Till very recently I believed to understand what is written in the Wikipedia entry on the exterior algebra: The bivector $u\wedge v$ represents the plane spanned by the vectors, "weighted" with a number, given by the area of the oriented parallelogram with sides $u$ and $v$.
But now I have doubts: In $\mathbb R^3$ the natural two-dimensional Hausdorff measure of the (non-oriented) parallelogramm $p(u,v)$ with sides $u$ and $v$ is not bilinear: If $u,v,w$ are the edges of a cube (say, the standard basis of $\mathbb R^3$) then $p(u,v)$ and $p(u,w)$ are squares (sides of the cube) of area ($2$-dimensional Hausdorff measure) $1$ but $p(u,v+w)$ is a rectangle inside the cube of area $\sqrt{2}$. This does not seem to be issue of orientation. (In $\mathbb R^2$ the geometric interpretation of bivectors seems plausible because of shearing.) What do I missunderstand?
I believe now that the point is that (at least some) alternating bilinear forms do not measure the area of an oriented parallelogram but the AREA OF A SHADOW of the parallelogram:
If $(b_1,\ldots,b_n)$ is a basis of the vector space we have the dual basis $(b_1^*,\ldots,b_n^*)$ in $X^*$ (so that $b_j^*(b_k)=\delta_{j,k}$). Then $b_j^* \wedge b_k^*$ should measure an oriented area of the projection (or the shadow) of an oriented parallelogramm into the plane spanned by $b_j,b_k$ (of course, the projection does not only depend on $b_j$ and $b_k$ but on the whole basis -- in the presence of a scalar product the orthogonal projection is a kind of canonical choice).
If one defines an oriented area in a plane by homogeneity in each component and the shearing property (i.e., the areas of the parallelogramms spanned by $(x,y)$, $(x+y,y)$ and $(x,y+x)$ are the same) one can prove alternating bilinearity and uniqueness up to constant factors. Because the projection is linear this implies the bilinearity of $b_j^*\wedge b_k^*$.
Hence $b_j^*\wedge b_k^*$ is determined by alternating bilinearity, by $b_j^*\wedge b_k^*(x,y)=b_j^*\wedge b_k^*(\pi(x),\pi(y))$ where $\pi$ is the projection, and the norming condition $b_j^*\wedge b_k^*(b_j,b_k)=1$.