} \ \Rightarrow & 	heta=\frac{\pi}{3} \end{array}$

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$, $l^2+m^2-n^2=0$ is

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