Understanding the definition of the cross product

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.

Here's the best way to think about the cross product, in my opinion. Let $b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}, c = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} \in \mathbb R^3$, and let $L:\mathbb R^3 \to \mathbb R$ be the function defined by $$ L(a) = \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} . $$ The function $L$ is linear, so there exists a vector $w \in \mathbb R^3$ such that $L(a) = a \cdot w$ for all $a \in \mathbb R^3$. This vector $w$ is called the cross product of $b$ and $c$, and is denoted $b \times c$. So the defining equation for the cross product is $$ \tag{$\heartsuit$}\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} = a \cdot (b \times c) $$ for all $a \in \mathbb R^3$.

The geometric interpretation of the determinant (it is the signed volume of the parallelepiped spanned by $a,b$, and $c$) leads directly to the geometric interpretation of the cross product. For example, the parallelepiped spanned by $b, b$, and $c$ is degenerate, so its volume is $0$. It follows that $$ b \cdot (b \times c) = 0. $$ Similarly, $c \cdot(b \times c) = 0$. So $b \times c$ is orthogonal to both $b$ and $c$.

What about the norm of $b \times c$? Let $A$ be the area of the parallelogram spanned by $b$ and $c$, and let $\theta$ be the angle between $b \times c$ and $a$. (Assume for simplicity that $0 \leq \theta \leq \pi/2$.) The volume $V$ of the parallelepiped spanned by $a, b$, and $c$ is \begin{align} V &= \text{"area of the base times the height"} \\ &= A \|a \| \cos(\theta). \end{align} Comparing this formula with $$ V = a \cdot (b \times c) = \| a \| \| b \times c \| \cos(\theta) $$ reveals that $\| b \times c \| = A$. So the norm of $b \times c$ is the area of the parallelogram spanned by $b$ and $c$.

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 $u \times v$.