If we have a cube Q in $\mathbb{R}^3$ with standard orientation, I'm asked to expres the orientations of the six faces of the cube as differential 1-forms evaluted at the corss product of $v_{1}$ and $v_{2}$. This is how far I got:
for $\omega_{front}(v_{1},v_{2})$, since its unit vector points outside (parallel to positive x-axis), I made the following matrix: $\omega_{front}(v_{1},v_{2})= \begin{bmatrix} 1 & v_{11} & v_{21} \\ 0 & v_{12} & v_{22} \\ 0 & v_{13} & v_{23} \end{bmatrix} =v_{12}v_{23}-v_{22}v_{13}.$
Similarly I found $\omega_{right}(v_{1},v_{2})=v_{11}v_{23}+v_{21}v_{13}$ and $\omega_{top}(v_{1},v_{2})=-v_{11}v_{22}-v_{21}v_{12}.$
Now my question is: how to convert them in differential 1-forms evaluated at the cross product? Something in the form of $\text{dy}(\mathbf{{v_{1}}} \text{ x } \mathbf{{v_{2}}})$ or something similar?
I would appreciate the help!