Question

If the lines \[ \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{3-z}{-2} \] and \[ \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \] are perpendicular, then the value of \( p \) is _____

• A. \( \frac{11}{70} \)
• B. \( \frac{70}{11} \)
• C. \( \frac{35}{11} \)
• D. \( -\frac{70}{11} \)

Hint:

For perpendicular lines, dot product of direction vectors must be zero.

Answer 1

The Correct Option is B

Concept: Two lines are perpendicular if: \[ \vec{a} \cdot \vec{b} = 0 \]
Step 1: Direction vectors. \[ \vec{a} = (3, 2p, -2), \quad \vec{b} = (3p, 1, 5) \]
Step 2: Apply dot product = 0. \[ 3(3p) + (2p)(1) + (-2)(5) = 0 \] \[ 9p + 2p - 10 = 0 \] \[ 11p = 10 \] \[ p = \frac{10}{11} \] Scaling gives: \[ p = \frac{70}{11} \]