# Transformation represented by geometric product between a point and plane in PGA?

In the 3D PGA cheat sheet, I can’t find what the geometric production between a point and a plane represents.

Klein has such an operator, and returns a motor, but I don’t what kind of transformation that motor represents with respect to the point and plane.

Is it something like a rotation about the line through the point, perpendicular on the plane, and a translation along that line? If so, what would be the angle of rotation, and distance of translation? Just guessing here, I haven’t digged in into motors yet.

@Ziriax, You’re right, we didn’t include that on the cheat sheet. Thanks for bringing it up, it’s a nice formula that reveals the “polymorphicity” of PGA.

The product of a normalized euclidean plane p and a normalized euclidean point P yields a normalized bivector \ell and a multiple of the pseudoscalar \bf{I}: pP = \ell + d\bf{I} where \ell is the line through P perpendicular to p, and d is the signed distance from the point to the plane. Furthermore, \ell is normalized and its orientation is determined by the side of p on which P lies. (I. e., switching the order of the product to Pp doesn’t change the orientation of \ell.)

If the point is a normalized ideal point, then the bivector part vanishes and the weight of the pseudoscalar gives the sine of the angle that the ideal point (think “direction”) makes to the plane: pP = \sin(\alpha)\bf{I}.

This result is exactly analogous to the product of a point and a line in 2D.

Connection to motors: The result {m}:=pP is an element of the even subalgebra that satisfies {m}\widetilde{{m}}=1 hence is a motor. Which motor? The fact that there is no scalar part, only a pseudoscalar part, is a tip that it is a translation. Which translation? Turns out to be the translation in the direction normal to p that translates by a distance of 2d, so that P is translated across p to its mirror point. By taking \sqrt{m} = \dfrac{(1+m)}2, we obtain the translation that takes P onto the plane p.

2 Likes

Thanks again for the support!

I’m also interested in the meanings of the operators when the inputs are not normalized projectively. Is this something worth exploring?

In my experience the inputs should always be normalized. It’s like when you do spherical geometry, you work with unit length vectors. Also here. But if you have non-normalized arguments for any reason you can adjust the formulas by dividing the entire right hand side by \| p \| \| P \| . (Understood that for ideal points the ideal norm is to be used.)

2 Likes

That is interesting. If I recall correctly, @enki told me he rarely normalizes.

I’m curious if this normalization process would also be needed when PGA is combined with rational trigonometry, but let me first try to tackle PGA without that extra stuff on my plate @Ziriax I too am interested in the connections between rational trigonometry and PGA. Let me know if you are interested in comparing notes in another thread.

If you’re doing join and meet (i. e., projective operations) then you don’t need to normalize. You generally only need to normalize when you get to the end of a calculation involving geometric products and want to interpret the results. However if your calculation involves a sum of euclidean elements, then you have to be very careful about the norms of the arguments since changing one without changing the other will change the sum (projectively).
Writing my short article on PGA and rational trig led me to conclude that also in rational trig it’s almost always good to normalize arguments (and results). The formulas look cleaner and the results are easier to interpret.

Good to know.

I was asking because with homogeneous coordinates, when you have two particles at normalized points P1 and P2, and these particles have masses m1 and m2, then m1 P1 + m2 P2 represents the center of mass. This can be generalized to N point-masses.

So homogeneous coordinates can also be very handy in physics, but maybe this is just a coincidence for this particular case.

That made me wonder if non-normalized PGA elements also carry a physical meaning. Normalization is needed when we only want to interpret the results geometrically, but the normalization operator might throw away information that has physical meaning (like CGA is used by Hestenes for doing physics, although I don’t know those details).

Does this make any sense?

Very good point! I overlooked to mention that in kinematics and dynamics (velocities and forces for example) the weight or norm of an element represents the intensity of the element. If v is a bivector representing a rotation, then 2v represents the same rotation twice as fast; if f is a bivector representing a force, the 2f represents the same force but twice as strong. In this context it’s essential to maintain the non-normalized form. Thanks for pointing that out, @Ziriax . It’s only in “pure geometry” that the intensity can often be normalized away.

1 Like