Question

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

Answer 1

$l+m+n=0$ $\Rightarrow l=-(m+n)$ -$\left(1\right)$

$l^2+m^2+n^2=0$ $\Rightarrow l^2=n^2-m^2$ -$\left(2\right)$

squaring $\left(1\right)$

$l^2=m^2+n^2+2nm=n^2-m^2$(from $\left(2\right)$)

$\Rightarrow 2m^2+2nm=0$

$\Rightarrow 2m(m+n)=0$

$\Rightarrow m=0$ or $m=-n$.

If $m=0$

$l=-n$

$l_1,m_1,n_1=(-n,0,n)$

If $m=-n$

$l=-(-n+n)$

$l=0$

$l_2,m_2,n_2=(0,-n,n)$

$\therefore$ Angle is $\frac{l_1{l}_2+m_1{m}_2+n_1{n}_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}=\cos \left(\theta \right)$

$\cos \theta =\frac{(-n)\left(0\right)+0(-nn(n)}{\sqrt{2n^2}\sqrt{2n^2}}$ $\Rightarrow \frac{n^2}{2n^2}$

$\cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$.

$\therefore Anglebetweenthelinesis\frac{\pi }{3}$.