4D cross product (SoME2)

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  • Опубліковано 7 вер 2024

КОМЕНТАРІ • 5

  • @APaleDot
    @APaleDot 2 роки тому +8

    I want to add that each component in the result of the "hypercross" that you've defined is really just the scalar triple-product of three vectors in a 3D space: t · (u × v)
    In 3D, this triple product represents the oriented volume of the parallelipiped spanned by the three vectors. This is directly analogous to the 2D "cross product" which is really just the formula for the oriented area of the parallelogram spanned by the two vectors in 2D.
    If anyone is interested and wants to learn more, look up the wedge product, or Geometric Algebra generally.

  • @sharppencil7777
    @sharppencil7777 2 роки тому

    very helpful.

  • @angeldude101
    @angeldude101 Рік тому

    If the 3D cross product takes 2 vectors and gives another vector, while the 2D cross product takes 2 vectors and gives a _scalar,_ why does the 4D cross product take _3_ inputs and give a vector? It seems like either the 2D cross product should either take only 1 input vector, or the 4D cross product take only 2 input vectors, and then give something that is to a vector what a vector is to a scalar.
    That said, both possibilities seem useful in different ways. One gives a vector orthogonal to a set of other vectors, while the other gives some sort of relation between two vectors, such as their relative orientation. In the language of geometric algebra, one is just the dual of the outer product of n-1 vectors, while the other is the outer product itself. This video used the latter definition for 2D, but the former definition for 4D. In 3D the two are equivalent.

    • @tucan7112
      @tucan7112  Рік тому

      when using the 4d cross product in 3d it was eqivalent to the 3d cross product being used in 2d and returning a scalar, and when used in 4d it returns a vector just like 3d cross product returns a vector in 3d. Thats atleast how i interpret it.

    • @angeldude101
      @angeldude101 Рік тому

      @@tucan7112 Yes. That case would be equivalent to taking the outer product of N vectors to result in a pseudo-scalar.