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

We have,

$l+m-n=0$ and $l^2+m^2-n^2=0$

$\begin{array}{l}\Rightarrow l+m=n\text{ and }{l}^2+m^2=n^2\\ \Rightarrow l^2+m^2=(l+m{)}^2\\ \Rightarrow 2lm=0\Rightarrow l=0\text{ or, }m=0.\end{array}$

If $l=0,$ Then

$l+m-n=0\Rightarrow m=n\therefore \frac{l}{0}=\frac{m}{1}=\frac{n}{1}$

If $m=0$, then

$l+m-n=0\Rightarrow l=n\therefore \frac{l}{1}=\frac{m}{0}=\frac{n}{1}$

Thus, the direction ratios of the lines are proportional to 0, 1, 1 and 1, 0, 1. Therefore, angle between them is given by

$\cos \theta =\frac{0\times 1+1\times 0+1\times 1}{\sqrt{0+1+1}\sqrt{1+0+1}}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$