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 equation can be written in the standard form as
$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}=\frac{6-z}{-5}$

The direction ratios of the lines are -3 ,$\frac{2p}{7}$, 2, and $\frac{-3p}{7}$, 1, -5 respectively.

Two lines with direction ratios, a1, b1, c1, and a2, b2, c2, are perpendicular to each other, if a1a2+b1b2+c1c2=0

∴(-3)$\bigg(\frac{-3p}{7}\bigg)+\bigg(\frac{2p}{7}\bigg)$(1)+2.(-5)=0

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

$\Rightarrow$ 11p=70

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

Thus, the value of P is $\frac{70}{11}$.