Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The value(s) of \(p\) so that the lines \(\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\) and \(\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles

Options:

\(\frac{7}{11}\)

\(\frac{70}{11}\)

\(\frac{11}{70}\)

\(\frac{11}{7}\)

Correct Answer:

\(\frac{70}{11}\)

Explanation:

For right angles \(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0\)