We are tasked with finding the perpendicular distance from the point $(-10, 10, 4)$ to the line given by the parametric equations:

$x(t) = -2 - 2t, \quad y(t) = 3 + 2t, \quad z(t) = 5 + 8t$

This line can be expressed as a point on the line $(-2, 3, 5)$ plus a direction vector $\mathbf{d} = (-2, 2, 8)$. We can calculate the perpendicular distance using the following formula:

$d = \frac{|\mathbf{AP} \cdot (\mathbf{d} \times \mathbf{AP})|}{|\mathbf{d}|}$

where:

• $\mathbf{A} = (-2, 3, 5)$ is a point on the line,
• $\mathbf{P} = (-10, 10, 4)$ is the given point,
• $\mathbf{d} = (-2, 2, 8)$ is the direction vector of the line,
• $\mathbf{AP}$ is the vector from $\mathbf{A}$ to $\mathbf{P}$, i.e., $\mathbf{AP} = \mathbf{P} - \mathbf{A}$.

Let's calculate this step by step.

1. Find $\mathbf{AP}$: $\mathbf{AP} = (-10, 10, 4) - (-2, 3, 5) = (-8, 7, -1)$

2. Cross product $\mathbf{d} \times \mathbf{AP}$:

   \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & 8 \\ -8 & 7 & -1 \end{vmatrix}$$ This will give the vector perpendicular to both $$\mathbf{d}$$ and $$\mathbf{AP}$$.

3. Magnitude of the cross product $\mathbf{d} \times \mathbf{AP}$.

4. Magnitude of $\mathbf{d}$.

5. Find the distance $d$.

Let me calculate this now.The perpendicular distance from the point $(-10, 10, 4)$ to the line is approximately $10.36$ units.

Would you like further details on any part of this process, or do you have any other questions?

Here are some related questions you might find useful:

1. How do we derive the formula for the distance from a point to a line?
2. How do cross products work, and why do we use them in this problem?
3. Can you explain how to compute the magnitude of vectors?
4. How is the direction vector of the line obtained from parametric equations?
5. What are other methods to find the distance from a point to a line in 3D space?

Tip: The cross product of two vectors results in a vector perpendicular to both of the original vectors, which is useful in finding areas and distances in 3D geometry.