Projected areas in n-dimensional space I have two planes in an $n$-dimensional space, and a parallelogram in one of the planes, defined by two vectors ($v_1$, $v_2$), with area $S$. I want to find the area of the parallelogram defined by the projections of the vectors onto the other plane ($v'_1$, $v'_2$).
I can obviously just obtain the projections and compute the area, but I'm after a "simpler" result. For example, in 3-dimensional space a single angle $\theta$ can be defined between the two planes, and the projected area is then $S'=S\cos\theta$. But in an $n$-dimensional space there are, in general, two principal angles between the planes, is there a similarly simple relation in this case? Intuitively, I believe that if the principal angles are $\theta_1$ and $\theta_2$, maybe the area is $S'=S\cos\theta_1\cos\theta_2$ (which would still be applicable to 3 dimensions, since then $\theta_2=0$). Am I right? Where can I find a reference for this (or the correct) result?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Denote your planes by $P$ and $P'$, let $B = (\Basis_{1}, \Basis_{2})$ and $B' = (\Basis_{1}', \Basis_{2}')$ be orthonormal (ordered, oriented) bases for the respective planes, and put
$$
a_{ij} = \Basis_{i} \cdot \Basis_{j}',\quad i, j = 1, 2.
$$
(The coefficient $a_{ij}$ is, if it matters, the cosine of the angle between $\Basis_{i}$ and $\Basis_{j}'$.)
Orthogonal projection $\Reals^{n} \to P'$ defines a map $\Pi:P \to P'$ given by
\begin{align*}
\Pi(\Basis_{j})
  &= (\Basis_{j} \cdot \Basis_{1}')\, \Basis_{1}' + (\Basis_{j} \cdot \Basis_{2}')\, \Basis_{2}' \\
  &= a_{j1} \Basis_{1}' + a_{j2} \Basis_{2}'.
\end{align*}
The scale factor for area under $\Pi$ is
$$
\det(a_{ij}) = (\Basis_{1} \cdot \Basis_{1}')(\Basis_{2} \cdot \Basis_{2}')
  - (\Basis_{1} \cdot \Basis_{2}')(\Basis_{2} \cdot \Basis_{1}')
$$
(or its absolute value if you ignore orientation): Indeed, the (unique) linear map $\sigma:P' \to P$ satisfying $\sigma(\Basis_{i}') = \Basis_{i}$ obviously preserves area, and the composition $\Pi\sigma:P' \to P'$ has matrix $(a_{ij})$ with respect to $B'$.
