I know how to use the cross product, I know what it means and how it relates to the dot product.

$$|a \times b| = ||a||b| \cdot \sin(\theta) \vec{n}|\\ a \cdot b = |a||b| \cdot \cos(\theta)$$

I also understand why and how you can calculate the area of two 3d vectors with the cross product.

What I don't understand where this is coming from

$$\displaystyle{(a_x, a_y, a_z) \times (b_x, b_y, b_z) = (a_y b_z - a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x) = \|a\| \|b\| \sin(\theta) \vec{n}}$$

I mean I could imagine that I could invent it myself with the following properties and a lot of trial and error.

$a \times b = c$

$a \cdot c = 0$

$b \cdot c = 0$

But I don't really have an intuition of why it works the way it does.


First notice that if we define $(u \wedge v ) \cdot w = \det (u,v,w)$ and let $i,j,k,l = 1,2,3$ then $$(e_i \wedge e_j) \cdot (e_k \wedge e_l) = \begin{vmatrix}e_i \cdot e_k & e_j \cdot e_k \\e_i \cdot e_l & e_j \cdot e_l\end{vmatrix}$$

Consequently $$(u \wedge v)\cdot (u \wedge v)=|u \wedge v |^2 = \begin{vmatrix}u \cdot u & v \cdot u \\ u \cdot v & v \cdot v\end{vmatrix} = |u|^2 |v|^2 (1 - \cos ^2 \theta) = A^2$$

where $\theta$ is the angle between $u $ and $v$ and $\{u,v, u \wedge v\}$ is a positive basis. As for the intuition $A$ is the area of the paralelogram generated by $u$ and $v$.

Note: Sometimes $u \wedge v$ can be written as $u \times v$.


Let $i_x, i_y, i_z$ be the basis vectors of a right-handed coordinate system. Now we can write \begin{equation} i_x \times i_y = i_z \quad i_y \times i_z = i_x \quad i_z \times i_x = i_y \end{equation} By the definition of vector product if the vectors are orthogonal the $sin (\theta) = 1$ while if the vectors are parallel $sin (\theta) = 0$ then \begin{equation} i_x \times i_x = 0 \quad i_y \times i_y = 0 \quad i_z \times i_z = 0 \end{equation}

Now consider the vectors $A$ and $ B$. By using the above expressions, we can easily express the vector product of two vectors $A$ and $B$ in terms of their components: \begin{equation} C = A \times B = (A_x i_x + A_y i_y + A_z i_z) \times (B_x i_x + B_y i_y + B_z i_z) = (A_y B_z-A_z B_y)i_x+(A_z B_x-A_x B_z)i_y+(A_x B_y-A_y B_x)i_z \end{equation} Which can be written by using the determinant of a matrix: \begin{equation} C = A \times B = det (\begin{bmatrix} i_x & i_{y} & i_{z} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{bmatrix}) \end{equation}


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