How to prove this Gram determinant Let $E$ be an Euclidian oriented vector space of dimension $3$ and $x,y,u,w \in E$.

How do we prove (without coodinates)
  $$ \det \begin{pmatrix}
\langle x,u \rangle & \langle x,w \rangle \\  
\langle y,u \rangle & \langle y,w \rangle \\
\end{pmatrix}
= \langle x \times y, u \times w \rangle$$

This equality is claimed in this proof of the double cross product formula.
Related : wiki claims that the Gram determinant is equal to : $Gram(x_1,\dots,x_n) = \| x_1 \wedge \dots \wedge x_n \|^2$

How is defined the norm on the exterior product, and how to prove this formula ?

 A: I'm not really sure if you can prove it or simply take it as definition of the scalar product. You can derive it from the definition of a scalar product on the tensor space.
Or just notice that the Gramian determinant 
$$
\det\begin{pmatrix}
\langle x,u \rangle & \langle x,w \rangle \\  
\langle y,u \rangle & \langle y,w \rangle \\
\end{pmatrix}
$$
is already symmetric in $x∧y$ and $u∧w$. Considering it as a function in $x$ and $y$,
$$
\phi(x,y)=\det\begin{pmatrix}
\langle x,u \rangle & \langle x,w \rangle \\  
\langle y,u \rangle & \langle y,w \rangle \\
\end{pmatrix}
$$
one notices that it is bilinear, anti-symmetric, and so (extendable as) a linear form on $\Lambda^2E$. Further,
$$
\phi(u,w)=\langle u,u \rangle \langle w,w \rangle - \langle u,w \rangle^2=\|u\|^2\|w-\lambda u\|^2
$$
where $λ=\frac{\langle u,w \rangle}{\|u\|^2}$, 
tells us that this quadratic form is the square of the area of the parallelogram spanned by $u$ and $v$, which proves its positive definiteness.
Of course, all this reasoning is greatly simplified by using the Binet-Cauchy-formulas for the coordinates of the vectors.
A: A proof using clifford algebra.
First, write the dot product of two cross products in terms of bivectors.
$$(x \times y) \cdot (u \times w) = (-i)(-i) (x \wedge y) \cdot (u \wedge w)$$
Then, break down the inner product of bivectors into a vector-bivector product and a vector-vector inner product.
$$(x \wedge y) \cdot (u \wedge w) = ([x \wedge y] \cdot u) \cdot w$$
nb. These are actually wedge products I'm using, not cross products.
Then, apply the BAC-CAB rule:
$$[x \wedge y] \cdot u = x (y \cdot u) - y (x \cdot u)$$
The result is
$$([x \wedge y] \cdot u) \cdot w = (x \cdot w)(y \cdot u) - (y \cdot w)(x \cdot u)$$
which is identical to the determinant you were asked to find (within the minus sign that I dropped at the beginning for clarity).
