Use of the mixed product for the Wikipedia description of Barycentric Co-ordinates In reading Wikipedia entry for Barycentric co-ordinates, the authors states without proof that for a vector $h \in \mathbb{R}^3$ and a basis $\{e,f,g \}$ that 
$h = \frac{1}{(e,f,g)} \cdot [(h,f,g)e + (e,h,g)f+(e,f,h)g]$ where $(e,f,g) = (e \times f) \cdot g$
If $h = a_1 e_1 + a_2 e_2 + a_3 e_3$, then this amounts to saying that
$h = \frac{1}{(e_1 \times e_2) \cdot e_3} \cdot [((h \times e_2)\cdot e_3)e_1 + ((e_1 \times h) \cdot e_3)e_2+((e_1 \times e_2) \cdot h) e_3]$
The above formula I have manually verified, but I'm wondering why this is true for any basis?

This is just an application of Cramer’s rule for solving systems of linear equations.  
Recall that the coordinates of a vector are the coefficients of the unique linear combination of basis vectors that produces the vector. In other words, the coordinates of $h$ are the solution to the equation $h_1e+h_2f+h_3g = h$, which we can write as $$\begin{bmatrix}e&f&g\end{bmatrix}\begin{bmatrix}h_1\\h_2\\h_3\end{bmatrix} = h.$$ Now, the scalar triple product $(a,b,c)=a\times b\cdot c$ is equal to the determinant of $[a\;b\;c]$, so by Cramer’s rule, $$h_1 = {(h,f,g)\over(e,f,g)}, h_2 = {(e,h,g)\over(e,f,g)}, h_3 = {(e,f,h)\over(e,f,g)}.$$
