Q:

Is |AxB| = |A||B|sin(v) a theorem or a definition?

For cross products in $\mathbb{R}^3$, we have $ \| u \times v \| = \| u\| \| v\| \sin \alpha $. But is that a theorem or a definition? I read somewhere that it actually was a theorem which was proved by aid of Lagrange's identity $ \| u \times v \|^2 = \| a \|^2 \| b \|^2 - (a \cdot b)^2 $ but that begs the question: Is Lagranges identity derived independently of $ \| u \times v \| = \| u\| \| v\| \sin \alpha $?

A:

You could define $u \times v$ to be the vector in $\mathbb{R}^3$ such that:

$|u \times v| = |u||v| \sin(\alpha)$
$u \times v$ is orthogonal to both $u$ and $v$
$u \times v$ is chosen to be in the direction given by the right-hand rule

This can be a bit cumbersome to work with. You need to know about the angle $\alpha$ between the vectors, and you need to perform this right-hand-rule thing that makes physics students embarrass themselves during exams. 
You can also define the cross product by setting 

$i \times j = - j \times i = k$
$j \times k = - k \times j = i$
$k \times i = - i \times k = j$
$i \times i = j \times j = k \times k = 0$

and extending by the distributive law. Here, $i,j,k$ denote the unit basis vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, respectively. 
If you see a definition involving a determinant, it's quickly shown to be the same as this second definition. It's also a bit easier to remember and compute with. 
This definition has the advantage of being algebraically easy to work with, but possibly a bit cumbersome to understand. The first definition is equivalent to the second, however, with a bit of thought. You might see the first definition used to introduce the concept, with the second definition possibly discussed later to make manipulations a bit easier.

