A its at (8,0,8), B its at (0,4,0), P its at (-6,10,6) and F goes from A to B

Given the coordinates:
- Point \( A = (8, 0, 8) \)
- Point \( B = (0, 4, 0) \)
- Point \( P = (-6, 10, 6) \)

The vector \( \mathbf{F} \) goes from point \( A \) to point \( B \), so:

\[
\mathbf{F} = \overrightarrow{AB} = B - A
\]

### Step-by-Step Solution:

1. **Calculate Vector \( \overrightarrow{AB} \):**

\[
\overrightarrow{AB} = (x_B - x_A, y_B - y_A, z_B - z_A) = (0 - 8, 4 - 0, 0 - 8)
\]

\[
\overrightarrow{AB} = (-8, 4, -8)
\]

2. **Magnitude of Vector \( \overrightarrow{AB} \):**

\[
|\overrightarrow{AB}| = \sqrt{(-8)^2 + 4^2 + (-8)^2}
\]

\[
|\overrightarrow{AB}| = \sqrt{64 + 16 + 64} = \sqrt{144} = 12
\]

3. **Unit Vector in the Direction of \( \overrightarrow{AB} \):**

\[
\mathbf{u}_{AB} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \left( \frac{-8}{12}, \frac{4}{12}, \frac{-8}{12} \right)
\]

\[
\mathbf{u}_{AB} = \left( -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right)
\]

4. **Vector from Point \( A \) to Point \( P \):**

\[
\overrightarrow{AP} = P - A = (-6 - 8, 10 - 0, 6 - 8)
\]

\[
\overrightarrow{AP} = (-14, 10, -2)
\]

5. **Projection of Vector \( \overrightarrow{AP} \) onto \( \overrightarrow{AB} \):**

The projection formula is:

\[
\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = (\overrightarrow{AP} \cdot \mathbf{u}_{AB}) \mathbf{u}_{AB}
\]

First, compute the dot product:

\[
\overrightarrow{AP} \cdot \mathbf{u}_{AB} = (-14) \left( -\frac{2}{3} \right) + 10 \left( \frac{1}{3} \right) + (-2) \left( -\frac{2}{3} \right)
\]

\[
\overrightarrow{AP} \cdot \mathbf{u}_{AB} = \frac{28}{3} + \frac{10}{3} + \frac{4}{3} = \frac{42}{3} = 14
\]

6. **Final Projection Value:**

\[
\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = 14 \cdot \left( -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right)
\]

\[
\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = \left( -\frac{28}{3}, \frac{14}{3}, -\frac{28}{3} \right)
\]

### Conclusion:

The projection of vector \( \overrightarrow{AP} \) onto the direction of \( \overrightarrow{AB} \) is:

\[
\left( -\frac{28}{3}, \frac{14}{3}, -\frac{28}{3} \right)
\]

Would you like further details on this solution, or have any questions?

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Here are some related questions for you:

1. How is the dot product used to find projections?
2. What is the geometric interpretation of a vector projection?
3. Can you calculate the angle between vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AP} \)?
4. How does the magnitude of a vector affect its unit vector?
5. What are some practical applications of vector projections in physics?

**Tip:** When working with vectors, always visualize them in 3D space to understand their relative directions and magnitudes better!

Learn how to calculate the projection of vector AP onto vector AB using dot products and vector projection formulas. This problem involves vectors in 3D space and is suitable for high school students.

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