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Q.
The straight lines whose dc’s are given by the $al + bm + cn = 0 $ and $fmn + gnl+ hlm = 0$ are perpendicular if
Three Dimensional Geometry
Solution:
From the given equation, we have
$ag \left(\frac{l}{m}\right)^{2}+\left(af +bg -ch\right)\left(\frac{l}{m}\right)+bf =0 $
or $agX^{2}+\left(af +bg -ch\right)X+bf =0 \left(X=l/ m\right)$
Let two value of $ l /m$ are $\frac{l_{1}}{m_{1}}$ and $\frac{l_{2}}{m_{2}}$
$\therefore $ Product of roots $=\frac{l_{1}l_{2}}{m_{1}m_{2}}=\frac{bf}{ag} $
$\therefore \frac{l_{1}l_{2}}{\frac{f}{a}}=\frac{m_{1}m_{2}}{\frac{g}{b}}=\frac{n_{1}n_{2}}{\frac{h}{c}}$, using symmetry
Now lines to be $ \bot$ if $l_{1} l_{2} +m_{1}m_{2}+n_{1}n_{2}=0$
$\Rightarrow k\left(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\right)=0$
$\Rightarrow \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$