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

Answer: (A)

$\frac{\pi }{3}$

Given $l^2=m^2+n^2$ and $l+m+n=0$

We know that $l^2+m^2+n^2=1$

So we get $l^2+l^2=1$

$\Rightarrow l^2=\frac{1}{2}$

$\Rightarrow l=\pm \frac{1}{\sqrt{2}}$

Let us consider $l=\frac{1}{\sqrt{2}}$

Since $l+m+n=0$ , we get $m+n=-l$

By substituting $l=-(m+n)$ in $l^2=m^2+n^2$ , we get $mn=0$

Put $n=0$ , then we have $m=-l=-\frac{1}{\sqrt{2}}$

put $m=0$ , then we have $n=-l=-\frac{1}{\sqrt{2}}$

So two possible directional cosines are $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0)$ and $(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})$

Therefore angle between them is ${\cos }^{-1}\left(\frac{\frac{1}{2}+0+0}{1}\right)={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$