The exterior product of two vectors is:
\(u\land v\)
This is anticommutative (alternating).
\(u\land v=-v\land u\)
This implies that:
\(u\land u=0\)
\((a+b)\land (c+d)=(a\land c)+(a\land d)+(b\land c)+(b\land d)\)
We can look at the exterior product in component-basis terms.
Consider \(2\)-dimenional vector space with the following vectors:
\(u=ae_1+be_2\)
\(v=ce_1+de_2\)
The exterior product is:
\(u\land v=(ae_1+be_2)\land (ce_1+de_2)\)
\(u\land v=(ae_1\land ce_1)+(ae_1\land de_2)+(be_2\land ce_1)+(be_2\land de_2)\)
\(u\land v=ac(e_1\land e_1)+ad(e_1\land e_2)+bc(e_2\land e_1)+bd(e_2\land e_2)\)
\(u\land v=ad(e_1\land e_2)-bc(e_1\land e_2)\)
\(u\land v=(ad-bc)(e_1\land e_2)\)
The exterior algebra is the algebra generated by the wedge product.
The term \(u\land v\) can be interpreted as the area covered by the parallelogram generated by \(u\) and \(v\).
As \(a\mathbf u\land b\mathbf v=ab \mathbf u\land \mathbf v\), we can see that scaling the length of one of the vectors by a scalar, we also increase the exterior product by the same scalar.
We can describe the exterior product of two vectors as \(\mathbf u\land \mathbf v\) or \(\mathbf v \land \mathbf u\).