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The problem asks for the scalar triple product of three vectors \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\). To solve this, we need to follow these steps:

1. **Find the cross product** of the vectors \(\mathbf{v} \times \mathbf{w}\).
2. **Compute the dot product** of \(\mathbf{u}\) with the result from the first step.

### Step 1: Compute \(\mathbf{v} \times \mathbf{w}\)

Given:
\[
\mathbf{v} = [-2, -4, 4], \quad \mathbf{w} = [-3, -5, -1]
\]
The cross product \(\mathbf{v} \times \mathbf{w}\) is computed using the determinant of the following matrix:
\[
\mathbf{v} \times \mathbf{w} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & -4 & 4 \\
-3 & -5 & -1
\end{vmatrix}
\]

### Step 2: Compute \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\)

With \(\mathbf{u} = [6, 5, 0]\), we can calculate the final scalar triple product once the cross product is obtained.

Let me calculate the result step-by-step.The scalar triple product \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) is \(74\).

Would you like more details on how the calculation was performed or any additional questions? Here are some related questions to expand on the topic:

1. What is the geometric interpretation of the scalar triple product?
2. How does the scalar triple product relate to the volume of a parallelepiped?
3. What are the properties of the cross product used in this computation?
4. Can the scalar triple product be zero? If so, under what conditions?
5. How does changing one of the vectors affect the result of the scalar triple product?

**Tip:** The scalar triple product gives the volume of the parallelepiped formed by the three vectors, and a positive or negative result indicates the orientation of the vectors in space.

Given the vectors u = [6, 5, 0], v = [-2, -4, 4], w = [-3, -5, -1], find the scalar triple product u · (v × w).

Solve the scalar triple product for the vectors u = [6, 5, 0], v = [-2, -4, 4], and w = [-3, -5, -1]. Learn how to compute the cross product and dot product to find the volume of a parallelepiped.

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