Question

The acute angle between the lines whose direction ratios are given by $l+m-n=0$ and $l^2+m^2-n^2=0$, is

• A. $0$
• B. $\pi /6$
• C. $\pi /4$
• D. $\pi /3$

Answer 1

The correct option is D $\pi /3$

We have
$l+m-n=0$
and $l^2+m^2+n^2=0$
$\Rightarrow l^2+m^2=n^2$ and $l+m=n$
$\Rightarrow l^2+m^2={(l+m)}^2$
$\Rightarrow 2lm=0$
$\Rightarrow l=0,m=0$
If $l=0$, then $m=n$
$\therefore \frac{l}{0}=\frac{m}{1}=\frac{n}{1}$
If $m=0$, then
$l=n$
$\therefore \frac{l}{1}=\frac{m}{0}=\frac{n}{1}$
thus, the dr's of the lines are proportional to $0,1,1$ and $1,0,1$
Therefore, angle $\theta$ between them is given by
$\cos \theta =\frac{0\times 1+1\times 0+|1\times 1|}{\sqrt{0+1+1}\sqrt{1+0+1}}=\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$