Question

Show that the lines whose d.c.s are given by $l+m+n=0,2mn+3ln-5lm=0$ are perpendicular to each other.

Answer 1

First relation

$l=(-m-n)..........................\left(1\right)$

putting the value of $l$ in the relation

$2mn+3(-m-n)n-5(-m-n)m=0$

$\begin{array}{c}5m^2+4mn-3n^2=0\\ 5{\left(\frac{m}{n}\right)}^2+4\left(\frac{m}{n}\right)-3=\mathrm{0..........}\left(2\right)\end{array}$

suppose $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are the direction cosines of the two line, then the roots of the equation (2) are $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$

nOW,

Product of the roots $=\frac{m_1}{n_1}.\frac{m_2}{n_2}=-\frac{3}{5}$

then,

$\frac{m_1{m}_2}{3}=\frac{n_1{n}_2}{-5}$ ----- $\left(3\right)$

Again from $\left(1\right),n=-1-m$ and putting this value of $n$ in $\left(2\right)$ equation, we get

$2m(-l-m)+3l(-l-m)-5lm=0$

and,

$3(\frac{l}{m}{)}^2+10(\frac{l}{m})+2=0$

$\frac{l_1}{m_1}.\frac{l_2}{m_2}=\frac{2}{3}$

then,

$\frac{l_1{l}_2}{2}=\frac{m_1{m}_2}{3}$

From $\left(3\right)$ and $\left(4\right)$

$\frac{l_1{l}_2}{2}=\frac{m_1{m}_2}{3}=\frac{n_1{n}_2}{-5}=k$ (consider.)

$l_1{l}_2+m_1{m}_2+n_1{n}_2=(2+3-5)k=0$

$k=0$

therefore lines are at right angle.