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Step-by-Step Solution
Step 1: Express Each Line in Parametric Form
First line:
Given
$ \frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3} $,
let the common parameter be $t$. Then:
$x = 4 + t,\quad y = -1 + 2t,\quad z = 0 - 3t.$
Second line:
Given
$ \frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5} $,
let the common parameter be $s$. Then:
$x = \lambda + 2s,\quad y = -1 + 4s,\quad z = 2 - 5s.$
Step 2: Identify Points and Direction Vectors on Each Line
From the first line (when $t=0$), a point on the line is:
$ (4, -1, 0). $
The direction vector is $ \vec{d_1} = (1, 2, -3). $
From the second line (when $s=0$), a point on the line is:
$ (\lambda, -1, 2). $
The direction vector is $ \vec{d_2} = (2, 4, -5). $
Step 3: Compute the Cross Product of the Direction Vectors
Compute
$ \vec{d_1} \times \vec{d_2}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{vmatrix}. $
Expanding this determinant:
$ \vec{d_1} \times \vec{d_2}
= \mathbf{i}(2 \cdot -5 - (-3 \cdot 4))
\;-\; \mathbf{j}(1 \cdot -5 - (-3 \cdot 2))
\;+\; \mathbf{k}(1 \cdot 4 - 2 \cdot 2). $
Simplify each component:
$ = \mathbf{i}(-10 + 12)
- \mathbf{j}(-5 + 6)
+ \mathbf{k}(4 - 4)
= 2\mathbf{i} - \mathbf{j} + 0\mathbf{k}
= (2, -1, 0). $
The magnitude is:
$ |\vec{d_1} \times \vec{d_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}. $
Step 4: Use the Shortest Distance Formula Between Two Lines
For two skew lines with points
$ \vec{a_1} $ on the first line and
$ \vec{a_2} $ on the second line, and direction vectors
$ \vec{d_1} $ and
$ \vec{d_2},$
the shortest distance $D$ between them is:
$ D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}. $
From Step 2, let
$ \vec{a_1} = (4, -1, 0)
\quad \text{and} \quad
\vec{a_2} = (\lambda, -1, 2). $
Then
$ \vec{a_2} - \vec{a_1} = (\lambda - 4, 0, 2). $
Dot product with
$ \vec{d_1}\times\vec{d_2} = (2, -1, 0) $:
$ (\lambda - 4, 0, 2)\cdot (2, -1, 0) = 2(\lambda - 4) + 0 \cdot (-1) + 2 \cdot 0 = 2\lambda - 8. $
Hence the distance is:
$ D = \frac{|2\lambda - 8|}{\sqrt{5}}. $
Step 5: Equate to the Given Shortest Distance to Find $\lambda$
The shortest distance is given as
$ \frac{6}{\sqrt{5}}. $
Therefore:
$ \frac{|2\lambda - 8|}{\sqrt{5}} = \frac{6}{\sqrt{5}}
\;\; \Longrightarrow \;\;
|2\lambda - 8| = 6. $
This leads to two linear equations:
1) $2\lambda - 8 = 6 \quad \Rightarrow \quad 2\lambda = 14 \quad \Rightarrow \quad \lambda = 7.$
2) $2\lambda - 8 = -6 \quad \Rightarrow \quad 2\lambda = 2 \quad \Rightarrow \quad \lambda = 1.$
Step 6: Sum of All Possible Values of $\lambda$
The possible values are $\lambda = 7$ and $\lambda = 1.$
Their sum is:
$7 + 1 = 8.$
Final Answer
The sum of all possible values of $\lambda$ is
$8.$
