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.

Answer 2

Step 1: Given lines.
L₁ : (x + 2)/−3 = (y − 2)/4 = (z − 5)/2
L₂ : (x + 2)/−1 = (y + 6)/2 = (z − 1)/0

Let us write them in parametric form.

For L₁:
x = −2 − 3t, y = 2 + 4t, z = 5 + 2t ...(1)
For L₂:
x = −2 − s, y = −6 + 2s, z = 1 ...(2)

Step 2: Direction vectors.
For L₁, direction vector d₁ = (−3, 4, 2).
For L₂, direction vector d₂ = (−1, 2, 0).

Step 3: Find the line of shortest distance between L₁ and L₂.
Let a₁ = (−2, 2, 5), a₂ = (−2, −6, 1).
The vector between points on the two lines is:
a₂ − a₁ = (0, −8, −4).

The direction of the line of shortest distance is along (d₁ × d₂).
Compute d₁ × d₂:
| i  j  k |
| −3 4 2 |
| −1 2 0 | = i(4×0 − 2×2) − j(−3×0 − 2×−1) + k(−3×2 − 4×−1)
= i(−4) − j(2) + k(−6 + 4) = (−4, −2, −2).
So direction of shortest distance line is n = (−4, −2, −2) or simplified as (2, 1, 1).

Step 4: Equation of the line of shortest distance.
Let the point on L₁ be A(−2 − 3t₁, 2 + 4t₁, 5 + 2t₁),
and point on L₂ be B(−2 − s, −6 + 2s, 1).
For the shortest distance, AB ⋅ (d₁ × d₂) = 0.

AB = B − A = (−2 − s + 2 + 3t₁, −6 + 2s − 2 − 4t₁, 1 − 5 − 2t₁) = (3t₁ − s, −8 + 2s − 4t₁, −4 − 2t₁).

Now, (d₁ × d₂) = (2, 1, 1). So,
(3t₁ − s)·2 + (−8 + 2s − 4t₁)·1 + (−4 − 2t₁)·1 = 0.
Simplify:
6t₁ − 2s − 8 + 2s − 4t₁ − 4 − 2t₁ = 0.
Combine like terms: (6t₁ − 4t₁ − 2t₁) + (−8 − 4) = 0 ⇒ 0 − 12 = 0 — contradiction?
Wait, check carefully:
6t₁ − 2s − 8 + 2s − 4t₁ − 4 − 2t₁ = (6t₁ − 4t₁ − 2t₁) + (−8 − 4) + (−2s + 2s) = 0 − 12 = −12 ≠ 0.
So our vector (2, 1, 1) must have wrong sign; let’s use (−2, −1, −1) instead.

Using (−2, −1, −1):
(3t₁ − s)(−2) + (−8 + 2s − 4t₁)(−1) + (−4 − 2t₁)(−1) = 0.
Simplify:
−6t₁ + 2s + 8 − 2s + 4t₁ + 4 + 2t₁ = 0 ⇒ (−6t₁ + 4t₁ + 2t₁) + (8 + 4) = 0 ⇒ 12 = 0.
That also gives no solution, meaning parallel directions of shortest distance line.

We take another geometric route.
Shortest distance line passes through points where it is perpendicular to both direction vectors.
So (a₁ + λd₁ − a₂ − μd₂) ⋅ d₁ = 0 and (a₁ + λd₁ − a₂ − μd₂) ⋅ d₂ = 0.

Compute step by step.
a₁ − a₂ = (0, 8, 4).
Now, (a₁ + λd₁ − a₂ − μd₂) = (0, 8, 4) + λ(−3, 4, 2) − μ(−1, 2, 0).
= (−3λ + μ, 8 + 4λ − 2μ, 4 + 2λ).

Dot with d₁ = (−3, 4, 2):
(−3λ + μ)(−3) + (8 + 4λ − 2μ)(4) + (4 + 2λ)(2) = 0.
Simplify: 9λ − 3μ + 32 + 16λ − 8μ + 8 + 4λ = 0 ⇒ 29λ − 11μ + 40 = 0 ...(i)

Dot with d₂ = (−1, 2, 0):
(−3λ + μ)(−1) + (8 + 4λ − 2μ)(2) + (4 + 2λ)(0) = 0.
Simplify: 3λ − μ + 16 + 8λ − 4μ = 0 ⇒ 11λ − 5μ + 16 = 0 ...(ii)

Solve (i) and (ii):
From (ii): 11λ = 5μ − 16 ⇒ λ = (5μ − 16)/11.
Substitute into (i):
29(5μ − 16)/11 − 11μ + 40 = 0 ⇒ (145μ − 464 − 121μ + 440)/11 = 0 ⇒ (24μ − 24)/11 = 0 ⇒ μ = 1.
Then λ = (5(1) − 16)/11 = −11/11 = −1.

Step 5: Find point on line of shortest distance.
Point on L₁ (λ = −1):
x₁ = −2 − 3(−1) = 1, y₁ = 2 + 4(−1) = −2, z₁ = 5 + 2(−1) = 3.
So point on L₁ is (1, −2, 3).
Point on L₂ (μ = 1):
x₂ = −2 − 1 = −3, y₂ = −6 + 2(1) = −4, z₂ = 1.

The line of shortest distance passes through (1, −2, 3) and (−3, −4, 1).

Its direction vector = (−3 − 1, −4 + 2, 1 − 3) = (−4, −2, −2) = 2(−2, −1, −1).

Equation of the line of shortest distance:
x = 1 − 2t, y = −2 − t, z = 3 − t.

Step 6: The point (−1, α, β) lies on this line.
Substitute x = −1 into x = 1 − 2t ⇒ −1 = 1 − 2t ⇒ 2t = 2 ⇒ t = 1.
Then y = −2 − 1 = −3 and z = 3 − 1 = 2.
Hence α = −3, β = 2.

Step 7: Compute (α − β)².
(α − β)² = (−3 − 2)² = (−5)² = 25.

Final Answer: 25