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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?

Martin Sleziak
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IremadzeArchil19910311
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2 Answers2

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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.

Gyu Eun Lee
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    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 $7$). – Travis Willse Jul 05 '15 at 06:39
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    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..." – plasmacel Mar 03 '17 at 11:50
  • @plasmacel that is the exterior product and not the cross product. – Mr X Apr 04 '21 at 21:56
  • Non-existence in other than 3 and 7 dimensions? You can compute a cross product in 2 dimensions, but the output is a single number rather than a vector. – Aaron Franke Jan 05 '22 at 18:03
  • @AaronFranke, I think that you'd define it so that there's one vector input and one vector output. you can choose either a + or - 90 degree rotation (just how is the 3D cross product it's arbitrarily defined with the left-hand rule, when it also could've worked with the right-hand rule). – Elliott Jan 17 '22 at 10:34
  • @Elliott That wouldn't fulfill the goal of having two input vectors. The definition I was thinking of is that we have two vectors as input and the output is a number representing the signed area of the parallelogram they form. This is the same as if you embedded the two 2D vectors in 3D space and looked at the Z coordinate of the output. https://user-images.githubusercontent.com/1646875/148265358-9065fef3-6c84-40bd-a9fa-f7bf9cb735f8.png – Aaron Franke Jan 18 '22 at 22:07
  • Hi @GyuEunLee please can you help with https://math.stackexchange.com/q/4396103/585488 – linker Mar 05 '22 at 15:39
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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/

ml0105
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    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) – Math Machine Dec 23 '18 at 18:03
  • 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). – Math Machine Dec 23 '18 at 18:11
  • 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 – Math Machine Dec 23 '18 at 18:18
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    Your link is dead – Elaskanator Dec 03 '21 at 00:16
  • You say it is feasible to compute a cross-product in four dimensions, how is this done? – Aaron Franke Apr 27 '22 at 01:02
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    The perpendicular operator, $a^⊥$, is equal to $<-a_y,a_x>$, which is the same as taking the determinant of the matrix whose first row is $a_x,a_y$ and whose second row is i,j. The cross product is the same as taking the determinant of the matrix whose first row is $a_x,a_y,a_z$, second row is $b_x,b_y,b_z$, and whose third row is i,j,k. The 4-D cross product is the same as taking the determinant of the matrix whose first row is $a_x,a_y,a_z,a_w$, second row is $b_x,b_y,b_z,b_w$, third row is $c_x,c_y,c_z,c_w$, and 4th row is i,j,k,l. – Math Machine Jan 15 '23 at 04:34
  • In addition, something I hadn't realized when I first made the comment, was that there's also a way to extend the two-input operation into 4 dimensions. In 2D, we get the perpendicular dot product, a⊥b, a scalar equal to their magnitudes multiplied by the sine of their angle. In 3D, we get the cross product axb, a vector whose magnitude is their magnitudes multiplied by the sine of their angle and which points perpendicular to a and b. In 4 or more dimensions, we get the wedge product, a∧b, a "bivector" whose magnitude is their magnitudes multiplied by the sine of their angle, and (continued) – Math Machine Jan 15 '23 at 04:37
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    which represents the plane parallel to both vectors. Just as scalars represent single points and only have one component, and vectors represent lines and have one component for each dimension, bivectors represent planes and have one component for each pair of dimensions (xy, xz, yz, etc.). Technically, you can take the wedge product in any number of dimensions, it's just easier to use scalars or vectors in 2 or 3 dimensions than to use bivectors. Furthermore, if bivectors are too much for you, you can simply compute the products of the magnitudes times the sine of the angle (continued) – Math Machine Jan 15 '23 at 04:40
  • through the formula ||a∧b|| = √(||a||²||b||²-(a⋅b)²). – Math Machine Jan 15 '23 at 04:42