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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(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 \(k1\) 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
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX @mukushla sorry but you guys are the only ones who answer my questions anymore

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh man im poor in this topic ...sorry.. :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's ok. I'll go to SE if no one here helps...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0Agreed. I'll ask in SE and see what they say.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in higher dimensions it's called wedge product I believe, no longer cross product.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0Also, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Hey 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/isthevectorcrossproductonlydefinedfor3d

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0sorry ... i don't find anything beyond 3d quite intuitive.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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
 3 years ago
Best ResponseYou've already chosen the best response.0well that is one way to picture it out ... except that you need hyperspace to visualize :((

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i watched a series of video on it. it was very nice ... but can't find it anywhere online.
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