What is the connection between 2-forms, determinants, and cross products in R^3?

[JonnyMaddox]

Hi, 2-forms are defined as

$du^{j} \wedge du^{k}(v,w) = v^{j}w^{k}-v^{k}w^{j} = \begin{vmatrix} du^{j}(v) & du^{j}(w) \\ du^{k}(v) & du^{k}(w) \end{vmatrix}$

But what if I have two concret 1-forms in $R^{3}$ like $(2dx-3dy+dz)\wedge (dx+2dy-dz)$ and then I calculte $(2dx-3dy+dz)\wedge (dx+2dy-dz)=-7dy \wedge dx +3dz \wedge dx - dy \wedge dz= 7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ (now a 2-form)
I know this is the same as the cross product between $(2,-3,1)^{T}$ and $(1,2,-1)^{T}$
What has this to do with the determinant? If I calculate this for 1-forms in $R^{2}$ like $(2dx+4dx)\wedge (3dx+9dy) = -18 dx\wedge dy +12 dx \wedge dy = 6 dx \wedge dy$ which equals the determinant of the vectors $(2,4)$ and $(3,9)$. But this is not like 1-forms in $R^{3}$. And is the geometric interpretation right that $7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ means that 7 is the part of the area projected to the xy axis, 3 the part projecte to the zx axis and 1 the part projected onto the yz axis? Or is it a different coordinate system like dxdy, dzdx and dydz ?
Now actually these are all functions of vectors like $v$ or $w$ as in the definition of 2-forms. What happens if they come into this picture?

 

[Greg Bernhardt]

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?