Question

If shortest distance between the lines \[ \frac{x+1}{\alpha}=\frac{y-2}{-2}=\frac{z-4}{-2\alpha} \quad \text{and} \quad \frac{x}{\alpha}=\frac{y-1}{1}=\frac{z-1}{\alpha} \] is $\sqrt{2}$, then find the sum of all possible values of $\alpha$.

Hint:

For skew lines, always use vector cross product to compute shortest distance.

Answer 1

Correct Answer: -6

Step 1: Identify points and direction vectors

Line 1:

(x + 1)/α = (y − 2)/(-2) = (z − 4)/(-2α)

A point on Line 1: A(−1, 2, 4)
Direction vector of Line 1: d₁ = (α, −2, −2α)

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Line 2:

x/α = (y − 1)/1 = (z − 1)/α

A point on Line 2: B(0, 1, 1)
Direction vector of Line 2: d₂ = (α, 1, α)

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Step 2: Formula for shortest distance between two skew lines

Shortest distance,

D = |(B − A) · (d₁ × d₂)| / |d₁ × d₂|

Given: D = √2

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Step 3: Compute vector (B − A)

B − A = (1, −1, −3)

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Step 4: Compute cross product d₁ × d₂

| i    j    k |
| α   −2   −2α |
| α    1    α |

d₁ × d₂ = (−2α + 2α)i − (α² − (−2α²))j + (α + 2α)k

d₁ × d₂ = (0, −3α², 3α)

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Step 5: Compute numerator

(B − A) · (d₁ × d₂) = (1, −1, −3) · (0, −3α², 3α)

= 0 + 3α² − 9α

= 3α(α − 3)

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Step 6: Compute denominator

|d₁ × d₂| = √[0 + 9α⁴ + 9α²]

= 3|α|√(α² + 1)

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Step 7: Apply distance condition

|3α(α − 3)| / [3|α|√(α² + 1)] = √2

|α − 3| / √(α² + 1) = √2

Square both sides:

(α − 3)² = 2(α² + 1)

α² − 6α + 9 = 2α² + 2

α² + 6α − 7 = 0

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Step 8: Solve quadratic equation

α = [−6 ± √(36 + 28)] / 2

α = [−6 ± 8] / 2

α = 1, −7

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Final Answer:

Sum of all possible values of α = 1 + (−7) = −6