Question

Let \( L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \) and \( L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \), where \( \alpha \in \mathbb{R} \), be two lines which intersect at the point \( B \). If \( P \) is the foot of the perpendicular from the point \( A(1, 1, -1) \) on \( L_2 \), then the value of \( 26 \alpha (PB)^2 \) is:

Hint:

For perpendicular distance problems, use the parametric equations of the lines and the formula for the distance from a point to a line to find the required distances.

Answer 1

Correct Answer: 216

First, find the point \( B \) where the lines \( L_1 \) and \( L_2 \) intersect. From the system of equations, we get: \[ 3\lambda + 1 = 2\mu + 2, \quad -\lambda + 1 = 2\mu + 2, \quad -1 = \alpha \mu - 4 \] Solving this system gives \( \lambda = 1, \mu = 1, \alpha = 3 \), so the point \( B \) is \( (4, 0, -1) \). Next, calculate the perpendicular distance from \( A(1, 1, -1) \) to \( L_2 \): \[ P(2\lambda + 2, 0, 3\mu - 3) \] The distance \( PB \) is calculated using the formula for the perpendicular distance: \[ 26 \alpha (PB)^2 = 26 \times 3 \times \left( \frac{144}{169} + \frac{324}{169} \right) = 216 \]