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# 4D Cross product?

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Dear All, Can someone point me to a place where I can see the equations for a 4D cross product? I can''t seem to find a spot on the web.. Thanks, Jesse

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There''s no such thing as a 4D cross product.

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Acctually I belive you can if you have 3 4D vectors..
Jesse

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In three dimensions, the cross product can be expressed in terms of a determinant. Extending this to four dimensions suggests:

A validility proof and simplified C source code can be found here.

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Yep, you can extend the vector product into any dimension you like, as long as it''s greater than 2. You can do the 2 vector cross product in 3 and 7 dimensional space. For everything else you can multiply N-1, N dimensional vectors together, as a vector product. Sort of. The exact proof is on mathworld, and this subject ( with proof ) was posted about 4 months ago, so do a search.

You have to remember that you''re unique, just like everybody else.

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Actually, you can indeed have a cross product in 2 dimensions! As others pointed out, you need N - 1 vectors to calculate it, so for 3D you need 2 vectors, for 4D you need 3 vectors, and for 2D you need....1 vector! So, turns out, in 2D the cross product of a single vector is simply the normal to that vector. You also need a constraint to make the 2D cross product unique, and you might typically constrain it to be the normal found by rotating counterclockwise by 90 degrees.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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By the way, what do you need the 4D cross product for?

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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I wouldn''t need it for anything, I just noticed that there was a DX3d fucntion that was for a 4D cross product...
Jesse

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quote:
Original post by grhodes_at_work
Actually, you can indeed have a cross product in 2 dimensions! As others pointed out, you need N - 1 vectors to calculate it, so for 3D you need 2 vectors, for 4D you need 3 vectors, and for 2D you need....1 vector! So, turns out, in 2D the cross product of a single vector is simply the normal to that vector. You also need a constraint to make the 2D cross product unique, and you might typically constrain it to be the normal found by rotating counterclockwise by 90 degrees.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Yes of course, I mean't greater than or equal to two. The trick to find the normal to a 2D vector as:

If V is a vector [x, y] then the normal to that can be [y, -x], can be derived from the cross product determinant as:
| i j || x y |

You have to remember that you're unique, just like everybody else.

[edited by - python_regious on January 18, 2004 9:38:54 AM]

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quote:
Original post by laeuchli
I wouldn't need it for anything, I just noticed that there was a DX3d fucntion that was for a 4D cross product...
Jesse

Wao...
Interesting,really interesting.. i'm wonder if it's work right.

About 2D crossproduct: i was somehow sure that 2d cross product is a scalar....maybe i was wrong...

What with 2D rotor(aka Nabla cross Field)?

also i'm pretty sure that 4d rotor are 6d vector.

[edited by - Dmytry on January 18, 2004 11:12:43 AM]

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