Question

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

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

Answer 1

The correct option is C $\frac{\pi }{3}$

We also have $l^2+m^2+n^2=1$

So that $l^2+m^2-n^2=0\Rightarrow 2n^2=1\Rightarrow n=\pm \frac{1}{\sqrt{2}}$

and $l+m+n=0\Rightarrow {(l+m)}^2=n^2=\frac{1}{2}=l^2+m^2$

$\Rightarrow 2lm=0$

Either $l=0$ or $m=0$, if $l=0,m+n=0$

$\Rightarrow m=-n=\pm \frac{1}{\sqrt{2}}$

So direction ratios of one of the lines are

$0,\pm \frac{1}{\sqrt{2}},\mp \frac{1}{\sqrt{2}}$

and if $m=0,l+n=0\Rightarrow l=-n=\pm \frac{1}{\sqrt{2}}$

So the direction ratios of the other line are $\pm \frac{1}{\sqrt{2}},0,\mp \frac{1}{\sqrt{2}}$

Thus the required angle is

${\cos }^{-1}[0\times (\pm \frac{1}{\sqrt{2}})+(\pm \frac{1}{\sqrt{2}})\left(0\right)+(\pm \frac{1}{\sqrt{2}})(\pm \frac{1}{\sqrt{2}})]$

$={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$