Find the distance between the point and the line given by the set of parametric equations. (Round your answer to three decimal places.)
(1, -1, 5); x = 2t, y = t − 3, z = 2t + 2

To find the distance between a point and a line given by parametric equations, we'll use the following steps:

1. **Express the line in vector form:**
   The parametric equations for the line are:
   \[
   x = 2t, \quad y = t - 3, \quad z = 2t + 2
   \]
   This can be written as:
   \[
   \mathbf{r}(t) = (2t, t - 3, 2t + 2)
   \]
   The direction vector \(\mathbf{v}\) of the line is given by the coefficients of \(t\):
   \[
   \mathbf{v} = \langle 2, 1, 2 \rangle
   \]
   A point on the line when \(t = 0\) is:
   \[
   \mathbf{r}_0 = (0, -3, 2)
   \]

2. **Find the vector from the point to a point on the line:**
   Let the given point be \(\mathbf{P} = (1, -1, 5)\). The vector from \(\mathbf{r}_0\) to \(\mathbf{P}\) is:
   \[
   \mathbf{P} - \mathbf{r}_0 = \langle 1 - 0, -1 - (-3), 5 - 2 \rangle = \langle 1, 2, 3 \rangle
   \]

3. **Compute the cross product of this vector with the direction vector:**
   The cross product \(\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)\) is:
   \[
   \mathbf{v} \times \langle 1, 2, 3 \rangle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}
   \]
   \[
   = \mathbf{i} \left(1 \cdot 3 - 2 \cdot 2\right) - \mathbf{j} \left(2 \cdot 3 - 2 \cdot 1\right) + \mathbf{k} \left(2 \cdot 2 - 1 \cdot 1\right)
   \]
   \[
   = \mathbf{i} (3 - 4) - \mathbf{j} (6 - 2) + \mathbf{k} (4 - 1)
   \]
   \[
   = \langle -1, -4, 3 \rangle
   \]

4. **Calculate the magnitude of the cross product:**
   \[
   |\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)| = \sqrt{(-1)^2 + (-4)^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}
   \]

5. **Compute the magnitude of the direction vector:**
   \[
   |\mathbf{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3
   \]

6. **Calculate the distance:**
   The distance \(d\) from the point to the line is given by:
   \[
   d = \frac{|\mathbf{v} \times (\mathbf{P} - \mathbf{r}_0)|}{|\mathbf{v}|} = \frac{\sqrt{26}}{3}
   \]
   \[
   d \approx \frac{5.099}{3} \approx 1.700
   \]

So, the distance between the point \((1, -1, 5)\) and the line is approximately \(1.700\) units.

Would you like more details or have any questions?

Here are some related questions:
1. How is the cross product of two vectors calculated?
2. What is the significance of the magnitude of a vector?
3. How can you derive the distance formula from a point to a line?
4. What are the parametric equations of a line in three-dimensional space?
5. How does the direction vector of a line relate to its parametric equations?
6. Can the distance between a point and a line ever be negative?
7. What role does the dot product play in geometry?
8. How do you find the shortest distance between two skew lines?

**Tip:** Always verify your vector calculations step by step to avoid errors, especially in cross products.

Learn how to find the distance between a point (1, -1, 5) and a line defined by parametric equations in 3D space. This detailed step-by-step solution covers vectors, parametric equations, the cross product, and the distance formula. Suitable for advanced high school students.

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