https://file.aiquickdraw.com/mathbot/file/17235635200335b3ecutd.jpg

Let's solve part 1(c) of the problem.

### Given:
- Two lines in \(\mathbb{R}^3\):
  \[
  \mathbf{r}_1(t) = (4, 1, 1) + t(3, 0, -1)
  \]
  \[
  \mathbf{r}_2(s) = (-1, 2, -5) + s(-3, 2, 2)
  \]

### Steps to Solve:

1. **Parametric Equations:**
   - \(\mathbf{r}_1(t)\) gives: 
     \[
     \mathbf{r}_1(t) = (4 + 3t, 1, 1 - t)
     \]
   - \(\mathbf{r}_2(s)\) gives: 
     \[
     \mathbf{r}_2(s) = (-1 - 3s, 2 + 2s, -5 + 2s)
     \]

2. **Perpendicular Vector (Found in part a):**
   - We calculate \(\mathbf{d} = \mathbf{r}_1(t) - \mathbf{r}_2(s)\) and find the values of \(t\) and \(s\) where \(\mathbf{d}\) is perpendicular to the direction vectors of the lines.

   The direction vectors are:
   - For \(\mathbf{r}_1(t)\): \(\mathbf{v}_1 = (3, 0, -1)\)
   - For \(\mathbf{r}_2(s)\): \(\mathbf{v}_2 = (-3, 2, 2)\)

   The vector perpendicular to both lines is found using the cross product:
   \[
   \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ -3 & 2 & 2 \end{vmatrix} = (2, 9, 6)
   \]

3. **Setting the Dot Product to Zero:**
   - The distance vector between any two points on the lines is given by:
     \[
     \mathbf{d} = (4 + 3t) - (-1 - 3s), (1) - (2 + 2s), (1 - t) - (-5 + 2s)
     \]
     Simplifying,
     \[
     \mathbf{d} = (5 + 3t + 3s, -1 - 2s, 6 - t - 2s)
     \]
   - We set the dot product \(\mathbf{d} \cdot \mathbf{n} = 0\) for both \(\mathbf{v}_1\) and \(\mathbf{v}_2\):
     \[
     \mathbf{d} \cdot \mathbf{v}_1 = (5 + 3t + 3s) \cdot 3 + (-1 - 2s) \cdot 0 + (6 - t - 2s) \cdot (-1) = 0
     \]
     Simplifying this,
     \[
     15 + 9t + 9s - 6 + t + 2s = 0
     \]
     \[
     10t + 11s + 9 = 0 \quad \text{(Equation 1)}
     \]
     For \(\mathbf{v}_2\),
     \[
     \mathbf{d} \cdot \mathbf{v}_2 = (5 + 3t + 3s) \cdot (-3) + (-1 - 2s) \cdot 2 + (6 - t - 2s) \cdot 2 = 0
     \]
     Simplifying this,
     \[
     -15 - 9t - 9s - 2 - 4s + 12 - 2t - 4s = 0
     \]
     \[
     -11t - 17s - 5 = 0 \quad \text{(Equation 2)}
     \]

4. **Solve the System of Equations:**
   - From Equation 1:
     \[
     10t + 11s = -9
     \]
   - From Equation 2:
     \[
     -11t - 17s = 5
     \]
   - Solve this system to find \(t\) and \(s\).

Let's solve these equations to find \(t\) and \(s\).The solution to the system of equations is:
\[
t = -2, \quad s = 1
\]

### Conclusion:
The values of \(t\) and \(s\) at which the minimum distance between the two lines is achieved are \(t = -2\) and \(s = 1\).

Would you like further details on this solution or have any additional questions?

### Relative Questions:
1. How do you find a vector perpendicular to two given vectors?
2. Why is the dot product used to find when two vectors are perpendicular?
3. What is the significance of the cross product in vector mathematics?
4. How do you derive parametric equations from a vector equation?
5. What methods can be used to solve a system of linear equations?
6. How does the minimum distance between two skew lines relate to their direction vectors?
7. Can you always find a unique solution for \(t\) and \(s\) in such problems?
8. How would the problem change if the lines were parallel?

### Tip:
Always check the direction vectors of the lines to determine if they are parallel, as it simplifies the problem significantly.

Learn how to find the minimum distance between two skew lines in 3D space with a step-by-step solution. This problem involves parametric equations, dot product calculations, cross product applications, and solving a system of equations. Suitable for advanced high school students and above.

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