Cross-product properties from abstract definition Given two $3$D vectors $\mathbf{u}$ and $\mathbf{v}$ their cross-product $\mathbf{u} \times \mathbf{v}$ can be defined by the property that, for any vector $\mathbf{x}$ one has $\langle \mathbf{x} ; \mathbf{u} \times \mathbf{v} \rangle = {\rm det}(\mathbf{x}, \mathbf{u},\mathbf{v})$.
From this a number of properties of the cross product can be obtained quite easily. It is less obvious that, for instance $|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - \langle \mathbf{u} ; \mathbf{v} \rangle ^2$, from which the norm of the cross-product can be deduced.
Is it possible to obtain these properties nicely (i.e. without dealing with coordinates), but with elementary linear algebra only (i.e. without the exterior algebra stuff, only properties of determinants and matrix / vector multiplication).
Thanks in advance!  
 A: Hint
$$|\mathbf{u}\times \mathbf{v}| = |\mathbf{u}|\cdot|\mathbf{v}|\cdot|\sin \alpha|$$
$$\langle \mathbf{u}, \mathbf{v}\rangle = |\mathbf{u}|\cdot|\mathbf{v}|\cdot|\cos \alpha|$$
where $\alpha$ is an angle between vectors $\mathbf{u}$ and $\mathbf{v}$
A: I went through some books and found something I am happier with. It comes from Euclidean and Non-Euclidean Geometry: An Analytic Approach, by Ryan (p.85, or something like this)
Essentially it goes as follows : $\mathbf{n} = \mathbf{u} \times \mathbf{v}$ is defined by the property that for every $\mathbf{x}$ one has $\langle \mathbf{x} ; \mathbf{n}\rangle = \det( \mathbf{x},\mathbf{u},\mathbf{v})$. 


*

*Antisymetry and linearity follow directly from the corresponding properties of the determinant.

*It is also easy to get $\langle \mathbf{u} ; \mathbf{v} \times \mathbf{w} \rangle = \langle \mathbf{u}\times \mathbf{v} ; \mathbf{w}\rangle$;

*Using linearity, and restricting in a clever way to the basis vectors one shows that $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})  = \langle \mathbf{u};\mathbf{w} \rangle \mathbf{v} - (\mathbf{u};\mathbf{v})\mathbf{w}$. This is the only part in which some "dirty" work is needed, but that is not too bad : using symmetry argument and linearity, one really needs very little computations. 

*Using the last, one gets $\langle \mathbf{u}\times \mathbf{v}; \mathbf{w}\times \mathbf{z}\rangle  = \langle\mathbf{u};\mathbf{w} \rangle \langle\mathbf{v};\mathbf{z} \rangle - \langle\mathbf{v};\mathbf{w} \rangle \langle \mathbf{u};\mathbf{z} \rangle $

*From this one gets the Lagrange identity, which, by the way, allows to get another proof of Cauchy - Schwartz
A: The attempt to prove
$|\mathbf{u} \times \mathbf{v}|^2 - |\mathbf{u}|^2 |\mathbf{v}|^2 + \langle \mathbf{u} ; \mathbf{v} \rangle ^2=0$,
with the use of formula 
$\mathbf{u} \times \mathbf{v}= \mathbf {S(u)v}$ where  $\mathbf {S(u)}$ is skew-symmetric  matrix.   
For simplification let  normalize $|\mathbf{u}|=1$.
The formula can be written:     
$\mathbf {(S(u)v)}^T \mathbf {S(u)v}-(\mathbf{v}^T \mathbf{v})( \mathbf{u}^T \mathbf{u}) +(\mathbf{v}^T \mathbf{u})(\mathbf{u}^T \mathbf{v})$=
$\mathbf {v^TS(u)^T}  \mathbf {S(u)v} - \mathbf{v}^T \mathbf{v} \mathbf{u}^T \mathbf{u} +\mathbf{v}^T  \mathbf{u}\mathbf{u}^T \mathbf{v}$=
 $\mathbf {-v^TS^2(u)}  \mathbf {v} -\mathbf{v}^T \mathbf{v} \mathbf{u}^T \mathbf{u} + \mathbf{v}^T  \mathbf{u}\mathbf{u}^T \mathbf{v}$=
$\mathbf {-v^T( uu^T-I)}  \mathbf {v} - \mathbf{v}^T \mathbf{v} \mathbf{u}^T \mathbf{u} +\mathbf{v}^T  \mathbf{u}\mathbf{u}^T \mathbf{v}$=
$ -\mathbf {v^T uu^T    v}+  \mathbf {v}^T \mathbf {v} - \mathbf{v}^T \mathbf{v} \mathbf{u}^T \mathbf{u} +\mathbf{v}^T \mathbf{u} \mathbf{u}^T \mathbf{v}$ =
$   \mathbf {v}^T \mathbf {v} - \mathbf{v}^T \mathbf{v} \mathbf{u}^T \mathbf{u}$=
$\mathbf {v}^T \mathbf {v} (1-\mathbf {u}^T \mathbf {u})= 0$.
So in this case it is fulfilled. I hope all steps are understood.
A: I believe I've found an elegant proof.

*

*Assume that $ \mathbf{u}\times \mathbf{v} \ne 0$. For $\mathbf{x} = \mathbf{u}\times \mathbf{v}$ we have
\begin{align}
\|\mathbf{u}\times \mathbf{v}\|^4 &= \langle \mathbf{u}\times \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle^2 \\
&=\det(\mathbf{u}\times \mathbf{v},\mathbf{u},\mathbf{v})^2\\
&= \det\left(\begin{bmatrix} \mathbf{u}\times \mathbf{v} \\ \mathbf{u} \\ \mathbf{v}\end{bmatrix}\begin{bmatrix} \mathbf{u}\times \mathbf{v} ,\mathbf{u} ,\mathbf{v}\end{bmatrix}\right)\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & \langle \mathbf{u}\times \mathbf{v}, \mathbf{u}\rangle & \langle \mathbf{u}\times \mathbf{v},\mathbf{v}\rangle \\ \langle \mathbf{u},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{u},\mathbf{u}\rangle & \langle \mathbf{u},\mathbf{v}\rangle \\ \langle \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{v},\mathbf{u}\rangle & \langle \mathbf{v},\mathbf{v}\rangle\end{vmatrix}\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & 0 & 0 \\ 0 & \|\mathbf{u}\|^2 & \langle \mathbf{u},\mathbf{v}\rangle \\ 0 & \langle \mathbf{u},\mathbf{v}\rangle & \|\mathbf{v}\|^2\end{vmatrix}\\
&= \|\mathbf{u}\times \mathbf{v}\|^2(\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2).
\end{align}
Dividing by $\|\mathbf{u}\times \mathbf{v}\|^2$ gives the result.


*Assume that $ \mathbf{u}\times \mathbf{v} = 0$. Then the $\det(\mathbf{x},\mathbf{u},\mathbf{v}) = 0$ for all vectors $\mathbf{x}\in\Bbb{R}^3$ so $\{\mathbf{u},\mathbf{v}\}$ are linearly dependent. Equality condition in Cauchy-Schwartz gives
$$\|\mathbf{u}\times \mathbf{v}\|^2 = 0 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2.$$
