Question

The angle between the straight lines whose direction cosine are given by $l+m+n=0$ and $fmn+gln+hlm=0$ is

Answer 1

The given relations are

$al+bm+cn=0$ and $fmn+gnl+hlm=0.$

Eliminating $n$ between them, we get

$(fm+gl)(-\frac{al+bm}{c})+hlm=0$

$\Rightarrow {agl}^2+(af+bg-ch)lm+bf{m}^2=0.$

$\Rightarrow ag{\left(\frac{1}{m}\right)}^2+(af+bg-ch)\frac{l}{m}+bf=0$ ...(1)

Let $\frac{l_1}{m_1},\frac{l_2}{m_2}$ be the roots of the above quadratic; then

$\frac{l_1}{m_1}.\frac{l_2}{m_2}=$ product of the roots $=\frac{bf}{ag}=\frac{f/a}{g/b},$

i.e., $\frac{l_1{l}_2}{f/a}=\frac{m_1{m}_2}{g/b}=\frac{n_1{n}_2}{h/c}$ by symmetry.

The lines will be perpendicular if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0,$

i.e., if $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0.$