We are tasked with finding the shortest distance between a point $\mathbf{x} = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix}$ and the line described by the parametric equation $\mathbf{r}(t) = t \begin{pmatrix} 0 \\ 4 \\ -2 \end{pmatrix}$. This involves two steps:

1. Finding the point on the line closest to $\mathbf{x}$.
2. Finding the shortest distance between this point and $\mathbf{x}$.

Step 1: Finding the Point on the Line Closest to $\mathbf{x}$

The parametric equation for the line is given as: $\mathbf{r}(t) = t \begin{pmatrix} 0 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 0 \\ 4t \\ -2t \end{pmatrix}.$ Any point on this line is of the form $\mathbf{r}(t) = \begin{pmatrix} 0 \\ 4t \\ -2t \end{pmatrix}$.

To find the point on the line closest to $\mathbf{x} = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix}$, we need to minimize the distance between $\mathbf{r}(t)$ and $\mathbf{x}$. The distance is the magnitude of the vector $\mathbf{x} - \mathbf{r}(t)$: $\mathbf{x} - \mathbf{r}(t) = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix} - \begin{pmatrix} 0 \\ 4t \\ -2t \end{pmatrix} = \begin{pmatrix} -2 \\ -1 - 4t \\ 8 + 2t \end{pmatrix}.$ The square of the distance is: $D^2(t) = (-2)^2 + (-1 - 4t)^2 + (8 + 2t)^2.$ Expanding this: $D^2(t) = 4 + (1 + 8t + 16t^2) + (64 + 32t + 4t^2).$ Simplifying: $D^2(t) = 4 + 1 + 8t + 16t^2 + 64 + 32t + 4t^2 = 69 + 40t + 20t^2.$ Now, to minimize the distance, we take the derivative of $D^2(t)$ with respect to $t$ and set it equal to zero: $\frac{d}{dt} D^2(t) = 40 + 40t = 0.$ Solving for $t$: $t = -1.$

Substitute $t = -1$ back into the equation for $\mathbf{r}(t)$ to find the point on the line: $\mathbf{r}(-1) = \begin{pmatrix} 0 \\ 4(-1) \\ -2(-1) \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 2 \end{pmatrix}.$ Thus, the point on the line closest to $\mathbf{x}$ is $\begin{pmatrix} 0 \\ -4 \\ 2 \end{pmatrix}$.

Step 2: Finding the Distance

Now, we find the distance between $\mathbf{x} = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix}$ and the closest point $\begin{pmatrix} 0 \\ -4 \\ 2 \end{pmatrix}$.

The distance is the magnitude of the vector: $\mathbf{x} - \mathbf{r}(-1) = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix} - \begin{pmatrix} 0 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 6 \end{pmatrix}.$ The magnitude of this vector is: $\text{Distance} = \sqrt{(-2)^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.$

Thus, the shortest distance from the point $\mathbf{x} = \begin{pmatrix} -2 \\ -1 \\ 8 \end{pmatrix}$ to the line is $7$.

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Would you like more details or explanations?

Here are 5 related questions:

1. How do you find the direction vector of a line from a parametric equation?
2. What is the geometric interpretation of minimizing the distance between a point and a line?
3. How would the method change if the line equation were given in a different form?
4. Can this approach be extended to find the distance from a point to a plane?
5. How would you find the distance if the point and line were in a different coordinate system?

Tip: When working with distances between points and lines, always verify that the vector used to compute distances is perpendicular to the line for the shortest path.