To find the cross product $\mathbf{u} \times \mathbf{v}$ of the given vectors:

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

we use the formula for the cross product of two 3-dimensional vectors: $\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ 4 & 0 & -2 \end{vmatrix}.$

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

Now we compute each minor determinant:

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

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

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

Thus, the cross product is: $\mathbf{u} \times \mathbf{v} = 0 \mathbf{i} - (-2) \mathbf{j} + 0 \mathbf{k} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}.$

Final Answer:

$\mathbf{u} \times \mathbf{v} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}.$

Direction of the Cross Product

The direction of the cross product is along the positive $y$-axis, as the resulting vector points in the $\mathbf{j}$ (or $y$-axis) direction.

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Related Questions

1. How do you find the magnitude of a cross product vector?
2. Why is the cross product of two parallel vectors zero?
3. How does the right-hand rule apply to determine the direction of the cross product?
4. What are some physical applications of the cross product in physics?
5. What is the difference between the cross product and the dot product?

Quick Tip

For 3D vectors, remember that the cross product is always perpendicular to the plane formed by the original two vectors.