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 \( -3 \).

Answer 2

Lines: \( \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} \) and \( \dfrac{x}{1}=\dfrac{y}{\alpha}=\dfrac{z-5}{1} \).
L\(_1\): point \(A(1,2,3)\), direction \( \vec d_1=\langle 2,3,4\rangle \).
L\(_2\): \( (x,y,z)=(t,\alpha t,5+t) \Rightarrow \) point \(B(0,0,5)\), direction \( \vec d_2=\langle 1,\alpha,1\rangle \).
Shortest distance \( D=\dfrac{5}{\sqrt6} \).

Use \( D=\dfrac{|\,\overrightarrow{AB}\!\cdot(\vec d_1\times\vec d_2)\,|}{\lVert \vec d_1\times\vec d_2\rVert} \), where \( \overrightarrow{AB}=\langle -1,-2,2\rangle \).
\[ \vec d_1\times \vec d_2= \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2&3&4\\ 1&\alpha&1 \end{vmatrix} =\langle 3-4\alpha,\,2,\,2\alpha-3\rangle . \] \[ \lVert \vec d_1\times \vec d_2\rVert=\sqrt{(3-4\alpha)^2+4+(2\alpha-3)^2} =\sqrt{20\alpha^2-36\alpha+22}. \] \[ \overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(-1)(3-4\alpha)+(-2)(2)+2(2\alpha-3)=8\alpha-13 . \] Hence \[ \frac{|\,8\alpha-13\,|}{\sqrt{20\alpha^2-36\alpha+22}}=\frac{5}{\sqrt6}. \] Squaring: \[ (8\alpha-13)^2=\frac{25}{6}\left(20\alpha^2-36\alpha+22\right) \;\Longrightarrow\; 29\alpha^2+87\alpha-116=0. \] \[ \alpha=\frac{-87\pm\sqrt{87^2+4\cdot29\cdot116}}{58} =\frac{-87\pm145}{58}\in\{1,\,-4\}. \] \[ \boxed{\text{Sum of all possible values of }\alpha=-3.} \]