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. $\pi/6$
• B. $\pi/2$
• C. $\pi/3$
• D. $\pi/4$

Answer 1

The Correct Option is C

$l+m+n=0 $ $l^{2}=m^{2}+n^{2} $ Now, $(-m-n)^{2}=m^{2}+n$ $\Rightarrow m n=0 $ $m=0 $ or $ n=0 $ If $m=0 $ then $ l=-n $ $l^{2}+m^{2}+n^{2}=1$ Gives $\Rightarrow n=\pm \frac{1}{\sqrt{2}}$ i.e. $\left(l_{1}, m_{1}, n_{1}\right) $ $=\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$ $\therefore \cos \theta=\frac{1}{2}$ $\theta=\frac{\pi}{3}$ If $ n=0 $ then $ l=-m $ $l^{2}+m^{2}+n^{2}=1 $ $\Rightarrow 2 m^{2}=1 $ $\Rightarrow m^{2}=\frac{1}{2} $ $\Rightarrow m=\pm \frac{1}{\sqrt{2}} $ Let $ m=\frac{1}{\sqrt{2}} $ $l=-\frac{1}{\sqrt{2}} $ $n=0 $ $\left(l_{2}, m_{2}, n_{2}\right) $ $=\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$