Find the angle between the lines whose direction cosines are given by the equation $l + m + n = 0$ and ${l^2} + {m^2} - {n^2} = 0$.

Eliminating $n$ from both the equations, we have

${l^2} + {m^2} - {(l - m)^2} = 0$
$\Rightarrow$ ${l^2} + {m^2} - {l^2} - {m^2} + 2ml = 0 \Rightarrow 2lm = 0$

$\Rightarrow$ $lm = 0 \Rightarrow ( - m - n)m = 0$

$\Rightarrow$ $(m + n)m = 0$
$\Rightarrow$ $m = - n \Rightarrow m = 0$

$\Rightarrow$ $l = 0,l = - n$

Thus, Dr's two lines are proportional to $0, - n,n$ and $- n,0,n$ i.e., 0,-1,1 and -1,0,1 .

So, the vector parallel to these given lines are $\overrightarrow {\rm{a}} = - \widehat {\rm{j}} + \widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = - \widehat {\rm{i}} + \widehat {\rm{k}}$

Now, $\cos \theta = \frac{{\overrightarrow {\rm{a}} \overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} |}}$

$= \frac{1}{{\sqrt 2 }} \cdot \frac{1}{{\sqrt 2 }} \Rightarrow \cos \theta = \frac{1}{2}$

$\therefore$ $\theta = \frac{\pi }{3}$