Question

Find the values of $p$ so the line $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Answer 1

The given equations can be written in the standard form as
$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$
The direction ratios of the lines are $-3,\frac{-3p}{7},2$ and $\frac{-3p}{7},1,-5$ respectively.
Two lines with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$, are perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$.
$\therefore$ $(-3)\left(\frac{-3p}{7}\right)+\left(\frac{-3p}{7}\right).\left(1\right)+\left(2\right).(-5)=0$
$\Rightarrow$ $\frac{9p}{7}+\frac{2p}{7}=10$
$\Rightarrow$ $11p=70$

$\Rightarrow p=\frac{70}{11}$