Question

Find the angle between the lines whose directions cosines are given by the equation l+m+n=0 and $l^2+m^2-n^2=0.$

Answer 1

Eliminating n from both the equations, we have
$l^2+m^2-(l-m{)}^2=0$
$\Rightarrow l^2+m^2-l^2-m^2+2ml=0\Rightarrow 2lm=0\Rightarrow lm=0\Rightarrow (-m-n)m=0[\because l=-m-n]\Rightarrow (m+n)m=0\Rightarrow m=-n\Rightarrow m=0\Rightarrow l=0,l=-n$
Thus Dr's two lines are proportional to 0, -n,n and -n,0,n i.e., 0,-1,1 and -1,0,1.
So, the vector parallel to these given lines are $\overrightarrow{a}=-\hat{j}+\hat{k}and\overrightarrow{b}=-\hat{i}+\hat{k}$
Now, $cos\theta =\frac{\overrightarrow{a}\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}=\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\Rightarrow cos\theta =\frac{1}{2}\therefore \theta =\frac{\pi }{3}[\because cos\frac{\pi }{3}=\frac{1}{2}]$