# Find the angle between the lines whose direction cosines are given by the equations $l+m+n=0,l^2+m^2-n^2=0$

Toolbox:
• Angle between two lines is $cos \theta= \large\frac{\overrightarrow {b_1}.\overrightarrow {b_2}}{|\overrightarrow {b_1}||\overrightarrow {b_2}|}$
Step 1:
Given that $l+m+n=0$ -----(1)=>$l+m=-n$
=>$-(l+m)=n$
and $l^2+m^2-n_2=0$-----(2)
Let us substitute for $'n'$ in equation (2) we get
=>$l^2+m^2-l^2-m^2-2ml=0$
or $2ml=0$
(ie) either $l = 0 \: or\: m=0$
Let us put $m=0$ in equation (1)
If $m=0$ then $l=-n$
direction ratios $(l,m,n)=(1, 0, -1)$
Let us put $l=0$ we get $m=-n$
direction ratios $(l,m,n)=(0, 1, -1)$
Step 2:
Let us find out $b_1.b_2$
$b_1.b_2=(1,0,-1).(0,1,-1)$
$=0+0+1$
$1$
$|b_1|=\sqrt{0^2+1^2+(-1)^2}=\sqrt2$
$|b_2|=\sqrt{0^2+1^2+(-1)^2}=\sqrt2$
Step 3:
Now substituting the above values in
$cos \theta= \large\frac{\overrightarrow {b_1}.\overrightarrow {b_2}}{|\overrightarrow {b_1}||\overrightarrow {b_2}|}$
$\cos \theta= \large\frac{1}{\sqrt 2\sqrt 2}=\frac{1}{2}$
=>$\theta=\large\frac{\pi}{3}$