Consider points A(8,4,−8),B(6,6,0),C(2,−2,−8). Find the distance from A to the line BC.

To find the distance from point \( A \) to the line defined by points \( B \) and \( C \), we can use the formula for the distance from a point to a line in 3D space. The formula for the distance \( d \) from point \( A \) to line \( BC \) is given by:

\[
d = \frac{|\vec{AB} \times \vec{AC}|}{|\vec{BC}|}
\]

where:
- \( \vec{AB} \) is the vector from \( A \) to \( B \),
- \( \vec{AC} \) is the vector from \( A \) to \( C \),
- \( \vec{BC} \) is the vector from \( B \) to \( C \),
- \( \vec{AB} \times \vec{AC} \) is the cross product of \( \vec{AB} \) and \( \vec{AC} \),
- \( |\vec{BC}| \) is the magnitude of vector \( \vec{BC} \).

### Step-by-Step Solution

1. **Calculate vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{BC} \):**

   \[
   \vec{AB} = B - A = (6 - 8, 6 - 4, 0 - (-8)) = (-2, 2, 8)
   \]
   \[
   \vec{AC} = C - A = (2 - 8, -2 - 4, -8 - (-8)) = (-6, -6, 0)
   \]
   \[
   \vec{BC} = C - B = (2 - 6, -2 - 6, -8 - 0) = (-4, -8, -8)
   \]

2. **Compute the cross product \( \vec{AB} \times \vec{AC} \):**

   Using the determinant formula for the cross product:
   \[
   \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 8 \\ -6 & -6 & 0 \end{vmatrix}
   \]

   Expanding this determinant:
   \[
   \vec{AB} \times \vec{AC} = \hat{i}(2 \cdot 0 - 8 \cdot (-6)) - \hat{j}(-2 \cdot 0 - 8 \cdot (-6)) + \hat{k}(-2 \cdot (-6) - 2 \cdot (-6))
   \]


Consider points A(8,4,−8), B(6,6,0), C(2,−2,−8). Find the distance from A to the line BC.

Explore a solution to finding the distance from point A(8,4,-8) to the line defined by points B(6,6,0) and C(2,-2,-8) in 3D space. This problem includes vector operations, the cross product, and vector magnitude calculations, ideal for high school students in Grades 10-12.

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