TITLE: How does computing the determinant of a matrix with unit vectors give you the Cross Product?
QUESTION [5 upvotes]: Say you had $(a_x,a_y,a_z)\times(b_x,b_y,b_z)$, you would set up a matrix like the following:
 
And the resulting would be your Cross Product or the coordinates of an orthogonal vector. My question is why? Why does forming it that way give you the magnitude of an orthogonal vector and how is it related to the $\sin(\theta)$ definition of Cross Product.

REPLY [3 votes]: The determinant of a $3\times3$ matrix can be viewed as the triple product of its columns (or rows):
$$
\begin{align}
\det\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3
\end{bmatrix}
&=
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
\times
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\cdot
\begin{bmatrix}
z_1\\
z_2\\
z_3
\end{bmatrix}\\
&=
\begin{bmatrix}
(x\times y)_1\\
(x\times y)_2\\
(x\times y)_3
\end{bmatrix}
\cdot
\begin{bmatrix}
z_1\\
z_2\\
z_3
\end{bmatrix}\tag{1}
\end{align}
$$
If we replace $\begin{bmatrix} z_1\\ z_2\\ z_3 \end{bmatrix}$ in $(1)$ by $\begin{bmatrix} \boldsymbol{i}\\ \boldsymbol{j}\\ \boldsymbol{k} \end{bmatrix}$, we get
$$
\begin{align}
\det\begin{bmatrix}
x_1&y_1&\boldsymbol{i}\\
x_2&y_2&\boldsymbol{j}\\
x_3&y_3&\boldsymbol{k}
\end{bmatrix}
&=
\begin{bmatrix}
(x\times y)_1\\
(x\times y)_2\\
(x\times y)_3
\end{bmatrix}
\cdot
\begin{bmatrix}
\boldsymbol{i}\\
\boldsymbol{j}\\
\boldsymbol{k}
\end{bmatrix}\\[6pt]
&=(x\times y)_1\boldsymbol{i}+(x\times y)_2\boldsymbol{j}+(x\times y)_3\boldsymbol{k}\\[18pt]
&=x\times y\tag{2}
\end{align}
$$