Let \( \mathbf{u}=\langle 0,-3,0\rangle \) and \( \mathbf{v}=\langle 0,3,0\rangle \). Compute \( |\mathbf{u} \times \mathbf{v}| \). Then sketch \( \mathbf{u}, \mathbf{v} \), and \( \mathbf{u} \times \mathbf{v} \). \( |\mathbf{u} \times \mathbf{v}|=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Given the vectors: \[ \mathbf{u} = \langle 0, -3, 0 \rangle \quad \text{and} \quad \mathbf{v} = \langle 0, 3, 0 \rangle \] **Step 1: Compute the Cross Product \(\mathbf{u} \times \mathbf{v}\)** The cross product of two vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) is given by: \[ \mathbf{u} \times \mathbf{v} = \langle u_2 v_3 - u_3 v_2,\ u_3 v_1 - u_1 v_3,\ u_1 v_2 - u_2 v_1 \rangle \] For \(\mathbf{u} = \langle 0, -3, 0 \rangle\) and \(\mathbf{v} = \langle 0, 3, 0 \rangle\): \[ \mathbf{u} \times \mathbf{v} = \langle (-3)(0) - (0)(3),\ (0)(0) - (0)(0),\ (0)(3) - (-3)(0) \rangle = \langle 0, 0, 0 \rangle \] **Step 2: Compute the Magnitude \(|\mathbf{u} \times \mathbf{v}|\)** The magnitude of the cross product is: \[ |\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + 0^2} = 0 \] **Conclusion:** \[ |\mathbf{u} \times \mathbf{v}| = 0 \] **Explanation:** The vectors \(\mathbf{u}\) and \(\mathbf{v}\) are **parallel** (in fact, they are scalar multiples of each other: \(\mathbf{v} = -\mathbf{u}\)). When two vectors are parallel, the sine of the angle between them is zero, resulting in a cross product magnitude of zero. This means there is no vector perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\) in this case. **Sketch:** 1. **Vector \(\mathbf{u}\)**: Points downward along the negative \(y\)-axis. 2. **Vector \(\mathbf{v}\)**: Points upward along the positive \(y\)-axis. 3. **Vector \(\mathbf{u} \times \mathbf{v}\)**: The zero vector, which has no direction and magnitude.  *(Replace with an actual sketch if needed)* **Final Answer:** \[ |\mathbf{u} \times \mathbf{v}| = 0 \]