Say you have vectors $v$ and $w$. Let there cross product be denoted by $x$ so that:
$$v \times w = x$$
According to Wikipedia:
$$x_x = v_yw_z - v_zw_y$$ $$x_y = v_zw_x - v_xw_z$$ $$x_z = v_xw_y - v_yw_x$$
This is equivalent to saying:
$$x_x = \left|\begin{matrix}v_y&v_z\\w_y&w_z\end{matrix}\right|$$ $$x_y = \left|\begin{matrix}v_z&v_x\\w_z&w_x\end{matrix}\right|$$ $$x_z = \left|\begin{matrix}v_x&v_y\\w_x&w_y\end{matrix}\right|$$
The determinant can be interpreted as the area spanned by the column vectors. Could you give me a geometric explanation of why the area of those above vectors give the coordinates of the cross product? Thanks!
EDIT: An interpretation for a 2x2 matrix is fine too
Let $P$ be the paralellogram spanned by the vectors $a$ and $b$ and let $\phi$ be the angle between $a$ and $b$. Then we have \begin{eqnarray*} (Area(P))^2 & = & |a|^2|b|^2(\sin(\phi))^2 \\ & = & |a|^2|b|^2-|a|^2|b|^2(\cos(\phi))^2 \\ & = &|a|^2|b|^2-(a\cdot b)^2 \\ & = & (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\ & = &(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^.2 \end{eqnarray*} So the area of $P$ is equal to the length of the vector $(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1) = a \times b$. With the coordinates you gave.