Question

The value of $p$, so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ intersect at right angle, is

• A. $\frac{10}{11}$
• B. $\frac{70}{10}$
• C. $\frac{10}{7}$
• D. $\frac{70}{9}$

Answer 1

Answer: (B)

Equation of the given lines can be written in the standard form as
$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{-\frac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$
$\because$ Lines are perpendicular to each other
$\therefore a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$.
$\Rightarrow \left(-3\right)\left(\frac{-3p}{7}\right)+\left(\frac{2p}{7}\right)\left(1\right)+\left(2\right)\left(-5\right)=0$
$\Rightarrow p=\frac{70}{11}$