One nice thing is that these generalizations keep the notion that the absolute value of the cross product measures a kind of volume. The one in R^2 has an absolute value that is the lenght of the segment defined by the original vector and the one in R^4 has an absolute value that is the volume of the parallelepiped defined by the 3 vectors, just like the one in R^3 measures the area of the parallelogram the 2 vectors define.
Interesting generalization. I usually think of the cross product as representing the area of the parallelogram between the two vectors, as well as the plane that the parallelogram lives in. This leads to a very different kind of generalization commonly called the wedge product. Your generalization reminds me of the Hodge star operator, if you're familiar with it.
This operation appears to take the outer product of n-1 nD vectors and then finds the dual. It is something I considered before learning of the outer product, but the outer product seems more general in this case. Edit: ×(x⃗) = x⃗*, ×(x⃗, y⃗) = (x⃗ ∧ y⃗)*, ×(x⃗, y⃗, z⃗) = (x⃗ ∧ y⃗ ∧ z⃗)*. More generally: ×(v⃗ᵢ, ...) = (⋀ v⃗ᵢ)*
There are 3 different definitions for the cross product. The most restrictive definition only exists in 3 and 7 dimensions. The other 2 depends on if you want to stick to a binary operation but an unequal output dimension or allow an n-ary operation and stick to the same dimension. 3 dimensions is the only one where all 3 produce the same result.
i'm supposing if you take a 4x4 matrix with two columns, one using (e) elements, one using unknowns, and the other two using variables, it should be possible to get a function that represents the plane perpendicular to the two vectors in 4 dimensions, similarly a column of (e) and a column of (unknown) given a vector in 3d should return a polynomial representing the plane perpendicular to it, and finally in 2d the determinant of that simple matrix ends up being (y, -x) which if you vary the values x and y can give you the whole plane
now i get it now i get why we use the determinant to calculate the cross product and now i get it why the vector perpendicular to a vector or a line is (-b,a) goshdayyum teachers dont teach us this kind of interesting stuff it seems like they care about finishing the curriculum fast and thats it thank you man💪
I wonder what was the logic behind the claim that only 3 dimensions and 7 dimensions have analogs to the cross product. I suppose that was specifically for binary operations, and I thought it had something to do with quaternions and octonions, but your presentation reminds me that there's no such thing as a single line of all vectors orthogonal to just two vectors in 7D space.
No, I just created the products myself using some constraints, which is how pure math works. This channel is for sharing the reflexions I had during my years as a math student.
One nice thing is that these generalizations keep the notion that the absolute value of the cross product measures a kind of volume.
The one in R^2 has an absolute value that is the lenght of the segment defined by the original vector and the one in R^4 has an absolute value that is the volume of the parallelepiped defined by the 3 vectors, just like the one in R^3 measures the area of the parallelogram the 2 vectors define.
Interesting generalization.
I usually think of the cross product as representing the area of the parallelogram between the two vectors, as well as the plane that the parallelogram lives in. This leads to a very different kind of generalization commonly called the wedge product. Your generalization reminds me of the Hodge star operator, if you're familiar with it.
This operation appears to take the outer product of n-1 nD vectors and then finds the dual. It is something I considered before learning of the outer product, but the outer product seems more general in this case.
Edit: ×(x⃗) = x⃗*, ×(x⃗, y⃗) = (x⃗ ∧ y⃗)*, ×(x⃗, y⃗, z⃗) = (x⃗ ∧ y⃗ ∧ z⃗)*. More generally: ×(v⃗ᵢ, ...) = (⋀ v⃗ᵢ)*
I read that you could only do the cross product for 3 and 7 dimensions
There are 3 different definitions for the cross product. The most restrictive definition only exists in 3 and 7 dimensions. The other 2 depends on if you want to stick to a binary operation but an unequal output dimension or allow an n-ary operation and stick to the same dimension. 3 dimensions is the only one where all 3 produce the same result.
i'm supposing if you take a 4x4 matrix with two columns, one using (e) elements, one using unknowns, and the other two using variables, it should be possible to get a function that represents the plane perpendicular to the two vectors in 4 dimensions, similarly a column of (e) and a column of (unknown) given a vector in 3d should return a polynomial representing the plane perpendicular to it, and finally in 2d the determinant of that simple matrix ends up being (y, -x) which if you vary the values x and y can give you the whole plane
Really cool, keep up the good work
Interesting, although a bit cumbersome imo. Still prefer exterior/wedge product as an generalization, it encapsulates more of the geometry :p
Thanks for this explanation
now i get it
now i get why we use the determinant to calculate the cross product
and now i get it why the vector perpendicular to a vector or a line is (-b,a)
goshdayyum teachers dont teach us this kind of interesting stuff it seems like they care about finishing the curriculum fast and thats it
thank you man💪
Really interesting!!!
cool videos man!!!!
I wonder what was the logic behind the claim that only 3 dimensions and 7 dimensions have analogs to the cross product. I suppose that was specifically for binary operations, and I thought it had something to do with quaternions and octonions, but your presentation reminds me that there's no such thing as a single line of all vectors orthogonal to just two vectors in 7D space.
can provide references or authors used this
No, I just created the products myself using some constraints, which is how pure math works. This channel is for sharing the reflexions I had during my years as a math student.
@@matekon2 it is simular to simplex ( hyper triangles tetrahedrons) and their volume computation
@@tomoki-v6o Ok thanks you. Making connections between ideas is always fun.
👍
Just use geometric algebra
Wow, so geometric algebra and algebraic geometry are two different things.
@@Mr.Nichan yes, GA =-AG
geometric algebra has a much simpler generalization
I was the 69th like
v × w = (v ∧ w)i