Question

Find the angle between the two straight lines whose direction cosines $l,m,n$ are given by $2l+2m-n=0$ and $mn+nl+lm=0$

Answer 1

$2l+2m-n=0\Rightarrow n=2l+2m-\left(i\right)mn+nl+lm=0\therefore m(2l+2m)+(2l+2m)l+lm=02lm+2m^2+2l^2+2ml+lm=02m^2+2l^2+5ml=0$

$2{\left(\frac{m}{l}\right)}^2+5\left(\frac{m}{l}\right)+2=0$ (Quadratic in $\frac{m}{l}$)

$\therefore \frac{m}{l}=\frac{-1}{2},-2\therefore \frac{m_1}{-1}=\frac{l_1}{2}\&\frac{m_2}{-2}=\frac{l_2}{1}$

If $\frac{m_1}{-1}=\frac{l_1}{2}=k$ then $m_1=-k,l_1=2k$ from $\left(i\right)$

$\therefore \frac{l_1}{2}=\frac{m_1}{-1}=\frac{n_1}{2}$

Similarly for $(m_2=-2k,l_2=k,n_2=-2)$

$\therefore \frac{l_2}{1}=\frac{m_2}{-2}=\frac{n_2}{-2}$

$\therefore l_1{l}_2+m_1{m}_2+n_1{n}_2=0$

$\therefore$ angle between two lines ${90}^{\circ }$.