If the direction cosines of two lines are given by $l+m+n=0$ and $m n-2 l m-2 n l=0$, then the acute angle…
If the direction cosines of two lines are given by $l+m+n=0$ and $m n-2 l m-2 n l=0$, then the acute angle between those lines is
$\frac{2 \pi}{5}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{60}$
Solution
$l+m+n=0 \Rightarrow l=-m-n \qquad$ ...(i)
$\begin{aligned}
& \text { and } m n-2 l m-2 n l=0 \\
& \Rightarrow m n+2(m+n) m+2(m+n) n=0 \\
& \Rightarrow(2 m+n)(m+2 n)=0 \Rightarrow n=-2 m \text { or } m=-2 n \\
& \Rightarrow l=m \text { or } l=n
\end{aligned}$
$\therefore$ Direction ratios of lines are $(1,1,-2)$ and $(1,-2,1)$, then angle between them is given by
$\cos \theta=\left|\frac{1-2-2}{\sqrt{6} \sqrt{6}}\right|=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$