The cross product can be generalized to arbitrary dimensions as done below or here. I'm trying to state and prove the general analogue (for arbitrary dimensions) of the equation
$$|a\times b| = |a||b||\sin \theta |$$
where $\theta$ is the angle between $a$ and $b$.
Definition: let $V$ be an inner product space of dimension $n$. The cross product of the vectors $v_1, \ldots ,v_{n-1}$ of $V$ is defined as the unique vector $v$ such that
$$\det (u \ v_1 \ldots v_{n-1}) = u\cdot v$$ for every $u\in V$. It is denoted as $[v_1 \ldots v_{n-1}]$.
Let $v_0, v_1, \ldots , v_{n-1}$ be vectors of an inner product space $V$ (of dimension $n+1$). It is easy to show that the cross product $[v_0, v_1, \ldots ,v_{n-1}]$ is perpendicular to each $v_i$, and has a magnitude equal to the $n$th dimensional "area" of the parallelotope with edges $v_0, v_1, \ldots , v_{n-1}$.
Let $A$ be the $(n-1)$ dimensional area of the parallelotope with edges $v_1, \ldots , v_{n-1}$. That is, let
$$A:=\sqrt{| \det G |}$$
where $G$ is the Gram matrix of the vectors $v_1, \ldots, v_{n-1}$ (i.e. $G_{i,j}=v_i\cdot v_j$). Then the $n$-dimensional area of the parallelotope with edges $v_0, v_1, \ldots , v_{n-1}$ (namely $|[v_0, v_1, \ldots ,v_{n-1}]|$) should be
$$A|v_0||\sin \theta |$$
where $\theta$ is the angle between $v_0$ and its projection $v_0'$ onto the space $\langle v_1, \ldots , v_{n-1} \rangle$, namely
$$v_0':=\sum_{i=1}^{n-1}\frac{v_0\cdot v_i}{|v_i|^2}v_i.$$
Is it true that
$$|[v_0 \ v_1 \ldots v_{n-1}]| = A|v_0||\sin \theta| \ ?$$
If true, how could one prove it? If it isn't true, what am I getting wrong in my attempt to generalize the result stated in the introduction?
My attempt at proving it: the expression $\sin \theta$ can be expanded by using the following definition of angle:
$$\theta := \arccos \left( \frac{v_0\cdot v_0'}{|v_0||v_0'|}\right)$$
together with the fact that $\sin (\arccos (x))=\sqrt{1-x^2}$. Now $A$ can be expanded in multiple ways (i.e. using Leibniz's definition of the determinant or the Laplace expansion). Finally, the $LHS$ can be expanded by noting that
$$[v_0 \ v_1 \ldots v_{n-1}] = \sum_{i=1}^{n+1}(-1)^{i+1}M_{1,i}e_i$$
where $\{ e_1, \ldots , e_{n+1}\}$ is a basis of $V$ and $M_{1,i}$ is the determinant of the submatrix obtained by removing the $i$th row and the $j$th column of the matrix
$$[1 \ v_0 \ v_1 \ldots v_{n-1}] .$$
I've been playing around with the resulting expression trying to force an equality, with no avail.
I hope the following is correct. I have not checked the details. We are working in an $n$-dimensional inner product space.
Definition 1: Cross Product
The cross product, $[v_1,\ldots,v_{n-1}]$, of $n-1$ vectors is the unique vector $v$ such that $\det(u,v_1,\ldots,v_{n-1}) = u\cdot v$.
Definition 2: Volume
The volume of the parallelotope spanned $m$ vectors $v_1,\ldots,v_m$ with $m\leq n$ is $A(v_1,\ldots,v_m) = \sqrt{|\det G|}$ with $G_{ij} = v_i\cdot v_j$.
Lemma 1: If $v_{m+1}$ is orthogonal to the subspace spanned by $v_{1},\ldots,v_m$ then
$$A(v_1,\ldots,v_{m+1}) = A(v_1,\ldots,v_m)|v_{m+1}|$$
Lemma 2: Given $n$ vectors $v_1,\ldots,v_n$ we have
$$A(v_1,\ldots,v_n) = |\det(v_1,\ldots,v_n)|$$
Definition 3: Angles Given vectors $v_1,\ldots,v_m$ we say that the angle $0\leq \theta \leq \pi/2$ is of $v_m$ with the space spanned by $v_1,\ldots,v_{m-1}$ is determined by
$$\cos(\theta) = \frac{v_{m}\cdot v^\top_{m}}{|v_{m}||v_{m}^\top|}$$
where $v_{m}^\top$ is the orthogonal projection of $v_{m}$ onto the subspace spanned by $v_1,\ldots,v_{m-1}$. We then put,
$$\sin(\theta) = \sqrt{1 - \cos(\theta)^2}$$
Lemma 3: Given vectors $v_1,\ldots,v_m$, and $v_m^\top + v_m^\perp = v_m$ the orthogonal decomposition of $v_m$ with respect to the subspace spanned by $v_1,\ldots,v_{m-1}$, we have
$$|v_{m}^\perp| = |v_{m}|\sin(\theta)$$
Theorem: Given $n-1$ vectors $v_1,\ldots,v_{n-1}$ we have
$$ |[v_1,\ldots,v_{n-1}]| = A(v_1,\ldots,v_{n-2})|v_{n-1}|\sin(\theta) $$
The proof then becomes a computation, which I show below.
$$|[v_1,\ldots,v_{n-1}]|^2 = \det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v_{n-1}^\perp + v_{n-1}^\top) = \det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v_{n-1}^\perp)$$
first equality is by definition and the second follows by multilinearity of the determinant (and recalling that $v_{n-1}^\top$ is spanned by the other vectors). By Lemma 2 we have
$$\det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v^\perp) = A(v_1,\ldots,v_{n-2},[v_1,\ldots,v_{n-1}],v^\perp)$$
and so
$$|[v_1,\ldots,v_{n-1}]|^2 = A(v_1,\ldots,v_{n-2},[v_1,\ldots,v_{n-1}],v^\perp) = A(v_1,\ldots,v_{n-2})|v^\perp||[v_1,\ldots,v_{n-1}]|$$
where the last equality follows by two successive applications of Lemma 1. The result follows by cancelling and Lemma 3.