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 \(\frac{70}{11}\).