4 ms·

Could you explain some more about why?

• ninepoints 3y ago
  The simplest counterexample is just to consider vectors in 2D. It is not possible in general to construct a third vector perpendicular to two other vectors unless those other vectors are collinear or anitcollinear. The notion of an exterior product, represented as a signed area subtended by the two vector arguments of the operator sits in 2 space nicely however. In general, signed areas and M dimensional signed volumes are embeddable in N dimensional subspaces for M < N. The notion of a signed volume is directly tied to both the exterior product and the determinant. In higher dimensions, the cross product is equally unhelpful, given that for two vectors, the set of mutually orthogonal vectors abiding by the right hand rule is often infinite.
  • ninepoints 3y ago
    (should be M <= N but HN isn't letting me edit the comment at this point)