To take the cross product of two vectors ($\vec{B} \times \vec{C}$) in Cartesian coordinates in general, we set up a special 3-by-3 matrix that has as its rows the Cartesian unit vectors ($\hat{x}$, $\hat{y}$, and $\hat{z}$), the components of the first vector ($B_x$, $B_y$, and $B_z$), and the components of the second vector ($C_x$, $C_y$, and $C_z$). The columns are organized by component. The determinant of this matrix will give us the cross product of the two vectors:

$$\vec{B} \times \vec{C} = det\begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ B_x & B_y & B_z \\ C_x & C_y & C_z \\ \end{vmatrix}$$

A useful way to remember how to take the determinant is given online and uses cross-multiplication. We apply that method to find the 2-by-2 determinants that can be computed.

$$\vec{B} \times \vec{C} = det\begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ B_x & B_y & B_z \\ C_x & C_y & C_z \\ \end{vmatrix} = \hat{x}\begin{vmatrix} B_y & B_z \\ C_y & C_z \\ \end{vmatrix} - \hat{y} \begin{vmatrix} B_x & B_z \\ C_x & C_z \\ \end{vmatrix} + \hat{z} \begin{vmatrix} B_x & B_y \\ C_x & C_y \\ \end{vmatrix}$$

We further cross-multiply to find the determinants of the 2-by-2 matrices,

$$\vec{B} \times \vec{C} = \hat{x}\left(B_yC_z - C_yB_z\right)- \hat{y} \left(B_x C_z - C_xB_z\right) + \hat{z} \left(B_xC_y - C_xB_y\right)$$

So, in general, the cross product in Cartesian coordinates is given by,

$$\vec{B} \times \vec{C} = \langle B_yC_z - C_yB_z, C_xB_z-B_x C_z, B_xC_y - C_xB_y\rangle$$