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

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.