Question

find the angle between the lines whose direction cosines satisfy the equation $\ell +m+n=0,{\ell }^2+m^2-n^2=0$

Answer 1

Consider the problem

Given

$l+m+n=0$ ---- (i)

$n=-(l+m)$

And

${\ell }^2+m^2-n^2=0$ ---- (ii)

Put value of $n$ in equation (ii)

$l^2+m^2-l^2-m^2-2ml=0$

$2ml=0$

Either $m=0$ or $l=0$

Now, first

put $m=0$ in (i)

then

$l=-n$

therefore

direction ratios $(l,m,n)=(0,1,-1)$

Now, find $b_1,b_2$

$b_1.b_2=(1,0,-1).(0,1,-1)=0+0+1=1$

$\begin{array}{c}|b_1|=\sqrt{0^2+1^2+{(-1)}^2}=\sqrt{2}\\ |b_2|=\sqrt{0^2+1^2+{(-1)}^2}=\sqrt{2}\end{array}$

Then

$\begin{array}{c}cos\theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\\ \cos \theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\\ \theta =\frac{\pi }{3}\end{array}$