Question

Find the value of \( p \) such 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 mutually perpendicular.

Hint:

For mutually perpendicular lines, their direction ratios should satisfy \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \).

Answer 1

To find \( p \), we extract the direction ratios of the two given lines. For the first line: \[ \frac{x - 1}{-3} = \frac{y - 14/7}{2p} = \frac{z - 3}{2}. \] The direction ratios are: \[ (-3, 2p, 2). \] For the second line: \[ \frac{x - 7}{-3p} = \frac{y - 5}{1} = \frac{z - 6}{-5}. \] The direction ratios are: \[ (-3p, 1, -5). \] Since the lines are perpendicular, the dot product of their direction ratios must be zero: \[ (-3)(-3p) + (2p)(1) + (2)(-5) = 0. \] \[ 9p + 2p - 10 = 0. \] \[ 11p = 10 \Rightarrow p = \frac{10}{11}. \]