Wedge product - from multilinear form to parallelogram

Question

On wikipedia, the motivational example for introducting the wedge product $\wedge$ uses $\textbf{v}= a\textbf{e}_1+b\textbf{e}_2$ and $\textbf{w}=c\textbf{e}_1 + d\textbf{e}_2$ in $\mathbb{R}^2$, where $\lbrace \textbf{e}_1, \textbf{e}_2\rbrace$ is the usual cartesian basis. From here, they derive that $\textbf{v} \wedge \textbf{w} = (ad-bc) \textbf{e}_1 \wedge \textbf{e}_2$, where $ad-bc$ is the area of the parallelogram spanned by $\textbf{v}$ and $\textbf{w}$.

However, the general defintion of the wedge product of multilinear forms is

$v \wedge w (x_1, \dots, x_{k+n} ) := \frac{1}{k!} \frac{1}{n!} \sum_{\pi \in \text{perm} (k+n)} \text{sgn} (\pi) (v \otimes w) (x_{\pi (1)} , \dots x_{\pi (k+n)} ) $,

where $v$ is $k$-form ($v \in \Lambda^k (M)$) and $w$ is an $n$-form ($w \in \Lambda^n (M)$. It follows then that $v \wedge w = v \otimes w - w \otimes v$, if both $v$ and $w$ are 1-forms. How does this relate itself to the motivational example?

Answer 1

You can generalize the area of (oriented) parallelogram to determinant of a matrix - doesn't the $ad-bc$ ring a bell? If $k+n=dim(M)$ you would get something very similar.

Answer 2

"different" wedge products?

It is the same operation, in the first example you apply it to two vectors, in the second -- to two 1-forms. But in the either case if you can write $w$ and $v$ in the basis of the corresponding space as $v= ae_1+be_2$, $w=ce_1 + de_2$ and define $$e_1\wedge e_2=e_1\otimes e_2-e_2\otimes e_1$$ then $$v\wedge w = v \otimes w - w \otimes v=\ldots= (ad-bc)e_1\wedge e_2$$ where I skipped some simple manipulations of opening brackets and cancelling symmetric tensor product.