Question

The angle between the straight lines whose direction cosines are given by $2l+2m-n=0,mn+nl+lm=0$, is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{2}$

Relations are $2l-2m-n=0\Rightarrow n=2l+2m$ and $mn+nl+lm=0$.
Eliminating $n$, we have $(m+l)(2l+2m)+lm=0$

$\Rightarrow 2l^2+5lm+2m^2=0$

$\Rightarrow (2l+m)(l+2m)=0$
When $2l+m=0$, we have from $2l+2m-n=0,n=m.$
Thus $\frac{l}{-1}=\frac{m}{2}=\frac{n}{2},$
$\Rightarrow$ d.c's of one line are proportional to $[-1,2,2]$.
Again, when $l+2m=0$, we have from $2l+2m-n=0,n=l$
$\therefore \frac{l}{2}=\frac{m}{-1}=\frac{n}{2},$
$\Rightarrow$ d.c's of the other line are proportional to $[2,-1,2]$.
Now, if $\theta$ be the angle between the two lines, then
$\cos \theta =\frac{-1.2+2.-1+2.2}{\sqrt{(1^2+2^2+2^2)}\sqrt{(2^2+1^2+2^2)}}=0,$
$\Rightarrow \theta =\frac{\pi }{2}$ or the two lines are at right angles to each other.