To solve this problem, let's compute the cross product of the vectors $\mathbf{u}$ and $\mathbf{v}$, which are given as:

$\mathbf{u} = \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 4 \\ 0 \\ -2 \end{bmatrix}$

The cross product $\mathbf{u} \times \mathbf{v}$ is calculated using the following determinant formula:

$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ 4 & 0 & -2 \end{vmatrix}$

Expanding the determinant, we have:

$\mathbf{u} \times \mathbf{v} = \mathbf{i} \begin{vmatrix} 0 & -1 \\ 0 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -1 \\ 4 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix}$

Calculating each part:

1. For the $\mathbf{i}$-component: $\begin{vmatrix} 0 & -1 \\ 0 & -2 \end{vmatrix} = (0)(-2) - (-1)(0) = 0$

2. For the $\mathbf{j}$-component: $\begin{vmatrix} 3 & -1 \\ 4 & -2 \end{vmatrix} = (3)(-2) - (-1)(4) = -6 + 4 = -2$

3. For the $\mathbf{k}$-component: $\begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix} = (3)(0) - (0)(4) = 0$

Putting it all together:

$\mathbf{u} \times \mathbf{v} = 0 \mathbf{i} - (-2) \mathbf{j} + 0 \mathbf{k} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}$

The resulting vector $\mathbf{u} \times \mathbf{v} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}$ is along the $y$-axis.

Answer: The direction of the cross product $\mathbf{u} \times \mathbf{v}$ is parallel to the y-axis.

Would you like further explanation on the cross-product calculation or have any questions?

Here are some related questions:

1. How is the direction of the cross product determined?
2. Why is the cross product perpendicular to the original vectors?
3. What are some applications of the cross product in physics and engineering?
4. How would the cross product differ if vectors were not coplanar?
5. Can the cross product result in a zero vector? Under what conditions?

Tip: The magnitude of a cross product $\mathbf{u} \times \mathbf{v}$ gives the area of the parallelogram formed by $\mathbf{u}$ and $\mathbf{v}$.