Question

The direction cosines of two lines satisfy $2l+2m-n=0$ and $lm+mn+nl=0.$ The angle between these lines is

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

Answer 1

Answer: (B)

Substituting $n=2\left(l+m\right)$ in $lm+mn+nl=0,$ we get
$lm+2{\left(l+m\right)}^2=0\Rightarrow 2l^2+5lm+2m^2=0$
$\Rightarrow \left(2l+m\right)\left(l+2m\right)=0$
Hence $m=-2l\Rightarrow n=-2l$ or $m=-\frac{l}{2}\Rightarrow n=l$
So D.C’s of the lines are $(l,-2l,-2l)$ or $\left(l,-\frac{l}{2},l\right)$
Hence, the angle between them is $90^{\circ }$