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To solve the given problem, we need to compute the magnitude of the cross product \(|\mathbf{u} \times \mathbf{v}|\), and identify the correct graph that represents vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{u} \times \mathbf{v}\).

### Given:
- \(\mathbf{u} = (2, 2, 0)\)
- \(\mathbf{v} = (2, 2, 2\sqrt{2})\)

### Step 1: Compute the Cross Product \(\mathbf{u} \times \mathbf{v}\)

The cross product of two vectors \(\mathbf{u} = (u_1, u_2, u_3)\) and \(\mathbf{v} = (v_1, v_2, v_3)\) is given by:

\[
\mathbf{u} \times \mathbf{v} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
\]

Substitute the values of \(\mathbf{u}\) and \(\mathbf{v}\):

\[
\mathbf{u} \times \mathbf{v} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2 & 0 \\
2 & 2 & 2\sqrt{2}
\end{vmatrix}
\]

Compute the determinant:

\[
\mathbf{u} \times \mathbf{v} = \mathbf{i} \left(2 \cdot 2\sqrt{2} - 0 \cdot 2\right) - \mathbf{j} \left(2 \cdot 2\sqrt{2} - 0 \cdot 2\right) + \mathbf{k} \left(2 \cdot 2 - 2 \cdot 2\right)
\]

Simplify the expression:

\[
\mathbf{u} \times \mathbf{v} = \mathbf{i} \cdot 4\sqrt{2} - \mathbf{j} \cdot 4\sqrt{2} + \mathbf{k} \cdot 0
\]

\[
\mathbf{u} \times \mathbf{v} = (4\sqrt{2}, -4\sqrt{2}, 0)
\]

### Step 2: Compute the Magnitude of the Cross Product

The magnitude of the cross product \(|\mathbf{u} \times \mathbf{v}|\) is calculated as:

\[
|\mathbf{u} \times \mathbf{v}| = \sqrt{(4\sqrt{2})^2 + (-4\sqrt{2})^2 + 0^2}
\]

Simplify further:

\[
|\mathbf{u} \times \mathbf{v}| = \sqrt{(16 \cdot 2) + (16 \cdot 2)}
\]

\[
|\mathbf{u} \times \mathbf{v}| = \sqrt{32 + 32}
\]

\[
|\mathbf{u} \times \mathbf{v}| = \sqrt{64} = 8
\]

### Step 3: Determine the Correct Graph

Now that we know \(\mathbf{u} \times \mathbf{v} = (4\sqrt{2}, -4\sqrt{2}, 0)\), we can see that the cross product lies in the xy-plane (since the z-component is 0). 

The correct graph should show \(\mathbf{u} \times \mathbf{v}\) in the xy-plane.

From the given choices:
- Graph **A** shows \(\mathbf{u} \times \mathbf{v}\) in the xy-plane.
- Graph **B** and **C** do not show the correct orientation of \(\mathbf{u} \times \mathbf{v}\).

Thus, the correct graph is **A**.

### Conclusion:
- \(|\mathbf{u} \times \mathbf{v}| = 8\)
- The correct graph is **A**.

Would you like more details or have any questions?

#### Further Related Questions:
1. What is the geometric meaning of the cross product of two vectors?
2. How do you determine the direction of the cross product using the right-hand rule?
3. Can the magnitude of a cross product ever be negative? Why or why not?
4. What are some real-world applications of the cross product in physics?
5. How does the cross product change if vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel?

#### Tip:
When visualizing vector cross products, always use the right-hand rule to determine the direction of the resulting vector!

Let u = (2,2,0) and v = (2,2,2√2). Compute |u × v|. Then sketch u, v, and u × v.

This problem focuses on computing the cross product of vectors u = (2, 2, 0) and v = (2, 2, 2√2). The solution explains how to compute the magnitude of the cross product using determinant formulas and visualize the result. Suitable for high school and introductory college students.

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