Question

Find the values of p so the line ${\displaystyle \frac{1-x}{3}=\frac{7y-14}{2}p=\frac{z-3}{2}}$ and ${\displaystyle \frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}}$ are at right angles.

Answer 1

The given equations can be written in the standard form as

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

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

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

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

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

⇒ ${\displaystyle \frac{11p}{7}=10}$

⇒ 11p = 70

⇒ p = ${\displaystyle \frac{70}{11}}$

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