Question

If the shortest distance between the lines $ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} $ and $ \frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1} $ is $ \frac{5}{\sqrt{6}} $, then the sum of all possible values of $ \alpha $ is:

• A. \( \frac{3}{2} \)
• B. \( \frac{-3}{2} \)
• C. \( 3 \)
• D. \( -3 \)

Hint:

For skew lines, the shortest distance formula involves finding the cross product of their direction ratios and the vector connecting points on the lines.

Answer 1

The Correct Option is D

We are given two lines in the space. 
Let the equations of the lines be in parametric form: 
1. \( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) Parametric equations: \[ x = 1 + 2t, \quad y = 2 + 3t, \quad z = 3 + 4t \] 
2. \( L_2: \frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1} \) Parametric equations: \[ x = s, \quad y = \alpha s, \quad z = 5 + s \] Now, we use the formula for the shortest distance \( D \) between two skew lines: \[ D = \frac{|(\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|} \] 
Where: 
- \( \vec{a}_1 = \langle 2, 3, 4 \rangle \) and \( \vec{a}_2 = \langle 1, \alpha, 1 \rangle \) are direction ratios of the lines. 
- \( \vec{b}_1 = \langle 1, 2, 3 \rangle \) and \( \vec{b}_2 = \langle 0, 0, 5 \rangle \) are points on the lines. 
The shortest distance is given by \( D = \frac{5}{\sqrt{6}} \), so we set the formula equal to this value and solve for \( \alpha \). 
After solving, we find that the possible value of \( \alpha \) is \( -3 \).