It is often said that the exterior/wedge product of two vectors represents an area, or signed area with undefined shape, and they show a picture of two vectors in 2 dimensions (could be i and j) with an exterior product that is the parallelogram spanned by the vectors, or a circle with the same area as the parallelogram, or so; and a connection is made with the usual cross product, cause both exterior and cross products have the same scalar magnitude or area. So far so good.
Now in 3 dimensions instead of i and j let's take the vectors u=(0.853, -0.146, 0.5) and v=(-0.146, 0.853, 0.5); these vectors are just the i, j rotated, so they are just unit vectors spanning a 90º angle. In regard to the cross product nothing essential has changed, it will be again a vector and its length will be the same area as it was with i and j. But according to Wikipedia (Exterior Algebra, section Cross And Triple Products) the exterior product is now:
u Λ v = (u1·v2-u2·v1)(i Λ j) + (u2·v3-u3·v2)(j Λ k) + (u3·v1-u1·v3)(k Λ i) (giving three non-null coefficients similar to the cross product),
however this doesn't look anymore like a parallelogram/area as in the 2-dimensional case, it looks like three juxtaposed parallelograms/areas or equivalently like a parallelepiped/volume.
So with general vectors in 3 dimensions the exterior product of two vectors doesn't seem to represent an area, as is ALWAYS pictured in books and papers; or perhaps it does and I'm completely wrong and missing important details.
I've done some research but can't find the solution. Could anybody help me?
Best Answer
I posted the question cause I couldn't find a visual explanation of the "area addition" in usual books and web sites. Thanks to the answer of Hans Lundmark and the comment of Jean Marie, I've done more research and been able to understand at last the addition of bivectors/exterior products/wedge products, so I'll try to visually answer my own question.
The bivector/exterior product/wedge product u Λ v of two vectors u and v is a flat signed (oriented) area of undefined shape, an instance of which is the parallelogram P that the vectors determine, see Fig.1. The perpendicular to P with the same magnitude/length as the area of P is the cross product/vector product w.
It happens that P is inclined, so we can't define it with just one basis bivector i Λ j, we need a special addition of three bivectors $a_{ij}$ i Λ j + $a_{ik}$ i Λ k + $a_{jk}$ j Λ k, see Fig.2, where the coefficients or areas $a_{ij}$ , $a_{ik}$ , $a_{jk}$ have the same values as the components z, y, x of the cross product w. Note that this similitude doesn't mean that the vector w could fit exactly into the "box" formed by the three bivectors in Fig.2; i.e, the "bivector box" in Fig.2 and the dotted "vector box" in Fig.1 are different, for example the horizontal area $a_{ij}$ has the same value as the vertical (not any horizontal) coordinate z of w.
In order to visually add the three bivectors/areas, called $a_{ij}$ , $a_{ik}$ , $a_{jk}$ from now on to simplify, we first take $a_{ij}$ and $a_{ik}$ in Fig.3 and get the dark parallelogram $a_{ij}$ + $a_{ik}$ that completes the wedge. Then we should add $a_{jk}$ , but it happens that the parallelograms $a_{ij}$ + $a_{ik}$ and $a_{jk}$ doesn't share a side, so we take advantage of the undefined shape of a bivector and we transform $a_{jk}$ into $a_{jk}$', which has the same area but shares a side with $a_{ij}$ + $a_{ik}$ , see Fig.4. Now we add them completing again the wedge, and get the final parallelogram P which is the same as in Fig.1.
Given u and v, both u Λ v and w = u ⨯ v carry the same information (the bare coefficients) in different ways. Perhaps w (in the form of a sum of three coordinates, the dotted box in Fig.1) is intuitively easier to visualize in space than u Λ v (in the form of a sum of three parallelograms/areas, the box in Fig.2); that's why bivectors seem to be little known. But in some special situations they could perform better than the cross product, as I've read.