MHT CET · Maths · Three Dimensional Geometry
If 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, then \(p=\)
- A \(\frac{70}{11}\)
- B \(\frac{11}{70}\)
- C \(\frac{-70}{11}\)
- D \(\frac{-11}{70}\)
Answer & Solution
Correct Answer
(A) \(\frac{70}{11}\)
Step-by-step Solution
Detailed explanation
Given lines can be written as
\(
\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{3 p}=\frac{y-5}{1}=\frac{z-6}{-5}
\)
As these lines are at right angles, we get
\(
\begin{aligned}
& \quad-(-3)\left(-\frac{3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \quad \frac{9 p}{7}+\frac{2 p}{7}-10=0 \\
& \therefore \quad \frac{11 p}{7}=10 \Rightarrow p=\frac{70}{11}
\end{aligned}
\)
\(
\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{3 p}=\frac{y-5}{1}=\frac{z-6}{-5}
\)
As these lines are at right angles, we get
\(
\begin{aligned}
& \quad-(-3)\left(-\frac{3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \quad \frac{9 p}{7}+\frac{2 p}{7}-10=0 \\
& \therefore \quad \frac{11 p}{7}=10 \Rightarrow p=\frac{70}{11}
\end{aligned}
\)
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