To calculate the cross product of the vectors given:

$\mathbf{A} = \begin{bmatrix} 8 \\ 8 \\ 6 \end{bmatrix} \quad \text{and} \quad \mathbf{B} = \begin{bmatrix} 5 \\ 6 \\ 4 \end{bmatrix},$

we can use the formula for the cross product of two vectors in $\mathbb{R}^3$, represented as:

$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 8 & 6 \\ 5 & 6 & 4 \end{vmatrix}$

Expanding the determinant, we get:

$\mathbf{A} \times \mathbf{B} = \mathbf{i} \cdot \begin{vmatrix} 8 & 6 \\ 6 & 4 \end{vmatrix} - \mathbf{j} \cdot \begin{vmatrix} 8 & 6 \\ 5 & 4 \end{vmatrix} + \mathbf{k} \cdot \begin{vmatrix} 8 & 8 \\ 5 & 6 \end{vmatrix}$

Calculating each minor:

1. For $\mathbf{i}$: $\begin{vmatrix} 8 & 6 \\ 6 & 4 \end{vmatrix} = (8 \cdot 4) - (6 \cdot 6) = 32 - 36 = -4$

2. For $\mathbf{j}$: $\begin{vmatrix} 8 & 6 \\ 5 & 4 \end{vmatrix} = (8 \cdot 4) - (6 \cdot 5) = 32 - 30 = 2$

3. For $\mathbf{k}$: $\begin{vmatrix} 8 & 8 \\ 5 & 6 \end{vmatrix} = (8 \cdot 6) - (8 \cdot 5) = 48 - 40 = 8$

Putting it all together:

$\mathbf{A} \times \mathbf{B} = -4 \mathbf{i} - 2 \mathbf{j} + 8 \mathbf{k}$

So, the cross product is:

$\mathbf{A} \times \mathbf{B} = \begin{bmatrix} -4 \\ -2 \\ 8 \end{bmatrix}$

Let me know if you need any more details!

Questions for Further Exploration:

1. What is the geometric interpretation of a cross product?
2. How does the cross product differ from the dot product in terms of properties?
3. Can cross products be used for vectors in higher dimensions?
4. Why does the cross product only apply to 3-dimensional vectors?
5. How does the direction of the resulting vector relate to the original vectors?

Tip:

Remember that the cross product of two vectors results in a vector that is perpendicular to both original vectors, which follows the right-hand rule.