The direction of a two-dimensional plane in high dimensional space The direction of a two-dimensional(2D) plane in three-dimensional space(Euclidean space) is defined by the normal vector to this plane. And any (N-1)-dimensional hypersurface in N-dimensional can use the similar definition.  
However, what is the best definition for the direction of a 2D plane in a high dimensional space? For example, a 2D plane in the four-dimensional space. The "normal vector" for this plane is not a vector, but also a "plane". The reason that I want to ask this question is I want to calculate the "angle" between two 2D planes in high dimensional space. Is there any definition for this? For example, one plane is constructed by v1 and v2, and another plane is constructed by v3 and v4, what is the "angle" between these two planes in N-dimensional?
 A: Exterior algebra fulfills this need. If two planes passing through the origin are generated by vectors $\boldsymbol{v}_1,\boldsymbol{v}_2$ and $\boldsymbol{v}_3,\boldsymbol{v}_4$, then $\boldsymbol{v}_1\wedge\boldsymbol{v}_2$  and $\boldsymbol{v}_3\wedge\boldsymbol{v}_4$  are blades representing the planes that they generate,
$$\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle=\begin{vmatrix}
\langle\boldsymbol{v}_1,\boldsymbol{v}_3\rangle &\langle\boldsymbol{v}_1,\boldsymbol{v}_4\rangle\\
\langle\boldsymbol{v}_2,\boldsymbol{v}_3\rangle &\langle\boldsymbol{v}_2,\boldsymbol{v}_4\rangle
\end{vmatrix}$$
is an inner product,
and 
$$\cos\theta=\frac{\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle}{\sqrt{\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_1\wedge \boldsymbol{v}_2\rangle
\langle \boldsymbol{v}_3\wedge \boldsymbol{v}_4 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle}}\text{.}$$
gives the angle between the planes.
