Two lines whose direction cosines are given by $a l+b m+c n=0$ and $f m n+g n l+h l m=0$ are perpendicular…
Two lines whose direction cosines are given by $a l+b m+c n=0$ and $f m n+g n l+h l m=0$ are perpendicular to each other if .........
$\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$
$\frac{f}{a}-\frac{g}{b}-\frac{h}{c}=0$
$\frac{f}{a}+\frac{g}{b}-\frac{h}{c}=0$
$\frac{f}{a}-\frac{g}{b}+\frac{h}{c}=0$
Solution
Let the direction cosines of lines are $l_1, m_1, n_1$ and $l_2, m_2, n_2$
Since, $a l+b m+c n=0$ and, $f m n+g n l+h m=0$
So, on eliminating ' $l$ ', we get
$
\begin{gathered}
f m n+(g n+h m)\left(-\frac{b m+c n}{a}\right)=0 \\
\Rightarrow \quad a f m n=b h m^2+c g n^2+(b g+c h) m n=0 \\
\Rightarrow \quad b h\left(\frac{m}{n}\right)^2+(b g+c h-a f)\left(\frac{m}{n}\right)+c g=0 \text { having } \\
\text { roots } \frac{m_1}{n_1} \text { and } \frac{m_2}{n_2} \text {, so } \\
\text { product of roots }=\frac{m_1 m_2}{n_1 n_2}=\frac{c g}{b h}
\end{gathered}
$
Similarly on eliminating ' $m$ ', we get
$
\begin{gathered}
g n l+(h l+f n)\left(-\frac{a l+c n}{b}\right)=0 \\
\Rightarrow \quad b g n l=a h l^2+c f n^2+(c h+a f) l n=0 \\
\Rightarrow a h\left(\frac{l}{n}\right)^2+(c h+a f-b g)\left(\frac{l}{n}\right)+c f=0 \text {, having roots } \\
\frac{l_1}{n_1} \text { and } \frac{l_2}{n_2} \text {, so product of roots } \frac{l_1 l_2}{n_1 n_2}=\frac{c f}{a h} .
\end{gathered}
$
$\because$ The lines are perpendicular.
$
\begin{array}{ll}
\therefore & l_1 l_2+m_1 m_2+n_1 n_2=0 \\
\Rightarrow & \left(\frac{c g}{b h}+\frac{c f}{a h}+1\right) n_1 n_2=0 \Rightarrow \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0
\end{array}
$