anonymous
  • anonymous
Is the vector cross product only defined for 3D? Wikipedia formally introduces it as, for vectors \(\vec a\) and \(\vec b\),\[\vec a\times \vec b=(||\vec a|| ||\vec b|| sin \Theta )\vec n\]Where \(\vec n\) is the unit normal vector. It also mentions that n is vector normal to the plane made by a and b, implying that a and b are 3D vectors. Wikipedia mentions something about a 7D cross product, but I'm not going to pretend I understand that. However, working off a definition I more or less invented (so I cannot tell how accurate it is), I can theorize a "cross product" that spans over any number
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
(cont.) of dimensions: Let the cross product operator be C, such that when C is applied to a set of vectors S, we get the result v, which is orthogonal (as defined by cross product) to each vector in S. Assuming I have the right idea, \(C(S) = \vec v\) implies that actually you need S to be of length \(k-1\) if it contains vectors of \(k\) dimensions, and that \(\vec v\) contains \(k\) dimensions. Am I onto a something or is this completely useless, and cross product was only ever meant for 3D (and apparently 7D)?
anonymous
  • anonymous
@experimentX @mukushla sorry but you guys are the only ones who answer my questions anymore
anonymous
  • anonymous
oh man im poor in this topic ...sorry.. :(

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anonymous
  • anonymous
It's ok. I'll go to SE if no one here helps...
anonymous
  • anonymous
I will just mind map down what I carry around in my head conceding the vector product. If you have a cross product of \(n\) vectors, then its cross product is defined in \(n+1\) space. I believe we agree on that. You can reverse that statement and obtain the same as you have mentioned above. To obtain a cross product in three dimensions you need two vectors. In 4 Dimensions you need 3 Vectors.
anonymous
  • anonymous
Agreed. I'll ask in SE and see what they say.
anonymous
  • anonymous
in higher dimensions it's called wedge product I believe, no longer cross product.
anonymous
  • anonymous
Well the wedge product is present in lower dimensions, too. I think it's something that's different but connected. I recall some seemingly simple formula from Wikipedia that obviously has layers of depth I can't fathom. It went like this: For a set of vectors V, \[C(V)=\frac{\Lambda(V)}{E(V)}\]Where the lambda is the wedge product and the E is the elementary algebra of the system, whatever that means.
anonymous
  • anonymous
Also, before I saw that formula, I thought wedge products can only be taken on differentially sized vectors, but I guess the fact that their magnitude is infinitesimal doesn't mean anything...
anonymous
  • anonymous
Hey I'd love to continue chatting but I g2g. Here's the link to the SE page if they say anything interesting: http://math.stackexchange.com/questions/185991/is-the-vector-cross-product-only-defined-for-3d
experimentX
  • experimentX
sorry ... i don't find anything beyond 3d quite intuitive.
anonymous
  • anonymous
@experimentX Well I did some playing around on MMA and it turns out that, at least the way I did it with the matrices, the magnitude of a 4D cross product gives the volume drawn out by the first three 4D vectors if their 4th component is zero (like the 3D cross product gives the area of the parallelogram of two 3D vectors with a third zero component)
experimentX
  • experimentX
well that is one way to picture it out ... except that you need hyperspace to visualize :((
anonymous
  • anonymous
Well yeah i think of hyperspace like an "internal shadow" where a higher value is a longer shadow, if that helps. Or you can do mass.
experimentX
  • experimentX
yeah i watched a series of video on it. it was very nice ... but can't find it anywhere online.

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