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[Phys-L] dividing by vectors



Rauber, Joel wrote:

I don't mean to imply that, if it were ordinary mutliplication,
"canceling" implies dividing both sides by C, that is the inverse of
multiplication, hitting the l.h.s of

C*A=C*B with C^(-1)

That's all I meant.

Of course it isn't ordinary multiplication . . . (hence the discussion)

In Clifford algebra, you _can_ divide by any nonzero vector.

Let (A B) represent the geometric product of vector (A) with
vector (B). As usual, the geometric product is represented
simply by juxtaposing the multiplicands, without any special
operator symbol such as · or * or ×.

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 the special case where (B) equals (C), we have:
For the full blown geometric product, (A C / C == A) for
all vectors (A) and (C) (excluding C=0 of course).

For bonus points, we can divide scalars by vectors.
For the dot product, (A·C / C == A_c), where (A_c) is the
projection of (A) along the (C) direction.

It is amusing and easy (5 lines or less) to verify this, starting
from the definition of dot product in terms of geometric product.
Actually, you can almost do it by inspection, since you
know the dot product with (C) is insensitive to any
component perpendicular to (C), i.e. (C·A == C·A_c) always.

Being able to divide by vectors is sometimes very convenient.
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