Question

Find the angles between the lines, whose direction cosines are give by the equation $l^2-m^2+n^2=0,l+m+n=0$

• A. $0$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

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

Lines are $l+m+n=0\Rightarrow -l=(m+n)$

and $l^2-m^2+n^2=0\Rightarrow l^2=m^2-n^2$

Solving them gives,

${(-(m+n))}^2=m^2+n^2+2mn=m^2-n^2\Rightarrow 2n(n+m)=0$

Now, For $n=0$

$m=-l$, so d.c's $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0)$

And for $n=-m$

$l=0$ so d.c's $(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$

Angle between the lines $\left|\cos \theta \right|=\left|\frac{1}{2}\right|\Rightarrow \theta =\frac{\pi }{3}$