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This question already has an answer here:

We know how to make cross product of three dimensional vectors. $$ \vec A \times \vec B = \vec C$$

Where :
$ \vec A = (A_i; A_j; A_k)$
$ \vec B = (B_i; B_j; B_k)$
$ \vec C = (C_i; C_j; C_k)$

$C_i = \left|\begin{matrix}A_j&A_k\\B_j&B_k\end{matrix}\right|$ $C_j = \left|\begin{matrix}A_k&A_i\\B_k&B_i\end{matrix}\right|$ $C_k = \left|\begin{matrix}A_i&A_j\\B_i&B_j\end{matrix}\right|$

But what about if we have four dimensional vectors?
Is it possible to make cross product of four dimensional vectors?
If it is possible, then tell me when it can be possible?

Let say we have two vectors:
$ \vec A = (A_i; A_j; A_k; A_l)$
$ \vec B = (B_i; B_j; B_k; B_l)$

Then how to compute a cross product of this two vectors? Will it again vector? $$ \vec A \times \vec B = \vec C$$

$ \vec C = (C_i; C_j; C_k; C_l)$
Then how to compute those coordinates?

We know that only square matrices have a determinant property!
In this case it might not be correct if we will wright...
$\color{red} {\text { $C_i = \left|\begin{matrix}A_j&A_k&A_l\\B_j&B_k&B_l\end{matrix}\right|$} C_j = \left|\begin{matrix}A_k&A_i&A_l\\B_k&B_i&B_l\end{matrix}\right| C_k = \left|\begin{matrix}A_i&A_j&A_l\\B_i&B_j&B_l\end{matrix}\right| C_l = \left|\begin{matrix}A_i&A_j&A_k\\B_i&B_j&B_k\end{matrix}\right|}$

So tell me how to solve this problem?

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marked as duplicate by Rahul, Yiyuan Lee, Guy, user63181, Semsem Mar 21 '14 at 9:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Here is an article that precisely formulates and demonstrates the non-existence of a cross product in Euclidean spaces of dimensions other than $3$ and $7$: http://www.jstor.org/stable/2323537. If you can't access the article through the link, the article reference is

Cross Products of Vectors in Higher Dimensional Euclidean Spaces

W. S. Massey

The American Mathematical Monthly

Vol. 90, No. 10 (Dec., 1983), pp. 697-701.

Since it's in the Monthly, it should be pretty readable to a well-prepared undergraduate.

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  • 1
    $\begingroup$ Some sources, e.g., Brown and Gray, define cross products to be multilinear maps that satisfy certain antisymmetry and orthogonality conditions; in particular, they need not take exactly two arguments (restricting to this, the bilinear case, leads to the restriction that the dimension of the vector space be $3$ or $7$). $\endgroup$ – Travis Willse Jul 5 '15 at 6:39
  • $\begingroup$ This answer is totally incorrect. The referenced article does prove the existence of the cross product in higher dimensions (it's just not a 2-ary operation). From page 700: "Analogously, in $R^n$ we can define a cross product $v_1 \wedge v2 \wedge \ldots \wedge v_{n-1}$ of any ordered $(n - 1)$ tuple of vectors by a similar process..." $\endgroup$ – plasmacel Mar 3 '17 at 11:50
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While it is feasible to compute a cross-product in four dimensions, the cross-product only has the orthogonality property in three and seven dimensional spaces. You should consider instead looking at Gram-Schmidt Orthogonalization to find orthonormal vectors.

http://www.math.hmc.edu/calculus/tutorials/gramschmidt/

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  • $\begingroup$ It is true, 2 vectors can only yield a unique cross product in 3 dimensions. However, you can yield a cross product between 3 vectors in 4 dimensions. You see, in 2 dimensions, you only need one vector to yield a cross product (which is in this case referred to as the perpendicular operator.). It’s often represented by $ a^⊥ $. Additionally, if you perform a dot product between $a^⊥$ and another vector, say, b, you yield something called a perpendicular dot product, represented as a⊥b. It’s anticommutative and is equal to the area of a parallelogram formed by a and b. (Continued) $\endgroup$ – Math Machine Dec 23 '18 at 18:03
  • $\begingroup$ Now, in 3 dimensions, you need two vectors to yield a unique cross product. Incedentally, the components of this product can be computed using said ⊥ products. $$(axb)•î=<a_y,a_z>⊥<b_y,b_z>$$ $$(axb)•j=<a_z,a_x>⊥<b_z,b_x>$$ $$(axb)•k=<a_x,a_y>⊥<b_x,b_y>$$. Said cross product will also have a meagnitude equal to the area of a parallelogram formed by a and b. Now, on another note, there’s something called a triple scalar product. The triple scalar between a,b, and c is equal to a•(bxc). This yields the volume of a parallelopiped formed by a, b, and c (continued). $\endgroup$ – Math Machine Dec 23 '18 at 18:11
  • $\begingroup$ Now, in four dimensions, if I wanted to find the 4d cross product between a, b, and c, that is a vector which is perpendicular to all three vectors, I would do essentially the same thing I did in the last comment. Assuming w is the fourth component, $$(axbxc)•î=<a_y,a_z,a_w>•(<b_y,b_z,b_w>x<c_y,c_z,c_w>)$$ $$ (axbxc)•j=<a_z,a_w,a_x>•(<b_z,b_w,b_x>x<c_z,c_w,c_x>)$$ etc. And, you guessed it, if I do a dot product between this and another vector, say d, I’ll get the hypervolume (known as the bulk) of a 4-d parallelogram formed by a, b,c,&d. I think you get the pattern by now $\endgroup$ – Math Machine Dec 23 '18 at 18:18

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