# Thread: Vector Product Fromula for two >3D Vectors Found

1. I found the formula coupled with a method to define the cross product of two vectors in > 3D.

It satisfies:

Bilinearity, orthogonality, uxu = 0, anti-commutivity and a generalised length and Jacobi condition.

I need someone's opinion - email me or post a message.

2.

3. Well what is it?

4. Will post the main result shortly.

5. above 3 dimensions? im looking forward to seeing this!

6. nothing too special about a a generalised cross product above three dimensions rich - in fact there are a host of such creatures and that is where the problem lies, the operation is no longer natural in any meaningful way.

I'm curious if Talanum1's construction satisfies the reverse implication for parallel vectors though i.e. if u x v = 0 then u is parallel to v.

7. Hey, well im new to this so!

I've never heard of a 4D cross product for example...

I mean, what IS a 4D vector, we cant visualize it right? So how could we safely assume that a cross product is valid?

How could a 4D vector be perpendicular to another? Unless you mean lineraly independent?

8. The generalisation of prependicularity in >3D is orthogonality. We cannot visualise it but we can test it for consistency. My definition reduces to the 3D case when the two vectors are in 3D and the unwanted fourth component is discarded.

I can show that this definition in 4D at least is the only generalisation of the 3D one based on even and odd permutations. The determinant is defined in terms of even and odd permutations.

I'll see if that is satisfied river_rat.

The short version follows:

Define operators:

uO_i evaluates to: 1 if i odd, 0 if i even

uE_i evaluates to: 1 if i even, 0 if i odd

uIO_nm as assigning the numbers obtained by method 1.1 in the same

order then evaluation to +1

and in the opposite order then evaluation to -1

For: [k] being the k'th unit vector, the formula is:

uxv = [k]{ uO_k uIO_nm u_nv_m + uE_k uIO_mn u_nv_m }

Where we sum over k (the numbr of dimensions) and over n and m as determined from the method.

Note the reverse mn in the second uIO. The order of u_n,v_m is important: the first vector on the left must

get the first index: in this case u gets the index n.

Method 1.1: vector component numbers:

1) for the numbers in dD of n,m of [1], start with the string:

2342256227822...

and if d is odd, end at d; if d is even write the string untill d and add a 2.

2) complete the triangle by following the order in the following triangle:

in 7D:

2 3

4 2
3 4

2 5
5 3
4 5

6 2
3 6
6 4
5 6

2 7
7 3
4 7
7 5
6 7

The triangle didn't come out right, but the columns can be placed next to each other to form a triangle.

Now the numbers for [1] are 23, 42, 34, 25, 53, 45 etc.

To get the numbers for [2]: replace 2 with 1 in the triangle of [1]
" [3]: " 3 with 2 " [2]

Continue untill the numbers for all of [k] is determined.

10. ok i have no idea what that means

so is this really valid?

11. question - why is the article for sale?

12. yeah thats what i wondered?

13. :?

It is for sale because I don't work at a university.

I can show that uxv = 0 implies u parallel to v. Will post this shortly.

14. :?

[1] is the 1'st unit vector.

[1] uO_1 uIO_nm v_nu_m = [1] uIO_nm v_nu_m (1)

since uE_1 = 0

[2] uE_2 uIO_nm v_nu_m = [2] uIO_nm v_nu_m (2)

since uO_2 = 0

Now the number in [] tells you what set of numbers to sum v_nu_m over.

[1]' s numbers are determined by the method. In 4D this is

23, 42, 34.

Expand (1):

[1] uIO_nm v_nu_m + [1] uIO_nm v_nu_m + [1] uIO_nm v_nu_m

+ [1] uIO_nm v_nu_m + [1] uIO_nm v_nu_m + [1] uIO_nm v_nu_m

Now plug in the numbers in order, one pair into each n,m and change uIO_mn into +1 if assigning them in the same order and into -1 if in the opposite order.

Do the same expansion for (2) and plug in the numbers:

31, 14, 43

in the same fasion. The numbers for even unit vectors are copied from the method in reverse order:

13 > 31, 41 > 14, 34 > 43

before plugging them in.

If you have the method for one odd and one even unit vector and you have the method for getting the numbers then you have the method for any unit vector.

I proved the properties as in previous messages in my article.

I can post a vew of the proprty proofs. Please indicate wich one is the most important.

15. 8)

The equivalent formulation (easier to use is):

uxv = [k]{ uO_k uIO_nm u_nv_m - uE_k uIO_nm u_nv_m }

and now you don't have to reverse the numbers of the even unit vectors before inserting them.

river_rat: The proof is easy: Form theorem quoted:

if vxu = 0 then u and v are liniarly dependent. Therefore v can be written as au, a>0, a real.

Now au is parrallel to u and any vector v parallel to au is just au with the n-tuple like [b,b,b,b..b] added to both the start and endpoint of au.

We have uxu = 0, auxu = 0 therefore au is parrallel to u,

now replace au with v in the last sentence.

16. The same formula can actually be stated in terms of a determinant (of sorts) that reduces to a sum of 3 by 3 determinants.

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