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

Step 1: Understanding the Concept:
Two lines are perpendicular if the dot product of their direction vectors $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is zero: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Step 2: Formula Application:
First, convert lines to standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$: Line 1: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$. Direction ratios: $d_1 = (-3, \frac{2p}{7}, 2)$. Line 2: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. Direction ratios: $d_2 = (-\frac{3p}{7}, 1, -5)$.
Step 3: Explanation:
Setting the dot product to zero: $(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$ $\frac{9p}{7} + \frac{2p}{7} - 10 = 0$ $\frac{11p}{7} = 10 \implies p = \frac{70}{11}$.
Step 4: Final Answer:
The value of $p$ is $70/11$.