It's possible to extend the well known cross product between two vectors in $\mathbb{R}^3$ to $n-1$ vectors in $\mathbb{R}^n$.

Let $\vec{v_1}, \vec{v_2}, \dots, \vec{v}_{n-1} \in \mathbb{R}^n$ and $\vec{e}_1 = (1, 0, 0, \dots, 0)^T, \vec{e}_2 = (0, 1, 0, \dots, 0)^T, \dots, \vec{e}_n = (0, 0, \dots, 0, 1)^T$ be the unit vectors of the standard basis in $\mathbb{R}^n$. Then we can define the a "cross product" in $\mathbb{R}^n$ by the following determinant:

$\vec{v}_1 \times \vec{v}_2 \times \dots \times \vec{v}_{n-1} = det\
\begin{pmatrix} 
v_{1,1} & v_{2,1} & \cdots & v_{n-1,1} & \vec{e}_1\cr
v_{1,2} & v_{2,2} & \cdots & v_{n-1,2} & \vec{e}_2\cr
\vdots & \vdots & \ddots & \vdots & \vdots\cr
v_{1,n} & v_{2,n} & \cdots & v_{n-1,n} & \vec{e}_n
\end{pmatrix}$

In theory it's easy to compute the determinant by cofactor expansion along the last row, but i'm wondering how one would do this in practice. Computing the $n$ minors on their own seems like a lot of overhead to me and i guess it's not very stable.

The only approach i've found on this topic is in the code of [qhull][1]. But unfortunately it's quite obfuscated and there aren't any helpful comments. I figured out that there the problem is adressed by solving the following system of equations:

$A * \vec{x} = \vec{b}$ <br/> with $A = \begin{pmatrix} v_{1,1} & v_{1,2} & \dots & v_{1,n}\cr v_{2,1} & v_{2,2} & \dots & v_{2,n}\cr\vdots & \vdots & \ddots & \vdots\cr v_{n-1,1} & v_{n-1,2} & \dots & v_{n-1,n}\cr 0 &0& \cdots & 1 \\end{pmatrix}$, $\vec{b} = \begin{pmatrix}0\cr0\cr\vdots\cr sgn(det(A)) \end{pmatrix}$ and $sgn(x) := \begin{cases} +1 & x \geq 0 \cr -1 & x < 0 \end{cases}$

It's clear that the first $n-1$ rows are enforcing $\vec{x}$ to be orthogonal to $\vec{v}_1, \dots, \vec{v}_{n-1}$ but i can't get the origin or meaning of the last row. Furthermore i guess this approach is problematic if $det(A) = 0$ (e.g. if vectors are parallel to a standard unit vector).

So my questions are:

 1. How is the last row deduced and what is its meaning?
 2. Are there other approaches that are faster and/or more stable?

  [1]: http://qhull.org/