Question

Let the point \((-1, \alpha, \beta)\) lie on the line of the shortest distance between the lines \[\frac{x + 2}{-3} = \frac{y - 2}{4} = \frac{z - 5}{2} \quad \text{and} \quad \frac{x + 2}{-1} = \frac{y + 6}{2} = \frac{z - 1}{0}.\] Then \((\alpha - \beta)^2\) is equal to ______.

Answer 1

Correct Answer: 25

Given two lines represented in their symmetric forms:

\( L_1 : \frac{x + 2}{-3} = \frac{y - 2}{4} = \frac{z - 5}{2}, \quad L_2 : \frac{x + 2}{-1} = \frac{y + 6}{2} = z - 1. \)

We are asked to find the coordinates \((\alpha, \beta)\) such that the point \((-1, \alpha, \beta)\) lies on the line of shortest distance between \(L_1\) and \(L_2\).

Step 1: Direction Ratios (DR) of the Lines

• For line \(L_1\), the direction ratios (DRs) are: \((-3, 4, 2)\).
• For line \(L_2\), the DRs are: \((-1, 2, 0)\).

Step 2: Point of Intersection

Let the points on the lines be given by: \( P(-3\lambda - 2, 4\lambda + 2, 2\lambda + 5), \quad Q(-\mu - 2, 2\mu - 6, 1). \)

The direction ratios of line \(PQ\) (joining \(P\) and \(Q\)) are: \[ (3\lambda - \mu, 2\mu - 4\lambda - 8, 2\lambda - 4). \]

Step 3: Condition for Perpendicularity

The line of shortest distance is perpendicular to both \(L_1\) and \(L_2\), so: \[ \text{DR of } PQ \cdot \text{DR of } L_1 = 0, \quad \text{DR of } PQ \cdot \text{DR of } L_2 = 0. \]

Solving these equations gives: \(\lambda = -1, \quad \mu = 1.\)

Step 4: Coordinates of the Point

Substituting the values of \(\lambda\) and \(\mu\) into the parametric equations: \[ \alpha = -3, \quad \beta = 2. \]

Step 5: Calculate \((\alpha - \beta)^2\)

\[ (\alpha - \beta)^2 = (-3 - 2)^2 = (-5)^2 = 25. \]

Therefore, the correct answer is 25.