[Phys-L] Re: dividing by vectors

[John Denker <jsd@AV8N.COM>]

I 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.


Bob Sciamanda wrote:

> In ordinary vector language, you are not "dividing by a vector".

I disagree.

> You are
> dividing by the magnitude squared of a vector.

That's part of it, but not the interesting part.

>  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?

I have no idea where that is coming from.  Let C be a vector.
Then C/C = 1.  I'm dividing a vector by a vector.  The quotient
is a scalar.  In contrast, if I divide C by any nonzero scalar,
the quotient is a vector.

Please explain again why dividing by a vector is the same as
dividing by a scalar.
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