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Prove the identity $$\mathbf{U}\times(\mathbf{V}\times\mathbf{W})=(\mathbf{U}\cdot\mathbf{W})\mathbf{V}-(\mathbf{U}\cdot\mathbf{V})\mathbf{W}$$ given three vectors $\mathbf{U},\mathbf{V}$ and $\mathbf{W}$ by a quaternion calculus.

I am quite unsure of what specifically the above question asks me to do. What is the meaning of quaternion calculus?

Any help or hint would be greatly appreciated. Thank you.

$\textbf{EDIT}$: Basically, my understanding is that we use quaternions to avoid the tedious derivations of composition of rotations which can be computed by applying the Rodrigues formula given an angle $\theta$ and a unit vector $\mathbf{u}$. Following the Hamilton rules we define the Quaternion algebra and then proceed to prove the correspondence between quaternions and rotations.

Basically, I don't understand the wording of the above question; I am not asking for a solution.

johnny09
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  • What do you know about quaternion? –  Sep 09 '18 at 23:15
  • @JohnMa Not much. I was introduced to the definition of Quaternions and their basic properties in the context of rotations as part of a Dynamics course. – johnny09 Sep 09 '18 at 23:17
  • It might be better if you describe more your knowledge about that, in particular those related to rotations, so that others can write an answer that you understand. –  Sep 09 '18 at 23:44
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    @JohnMa I have edited my question. Not sure if my EDIT does it better but I appreciate your suggestion. – johnny09 Sep 09 '18 at 23:58
  • Is $\times$ the cross product? Are $U,V,W$ 3-vectors? If so, you can treat them as the imaginary components of a quaternion. – user3658307 Sep 25 '18 at 03:35
  • @user3658307 Yes, $\times$ is the cross product. From a rotational perspective I'd say it makes sense to consider only 3-dimesional vectors. I am unsure of the phrase "quartenion calculus". Isn't it a little bit tedious to do what you suggest? – johnny09 Sep 26 '18 at 00:14
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    Hmm, I thought it might fall out of the definition of the Hamilton product, but it's not so obvious. Incidentally, there is also this (unanswered) question. Honestly, I think "calculus" here should probably just be algebra ... unless derivatives of quaternions could somehow be used... – user3658307 Sep 26 '18 at 04:54

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I wrote a derivation on the page (mentioned by user3658307) Is there a way to prove vector triple product from quaternion multiplication? .

As to your question: "What is the meaning of the quaternion calculus?"

First, the word "calculus" here does not refer to any infinitesimal calculus, as one learns in a calculus course. It means a way of calculating. (That is also the origin of the more common use of the word "calculus").

The quaternions are a 4-dimensional algebra over the reals, basically real 4-vectors together with multiplication operation, so that the product of two quaternions is again a quaternion. It is a 4-dimensional extension of the complex numbers. (There is no 3-dimensional extension.)

Multiplication by quaternions describe orthogonality-preserving transformations in 4-space. This makes them the best way of describing certain important physical processes.

Multiplication by a quaternion can be regarded as a change of orientation in 3-space: imagine a pilot of an aircraft, which can perform a maneuver involving rotation in both pitch and yaw axes. Such a maneuver changes the pilot's orientation. A change of orientation is more than a simple rotation -- and it isn't very well represented even by two simple rotations.

Any orthogonality-preserving transformation of geometrical 4-space (that is, any element of O(4) ) can be written as one quaternion multiplication on the left, and another multiplication on the right.

A specific use of quaternions is to describe solid rotations in 3-space. (Any rotation of 3-space is just a multiplication on the left by a quaternion, and on the right by the inverse of the same quaternion.)

In these applications, it is true that the quaternion calculations take much simpler form than the equivalents involving multiple three-dimensional calculations, but more importantly, the three-dimensional analogs introduce nasty singularities, which the more natural quaternion calculations do not have. Those singlularites have bad practical consequences. Search for "gimbal lock".

The quaternions are complicated, because orthogonal transformations of four dimensions are complicated. But they provide a faithful reflection of those 4-space transformations --- that is, they are no more and no less complicated than orthogonal transformations of 4-space.

Steve White
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    "4-dimensional extension of the complex numbers" could be misleading. It is extension of the complex numbers, and it is 4-dimensional over $\mathbb R$, but the particular wording might be understood that it is 4-dimensional over $\mathbb C$. You do write that these are real vectors, but I would suggest making it stand out bit more. – Ennar Aug 27 '21 at 15:20
  • @Ennar OK. I was thinking in terms of geometry, but I take your point. – Steve White Aug 27 '21 at 15:36