# Prove that the lines $x=py+q;z=ry+s$ and $x=p'y+q';z=r'y+s'$ are perpendicular if $pp'+rr'+1=0.$

Toolbox:
• If two lines are perpendicular, then $a_1a_2+b_1b_2+c_1c_2=0$ where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are the direction ratios of the two lines.
Given $x=py+q\; ;\;z=ry+s$ and
$x=p'y+q'\; ;\;z=r'y+s'$
The given equation are not in symmetrical.
Let us first put them in symmetrical form.
Equation of the first line are
$x=py+q\; ;\;z=ry+s$
This can be written as
$\large\frac{x-b}{a}$$=y,\large\frac{z-d}{c}$$=y$
or $\large\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}$-----(1)
Similarly $x=p'y+q'\; ;\;z=r'y+s'$ can be written as,
or $\large\frac{x-b'}{a}=\frac{y-0}{1}=\frac{z-d'}{c'}$
Let $\overrightarrow {n_1}$ and $\overrightarrow {n_2}$ be the parallel vector to line $L1$ and $L2$ respectively
$\overrightarrow {n_1}=a \hat i+\hat j+c \hat k$ and $\overrightarrow {n_2}=a'\hat i+\hat j+c'\hat k$
we know $\overrightarrow {n_1}.\overrightarrow {n_2}=0$ if $\overrightarrow {n_1}$ is $\perp$ to $\overrightarrow {n_2}.$
$(ie)(a \hat i+\hat j+c \hat k).(a'\hat i+\hat j+c'\hat k)=0$
Apply dot product rule we get
$aa' +1+cc'=0$
Hence proved.