JD wrote:
| In Clifford algebra, you _can_ divide by any nonzero vector.
| . . .
| Define (A B / C) to be just (A B C / (C·C)) where (C·C) is a
| scalar, and division by scalars is already well defined.
In ordinary vector language, you are not "dividing by a vector". You are
dividing by the magnitude squared of a vector. You are dividing by a scalar
and changing the language so that you can say you are dividing by a vector.
What does this add besides confusion?