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

Answer: (C)

$\pi/3$

Answer 2

$l+m+n=0$ $l^{2}=m^{2}+n^{2}$ 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 $m=\frac{1}{\sqrt{2}}$ $l=-\frac{1}{\sqrt{2}}$ $n=0$ $\left(l_{2}, m_{2}, n_{2}\right)$ $=\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$