The angle between the lines whose direction cosines satisfy the equat…

We have, $ l + m + n = 0 $ ............(i)

and, $ l^2 = m^2 + n^2 $ ...........(ii)

$∴ (-m -n)^2 = m^2 +n^2 $            [On eliminating l ]

$⇒ 2mn = 0 ⇒ m = 0 \, or \, n = 0 $

 Now, 

$m = 0 ⇒ l + n = 0 $ and $l^2 = n^2 $  [Putting m = 0 in (i) & (ii)]

$⇒ l = -n $

Thus, the direction ratios of one of the two lines are proportional to $-n, 0, n \, or \, -1, 0, 1

When $n = 0,$

$l + m + n = 0 $ and $ l^2 = m^2 + n^2 $

$⇒ l + m = 0 $ and $l^2 = m^2 ⇒ l = -m $

 Thus, the direction ratios of one of the two lines are proportional to $-m, m , 0 \, or \, -1, 1, 0.$

Let $\theta $ be the angle between the given lines. Then,

$cos \theta = \frac{-1×(-1)+0×1+1×0}{\sqrt{(-1)^2+0^2+1^2}\sqrt{(-1)^2 +1^1 + 0}}=\frac{1}{2}⇒\theta = \frac{\pi}{3}$