Q55 / 120 · VITEEE 2009 · Mathematics
The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$, $l^{2}+m^{2}-n^{2}=0$ is
- A$\frac{\pi}{6}$
- B$\frac{\pi}{4}$
- Correct answerC$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
Solution
$$
\begin{array}{l}
l+m+n=0, \Rightarrow \\
\text { and } l^{2}+m^{2}-n^{2}=0 \\
\therefore(-m-n)^{2}+m^{2}-n^{2}=0 \\
\Rightarrow 2 m^{2}+2 m n=0 \\
\Rightarrow 2 m(m+n)=0 \\
\Rightarrow m=0 \text { or } m+n=0 \\
\text { If } m=0, \text { then } l= \\
\therefore \frac{l_{1}}{-1}=\frac{m_{1}}{0}=\frac{n}{1}
\end{array}
$$
and if $m+n=0 \Rightarrow m=-n$, then $l=0$
$$
\begin{array}{l}
\therefore \frac{l_{2}}{0}=\frac{m_{2}}{-1}=\frac{n_{2}}{1} \\
\text { ie., }\left(l_{1}, m_{1}, n_{1}\right)=(-1,0,1) \\
\text { and }\left(l_{2}, m_{2}, n_{2}\right)=(0,-1,1) \\
\therefore \cos \theta=\frac{0+0+1}{\sqrt{1+0+1} \sqrt{0+1+1}}=\frac{1}{2} \\
\Rightarrow \theta=\frac{\pi}{3}
\end{array}
$$
Hence, the correct answer is option: 'C'.