Question

Find the value 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 as

$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{-\frac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$

If the given lines are at right angles, then,

$-3(-\frac{3p}{7})+\frac{2p}{7}\left(1\right)+2(-5)=0$

$\frac{9p}{7}+\frac{2p}{7}-10=0$

$\frac{11p}{7}=10$

$11p=70$

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