Wikipedia introduces the vector product for two vectors $\vec a$ and $\vec b$ as
$$
\vec a \times\vec b=(\| \vec a\| \|\vec b\|\sin\Theta)\vec n
$$
It then mentions that $\vec n$ is the vector normal to the plane made by $\vec a$ and $\vec b$, implying that $\vec a$ and $\vec b$ are 3D vectors. Wikipedia mentions something about a 7D cross product, but I'm not going to pretend I understand that.
My idea, which remains unconfirmed with any source, is that a cross product can be thought of a vector which is orthogonal to all vectors which you are crossing. If, and that's a big IF, this is right over all dimensions, we know that for a set of $n-1$ $n$-dimensional vectors, there exists a vector which is orthogonal to all of them. The magnitude would have something to do with the area/volume/hypervolume/etc. made by the vectors we are crossing.
Am I right to guess that this multidimensional aspect of cross vectors exists or is that last part utter rubbish?
Best Answer
Yes, you are correct. You can generalize the cross product to $n$ dimensions by saying it is an operation which takes in $n-1$ vectors and produces a vector that is perpendicular to each one. This can be easily defined using the exterior algebra and Hodge star operator http://en.wikipedia.org/wiki/Hodge_dual: the cross product of $v_1,\ldots,v_{n-1}$ is then just $*(v_1 \wedge v_2 \cdots \wedge v_{n-1}$).
Then the magnitude of the cross product of n-1 vectors is the volume of the higher-dimensional parallelogram that they determine. Specifying the magnitude and being orthogonal to each of the vectors narrows the possiblity to two choices-- an orientation picks out one of these.