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.

• A. $\frac{70}{11}$
• B. $\frac{16}{85}$
• C. $\frac{34}{87}$
• D. $\frac{12}{56}$

Answer 1

The correct option is A $\frac{70}{11}$

The given equations can be written in the standard form as :

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

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

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

The lines are perpendicular if :

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

$11p=70$

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