Find the shortest distance from the point x––=⎛⎝⎜−2−118⎞⎠⎟ to the line r–=t⎛⎝⎜04−2⎞⎠⎟.

First, find the point on the line r– that is closest to x––.

 	
-2
 
 	
-1
 
 	
8
 
 ⎛⎝⎜−2−18⎞⎠⎟
Now find the distance from x–– to the point you just found.

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 \(7\).

---

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.

Find the shortest distance from the point x––=⎛⎝⎜−2−118⎞⎠⎟ to the line r–=t⎛⎝⎜04−2⎞⎠⎟. First, find the point on the line r– that is closest to x––. Now find the distance from x–– to the point you just found.

This page explores the step-by-step process of finding the shortest distance between the point (-2, -1, 8) and the line r(t) = t(0, 4, -2). Learn key concepts from linear algebra, such as minimizing quadratic functions, using projection theorem, and applying vector geometry for accurate distance calculations. Suitable for undergraduate students studying linear algebra.

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