The angle between two lines whose direction cosines are given by the equation $I+m+n=0, l^2+m^2+n^2=0$ is
Answer & explanation
Correct answer: option 2
EIiminating n between the two reIations, we have
l2 + m2 – (I + m)2 = 0 or 2Im = 0 ⇒ either l = 0 or m = 0
if l = 0, then m + n = 0 i,e. m = – n
$\Rightarrow \frac{l}{0}=\frac{m}{1}=\frac{n}{-1}$, giving the direction ratios of one line.
If m = 0, then I + n = 0 i.e. I = – n
$\Rightarrow \frac{l}{1}=\frac{m}{0}=\frac{n}{-1}$, giving direction ratios of the other lines.
The angIes between these lines is
$\cos ^{-1}\left\{ \pm \frac{0.1+1.0+(-1)(-1)}{\sqrt{0^2+1^2+(-1)^2} \sqrt{1^2+0^2+(-1)^2}}\right\}=\cos ^{-1}\left( \pm \frac{1}{2}\right)=\frac{\pi}{3}$ or $\frac{2 \pi}{3}$