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Q.
The lines whose direction cosines are given by the relations $a l+b m+c n=0$ and $m n+n l+l m=0$ are
TS EAMCET 2018
Solution:
Given relations
$a\, l+\,b \,m + c \,n=0\,.....(i)$
and $m \,n + n \,l + l m=0\,.....(ii)$
From relations Eqs. (i) and (ii),
$\left(\frac{a l+b m}{-c}\right)(l+m)+l m=0$
[by eliminating ' $n^{\prime}$ ]
$\Rightarrow a l^{2}+(a+b-c) l m+b m^{2}=0$
$\Rightarrow a\left(\frac{l}{m}\right)^{2}+(a+b-c)\left(\frac{l}{m}\right)+b=0 \ldots(iii)$
Let roots of Eq. (iii) is $\frac{l_{1}}{m_{1}}$ and $\frac{l_{2}}{m_{2}}$,
$\therefore \frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{b}{a} \ldots(iv)$
Similarly, $\frac{l_{1} l_{2}}{n_{1} n_{2}}=\frac{c}{a}\ldots(v)$
So, $\frac{l_{1} l_{2}}{\frac{1}{a}}=\frac{m_{1} m_{2}}{\frac{1}{b}}=\frac{n_{1} n_{2}}{\frac{l}{c}} [$ from Eqs. (iv) and (v) ]
If lines whose direction cosines given by the relation Eqs. (i) and (ii) are perpendicular if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$