# Find the values of p so the line

Question:

Find the values of $p$ so the line $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and

$\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Solution:

The given equations can be written in the standard form as

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

The direction ratios of the lines are $-3, \frac{2 p}{7}, 2$ and $\frac{-3 p}{7}, 1,-5$ 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 $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

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

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

$\Rightarrow 11 p=70$

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

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