Question

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is :

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

Answer 1

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

As, $l+m+n=0$
$\Rightarrow (m+n{)}^2=l^2$
$\Rightarrow m^2+n^2+2mn=(m^2+n^2)(\because l^2=m^2+n^2)$
$\Rightarrow mn=0$
Now, $l^2+m^2+n^2=1$
$\Rightarrow l^2+l^2=1(\because l^2=m^2+n^2)$
$\Rightarrow l=\pm \frac{1}{\sqrt{2}}$
case-1:
if $l=-(m+n)=\frac{1}{\sqrt{2}}$
$\Rightarrow m+n=-\frac{1}{\sqrt{2}}$
$\therefore m-n=\frac{1}{\sqrt{2}}(\because mn=0)$
Thus, $m=0,n=-\frac{1}{\sqrt{2}}$
Direction cosines are $(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})...\left(1\right)$
case-2:
if $l=-(m+n)=-\frac{1}{\sqrt{2}}$
$\Rightarrow m+n=\frac{1}{\sqrt{2}}$
$\therefore m-n=\frac{1}{\sqrt{2}}(\because mn=0)$
Thus, $m=\frac{1}{\sqrt{2}},n=0$
Direction cosines are $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)...\left(2\right)$
From equation (1) and (2)
$\Rightarrow |-\frac{1}{2}+0+0|=\sqrt{\frac{1}{2}+0+\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}+0}\cdot \cos \theta$
$\Rightarrow \cos \theta =\frac{1}{2}$
Since, angle is acute so,
$\theta =\frac{\pi }{3}$