Consider the two lines in ℝ3: 𝒓𝟏(𝑡)=(4,1,1)+𝑡(3,0,−1) and 𝒓𝟐(𝑠)=(−1,2,−5)+𝑠(−3,2,2).
a. Find a vector perpendicular (i.e., normal) to both 𝒓𝟏(𝑡) and 𝒓𝟐(𝑠).
b. Use a scalar projection to find the minimum distance between the two lines.
c. Find the values of t and s at which this minimum distance is achieved.

### Given Lines in ℝ³:
1. \( \mathbf{r}_1(t) = (4, 1, 1) + t(3, 0, -1) \)
2. \( \mathbf{r}_2(s) = (-1, 2, -5) + s(-3, 2, 2) \)

### Step-by-Step Solution:

#### Part (a): Find a vector perpendicular (normal) to both \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(s) \).

- The direction vectors of the lines are:
  - \( \mathbf{d}_1 = (3, 0, -1) \) for \( \mathbf{r}_1(t) \)
  - \( \mathbf{d}_2 = (-3, 2, 2) \) for \( \mathbf{r}_2(s) \)

- A vector normal to both \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(s) \) can be found using the **cross product** of the direction vectors \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \):
  \[
  \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2
  \]

Let’s calculate the cross product.

\[
\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ -3 & 2 & 2 \end{vmatrix}
\]

I will calculate the cross product now.The vector perpendicular (normal) to both \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(s) \) is \( \mathbf{n} = (2, -3, 6) \).

#### Part (b): Use a scalar projection to find the minimum distance between the two lines.

- The minimum distance between two skew lines can be found using the formula:
  \[
  d = \frac{|(\mathbf{r}_2(0) - \mathbf{r}_1(0)) \cdot \mathbf{n}|}{|\mathbf{n}|}
  \]
  where \( \mathbf{r}_1(0) = (4, 1, 1) \) and \( \mathbf{r}_2(0) = (-1, 2, -5) \), and \( \mathbf{n} \) is the normal vector from part (a).

Let’s compute the difference \( \mathbf{r}_2(0) - \mathbf{r}_1(0) \) and the scalar projection now.The minimum distance between the two lines is \( 7 \) units.

#### Part (c): Find the values of \( t \) and \( s \) at which this minimum distance is achieved.

To find the values of \( t \) and \( s \) where the minimum distance occurs, we solve the system of equations that ensures the direction vector connecting the two points on the lines is parallel to the normal vector.

The vector from a point on \( \mathbf{r}_1(t) \) to a point on \( \mathbf{r}_2(s) \) can be expressed as:
\[
\mathbf{r}_2(s) - \mathbf{r}_1(t) = \left( (-1 + s(-3)) - (4 + t(3)), (2 + s(2)) - (1 + t(0)), (-5 + s(2)) - (1 + t(-1)) \right)
\]
This vector must be parallel to the normal vector \( \mathbf{n} \). Let’s solve for \( t \) and \( s \) using this condition.

Let me calculate that next.It seems there might be a slight issue with directly solving this system symbolically. I will take a different approach to compute the parameters \( t \) and \( s \) for the points of closest approach between the lines, using vector algebra instead.

I'll resolve this numerically.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can guide you through the steps to solve it manually. Let me know if you'd prefer that or have any other questions!

Learn how to find a vector perpendicular to two lines in ℝ³ and calculate the minimum distance between them using scalar projection. Detailed step-by-step solution with vector algebra techniques. Suitable for advanced high school students and beyond.

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