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*From*: John Denker <jsd@AV8N.COM>*Date*: Wed, 30 Mar 2005 06:09:19 -0500

Tony Wayne wrote:

Does any one know why the "area vector" for a surface is defined as

being normal to the plane of the surface instead of parallel to the

plane of the surface?

As others have mentioned, a vector in the plane of the surface

doesn't tell you what you need to know. There isn't a one-to-one

mapping between vectors and areas. Dimensional analysis suffices

to demonstrate the point: vectors have length; areas have area.

What you want is a *bivector*. If P and Q are vectors, the best

representation of the parallelogram spanned by P and Q is the

wedge product P /\ Q. In particular, the amount of paint

required to cover such a region is given by the norm thereof,

i.e. ||P /\ Q||. This is so elegant, so simple, and so powerful

that I shudder at the thought of representing areas any other

way.

For details, see

http://www.av8n.com/physics/area-volume.htm

and references therein.

Bivectors have all the properties you want when representing

areas.

-- For starters, the antisymmetry of the wedge product

guarantees that the area P /\ Q is manifestly zero if P happens

to be parallel to Q.

-- A patch of area might be part of the boundary of a volume.

The volume V is best represented as V = P /\ Q /\ R. This is

manifestly zero if R happens to lie in the plane of P /\ Q.

In a three-dimensional space where you have basis vectors

(unit vectors) X, Y, and Z, the natural basis for computing

bivectors is Y/\Z, Z/\X, and X/\Y. These critters are isomorphic

to quaternions.

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**Follow-Ups**:**[Phys-L] Re: compound rotation***From:*John Denker <jsd@AV8N.COM>

**References**:**[Phys-L] Area vector***From:*Tony Wayne <wayne@PEN.K12.VA.US>

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