Geometric Algebra in 3D - The Vector-Bivector Product (Part 2)

[@NoNameEntered]

Thank you so much for these videos!!! It's been hard to find good material explaining GA clearly and your series is by far the best I've found! Thanks to you I've been able to build my own GA library from scratch and I'm shocked at how elegant it all is. Once you go GA you cannot go back - doing geometry without GA is like programming in assembly language!

[@nrr8532]

Ditto, exactly my case as well. @Mathoma, congratulations and many thanks for such a wonderful job!

[@harryhirsch2024]

These lessons are the best I have ever heard. Thank you so much.

[@jamesmarlar4231]

Thanks for these informative and interesting presentations. Keep these coming.

[@alecchalmers3744]

Couldn't the unit vector products at 7:00 have been simplified with the use of the Minkowski metric which defines their products as 1 or -1 (or 0).

[@pronounjow]

This stuff is so cool! It really makes me think and focus!

[@fredspiessens2292]

This was very interesting and very well explained, thanks!

[@erickweil4580]

The wedge product of two vectors produce a three component bivector in 3D, and when talking about geometric interpretations it is considered both a oriented area path and the perpendicular vector against the two whom produced the bivector (considering two linearly idependent ones). So in G(4), it is the Geometric Algebra in 4 Dimensions, if I calculate the wedge of three vectors, A ^ B ^ C, I get a 4 component trivector, that is a oriented patch of volume, and is it correct to say that this can be also the perpendicular vector against the three vectors A, B and C? is this in a way the 4D analog of a Cross product?

[@janplaza6741]

I agree with all the formulas derived in this segment of the series, but I do not follow why in the general case, the projection of a is rotated by 90 deg. Multiplying the projection by B results in the rotation by some angle, but why by 90 deg? Thank you for this excellent presentation of GA.

[@darksecret965]

cus in a plane, a bivector acts like i, or an orthogonal rotator, check G(2) for more info

[@BlackEyedGhost0]

The fact that the bivector rotates the parallel part seemed odd to me at first, but then I realized that it's the same as if you take the cross product with the vector perpendicular to the bivector. It's so much more natural to think of magnetic fields as bivectors than as vectors.

[@cristian-bull]

I don't think your comparison of the Bivector rotation with the cross product holds true. If you have a vector parallel to a Bivector, and cross product it with the vector orthogonal, the rotation will always be 90 degrees, however, the angle of rotation given by the dot product depends on the angle between the vectors that compose the Bivector, its not always 90 degrees. I also though of magnetic fields, but in a different way: If the Bivector is an area, and the Magnetic field is a vector, the wedge product gives the volume of the product of the Area with the component of the magnetic field orthogonal to such area. That would be equal to the magnetic flux I think.

[@BlackEyedGhost0]

@@cristian-bull I didn't say anything about a vector parallel to the bivector. R3 Bivectors are represented by 3 real numbers, just like a vector. The pseudovector represented by those numbers is a vector perpendicular to the bivector. So if you take an arbitrary vector and "dot" it with a bivector, it's the same as taking that same arbitrary vector and crossing it with the vector perpendicular to the bivector. Both give you the same vector. That is, a.B = axb, where b is the vector represented by the same numbers as the bivector B.

[@finweman]

I don't think I caught this on the video but it seems like to actually find the projection of a vector, 'a', onto a plane, 'B', I've worked out that you need to do B dot(a,B). To find the perpendicular part of vector 'a' to a plane, you do B wedge(a,B). (B is a unit bi-vector.) Is that right?

[@Math_oma]

+Marco Dalla Gasperina There should be a B^(-1) in there somewhere, like the vector formulas.

[@finweman]

Thank you. I missed a transposition. a-perp = -B a^B.

[@Math_oma]

+Marco Dalla Gasperina But I mean the inverse bivector, B^-1 or 1/B not merely the negative bivector.

[@finweman]

Thanks. I had assumed that B was a unit bi-vector. I still think the minus sign is appropriate though... or did I miss something else?

[@Math_oma]

+Marco Dalla Gasperina I would check your derivation again. Remember that when the vector is in the plane of the bivector, they anticommute and when the vector is orthogonal to the plane they commute.

[@adamz8314]

this was confuse for me,first Geometric Product of any vector consist of linear combination of (scalar+Vector+Bivector +Trivector)with itself is should be equal =1,now i see is not case also same applied for Geometric Product of 2 vector consist of linear combination of (Scaler +Vector+Bivector +Trivector) are always anti commutative.

[@arminth4117]

Hello I think, as in part 1, your definitions for wedge and dot and swapped again. If A || B then the bivector component would be 0 and we are left with the scalar component, so AB = BA. If A _ | _ B then the scalar component is 0 and AB = -BA inherited from the wedge product component's antisymmetry. Peace and thanks again for the videos.

[@stellarwind72]

After understanding bivectors, the Lorentz force law makes a lot more sense. F = q(E + v × B). The v × B is really a vector-bivector dot product (Magnetic field is a bi-vector). So the Lorentz force law should really be written as F = q(E + v ⋅ B).

[@housamkak8005]

A key point that wasn't really obvious to me or very clear even though it is trivial: "Whenever there is a negative sign that results from swapping two quantities, there is a corresponding geometric interpretation of switching the sense of rotation." I concluded this from thinking about why really the bivector and vector dot product anticommutes. The position of the bivector dictates the e1e2 (i=1,j=2 as example) position in the equation. After the projection of the vector on the bivector, if the bivector is on the right, e1e2 is on the right, which tells us we are going from e1 to e2, which is an anticlockwise rotation. if the bivector is on the left, e1e2 is on the left, which tells us we are going from e2 to e1, which is a clockwise rotation. Take this e1 and e2 system as reference: e2 | | | _ _ _ _ _ e1 If it doesn't matter who goes first, then things commute. Always think whether order in things matter. For example, why does the dot product of two vectors commute? Ask the question, If I visually ask whether who goes first, v1 or v2, it doesn't really matter, as they are above each other. Another example, why does the order of wedge product of vector and bivector commutes? Because simply if I sweep the bivector B over its normal vector for example, I will get a volume. Less intuitive but serves the argument, If I sweep the vector, keeping it perpendicular, in both directions of the bivector, first sweeping in the first direction and then in the other direction, I will get the same volume.

[@jamesmarlar4231]

btw, your volume settings are too low.

[@ARBB1]

Thanks.

[@wsdftgr5052]

Thanks a lot, I just got confused between the following two parts: part1: kzbin.info/www/bejne/iJOum5SGedR6otk part2: kzbin.info/www/bejne/iJOum5SGedR6otk when we have a bivector (for example e1e2), we must reverse it as a block or we can treat the e1 and e2 inside it as individual vectors as in part 2

[@bonbonpony]

Furries are gonna love this part. Lots of uwu :J

[@wolftribe66]

this is an angry comment

[@rasp1628]

this is an sad comment

[@gyorgyo7597]

@@rasp1628 this is a silly comment

[@Minderance]

this is a late comment