Question

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0, l^{2}+m^{2}-n^{2}=0$ is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

$\frac{\pi}{3}$

Answer 2

Given, $l+m+n=0$
$\Rightarrow l=-m-n$
and $l^{2}+m^{2} n^{2}=0$
$\therefore(-m-n)^{2}+m^{2}-n^{2}=0$
$\Rightarrow 2 m^{2}+2 m n=0$
$\Rightarrow 2 m(m+n)=0$
$\Rightarrow m=0$ or $m+n=0$
If $m=0$, thenl $=-n$
$\therefore \frac{l_{1}}{-1}=\frac{m_{1}}{0}=\frac{n}{1}$
and if $m+n=0$
$\Rightarrow m=-n$, then $l =0$
$\therefore \frac{l_{2}}{0}=\frac{m_{1}}{-1}=\frac{n_{2}}{1}$
ie.,$\left(l_{1}, m_{1}, n_{1}\right)=(-1,0,1)$
and $\left(l_{2}, m_{2}, n_{2}\right)=(0,-1,1)$
$\therefore \cos \theta=\frac{0+0+1}{\sqrt{1+0+1} \sqrt{0+1+1}}=\frac{1}{2}$
$\Rightarrow \theta=\pi$