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Tamás Zahola (11 months ago)

This looks just like rotation via quaternions 🤔

Alan Macdonald (11 months ago)

Yes, this is indeed quaternion multiplication. I use the term in my book. If all you ever want to do is rotate vectors in 3D, then all you need is quaternions. But as I indicate at the end of the Geometric Algebra 4 video, geometric algebra can also rotate a plane in 3D
and higher dimensional objects in higher dimensional nD. I also show there that geometric algebra can project and reflect high dimensional objects in nD. And this is just the beginning. Slide 3 of the Geometric Algebra 0 video describes the vast scope of geometric algebra and calculus. Rotation of vectors with quaternions is just a tiny part of this *unified* framework.

and then i said (1 year ago)

2:06 My goodness gracious! That tickles!

mathIsART (1 year ago)

I know, the beauty is too much for my body to handle

Mathoma (1 year ago)

+and then i said
lol

DaMaoHuDong大毛忽洞 (2 years ago)

Thank you!!

Nancy Jazmín (2 years ago)

Thank you!!

720SouthCalifornia (2 years ago)

My thoughts toward a geometric intuition of composing rotations:
If we leave the angles ambiguous as alpha/2 and beta/2 on slide 29, the second to last expression will be composed of 4 terms with scalars cos(a/2)cos(b/2), sin(a/2)cos(b/2), cos(a/2)sin(b/2), sin(a/2)sin(b/2). To normalize and find the composed angle of rotation, the cos*cos (non-rotational/scalar) term will form the cosine edge of a unit triangle with the sum of the three remaining (rotational/bivector) terms as the sine edge, corresponding to the sine of the total rotation. The angle sum identity in the svg linked below helps visualize the respective magnitudes of each term, using the corresponding edges (and substituting a/2 and b/2 for a and b). We can picture the three rotational terms as follows. If we let either of the angles go to 0, the sin terms of that angle drop out and the equation returns to a single rotation in the other plane. Otherwise if we rotate pi/2 in both planes, both the cos terms go to 0 and the sin*sin term to -1, this corresponds to totally rotating on both axes. This sin*sin term is the composed rotational component and its bivector/plane of rotation is the product of the two planes. The sum of these three scaled rotational components gives the resulting plane.
In your example the non-rotational component is the scalar 1/2, the e2e3 term is the cossin and corresponds to the e1 component of the axis, the e1e2 is the sincos corresponding to e3 axis component, and the e1e3 is the sinsin component, which gives the negative -e2 axis component, all in proportion according to the edges of the angle sum identity with a=b=pi/4. To visualize the final axis changing as we go through the second rotation, it starts at the original (sin*cos) e3 axis. This axis will be weighted less and less as we rotate, in proportion to cos(b/2). The other two axes e1 and -e2 will be weighted increasingly at the same rate in proportion to sin(b/2). (since cos(a/2) = sin(a/2) = 1/sqrt(2) in this case)
Thanks a ton for these videos im looking forward to reading your books.
https://upload.wikimedia.org/wikipedia/commons/thumb/0/03/TrigSumFormula.svg/350px-TrigSumFormula.svg.png
coscos = OA
cossin = AQ
sincos = RP
sinsin = RQ
||OP|| = ||OQ|| = 1

Alan Macdonald (2 years ago)

+720SouthCalifornia I don't understand your figure. In it, the angles alpha and beta are in a common frame. but in the video they are not.

Paul Frischknecht (2 years ago)

Can I argue as follows to obtain the axis of rotation from the expression for the rotation plane: e3 is the vector perpendicular to e1e2 (it is the outer product of these two vectors), e1 to e2e3, -e2 to e1e3.
The axis perpendicular to the plane e1e2+e2e3+e1e3 is thus the sum of these three vectors, e3 + e1 - e2.
?

Alan Macdonald (2 years ago)

+Paul Frischknecht Yes, you are right. Slide 34 in the next video spells it out.

Paul Frischknecht (2 years ago)

Thanks. I almost missed the fact that the exponential is for theta/2, so only half the angle (pi/3 = 60°).
After plotting the rotation for the example vector {1,0,0} I find it easy to see why the composed rotation is 120° about the axis {1,-1,1}.
In mathematica:
a = {1, 0, 0};
b = {0, 1, 0};(*rotate a 90\[Degree] ccw in xy plane*)
c = {0, 0, 1};(*rotate b 90\[Degree] ccw in yz plane*)
ax = {1, -1, 1};
o = {0, 0, 0};
f = Arrow[{o, #}] &;
Graphics3D[{Red, [email protected], Green, [email protected], Blue, [email protected], Black, [email protected]}]

Christian Caballero (3 years ago)

First of all thank you very much for this explanation Professor Macdonald, on a second note, for me it's not that clear the development of the rotation formula on slide 26, in particular, why the exponential changes signs when is moved to the left of the parallel part of u but it doesn't change signs for the perpendicular part of u. You mention that you haven't said explicitly why the exponential behaves like that, but the rest of the exposition is perfectly clear until this point and since the rest of the video is sustained on this formula it kind of clouds the exposition. My guess is that it has to do because the exponential commutes or anticommutes depending on whether the vector lies on the same plane as the bivector or if it's orthogonal.
Again thank you very much for making these videos, there's not much material on GA on Youtube yet, I'm very much looking forward on your series on Geometric Calculus!

Alan Macdonald (3 years ago)

+Christian Caballero Your guess is right. Expand the exponentials as defined on Slide 23. Use the definition of the geometric product on Slide 24 to see that i commutes with uperp and anticommutes with uparallel.

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