question on unit bivector in 3d space

[American Citizen <website.reader3@gmail.com>]

To all:

After trying to download broken software for Clifford algebra, and 
finding out that the discussion group for bivector.net is disabled, and 
finding out that almost nothing exists for linux except very old CLICAL 
software running on MSDOS, the past 2 or 3 days, I am turning to this 
group for help.

The question is "what is a unit 3d bivector?"

I use @ to denote geometric product of two vectors a,b in 3d space (but 
we must keep track of the basis vectors (e1),(e2) and (e3) too.

 a @ b = [ r(e1) + s(e2) + t(e3) ] * [ x(e1) + y(e2) + z(e3) ]

multiplying distributively:

       =  rx(e1e1) + ry(e1e2) + rz(e1e3) + sx(e2e1) + sy(e2e2) + 
sz(e2e3) + tx(e3e1) + ty(e3e2) + tz(e3e3)

We have in Clifford Algebra that e1e1 = 1 (scalar) and similarly, e2e2 = 
e3e3 = 1

Also the order of the bivectors e1e2, e1e3 and e2e3 is very important, 
they must be not be confused that e1e2 = e2e1, as they are defined as 
e1e2 + e2e1 = 0 so order is important, they're anticommunicative.

I prefer to present the final 3d bivector in terms of e1e2, e2e3, e3e1 
and you can grasp the cyclic nature from that order.

Continuing to reduce:

   =  rx*1 + ry(e1e2) + rz(e1e3) + sx(e2e1) + sy*1 + sz(e2e3) + 
tx(e3e1) + ty(e3e2) + tz*1

ordering the e1,e2,e3 products to match e1e2, e2e3 and e3e1

   =  rx*1 + ry(e1e2) - rz(e3e1) - sx(e1e2) + sy*1 + sz(e2e3) + 
tx(e3e1) - ty(e2e3) + tz*1

Grouping scaler and bivectors components.

       scalar part      bivector(e1e2)   bivector(e2e3)  bivector(e3e1)

   = ( rx + sy + tz ) + (ry-sx)(e1e2)  +  (sz-ty)(e2e3) + (tx-rz)(e3e1)
        dot product

Assuming I did everything okay by hand.

So the 3d bivector part of the geometric product is

   =  (ry-sx)(e1e2)  +  (sz-ty)(e2e3) + (tx-rz)(e3e1)

But what is the magnitude of this bivector? And how do we create a unit 
3d bivector?

Do we divide by sqrt((ry-sx)^2 + (sz-ty)^2 + (tx-rz)^2) = magnitude ???  
Is that the magnitude?

The available information I have looked at on the internet the past 3 
days does NOT say anything about this, all that is written is |B|.

Can anyone of you help out?

- Randall

P.S.:  For 2d vectors a,b, the unit bivector e1e2 (only 1 exists) 
becomes ( (a~*b) - (b~*a) a / (a[1]*b[2]-b[1]*a[2]) as the magnitude is 
(a[1]*b[2]-b[1]*a[1]), and the scalar part is the a dot b => a*b~.