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I am currently in my last year of high school and I am looking at how quaternions are used for 3D animation for my math individual investigation and got to the section where a point p=[0,(x,y,z)] is rotated by angle θ about a unit axis n with a quaternion q=[w, v]=[cos(θ/2),n sin(θ/2)] and its inverse q^(-1)=[w, -v]=[cos(θ/2),-n sin(θ/2)] using:

p'=qpq^(-1)

The formula for the new point p' is then:

p'=p+2w(v x p)+2(v x (v x p))

I've been trying and using the standard definition of quaternion multiplication:

[w_1,v_1 ][w_1,v_2 ]=[w_1 w_2-v_1∙v_2,w_1 v_2+w_2 v_1+v_2×v_1 ]

But I can't seem to get to that new point formula.

How is it done algebraically?

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  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Dec 29 '17 at 0:12
  • $\begingroup$ Where are you getting stuck? If you show your work, someone is more apt to show you where you might be going wrong. $\endgroup$ – amd Dec 29 '17 at 0:21
  • $\begingroup$ Hint: somewhere along the way, you’ll need to use double-angle trig identities and the “BAC-CAB identity” for the cross product of three vectors. $\endgroup$ – amd Dec 29 '17 at 0:57
  • $\begingroup$ One other thing that might be causing you trouble is that you’ve got a sign error in your target expression for $p'$. If you work out what it gives for $\vec n = (0,0,1)$ and $\vec p = (x,y,0)$, you’ll find that it represents a rotation through $-\theta$ instead of $\theta$. $\endgroup$ – amd Dec 30 '17 at 0:27
  • $\begingroup$ You can find a derivation here. $\endgroup$ – amd Jan 4 '18 at 1:49
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$\newcommand{\veca}{\boldsymbol{a}}\newcommand{\vecb}{\boldsymbol{b}}\newcommand{\vecc}{\boldsymbol{c}}\newcommand{\vecq}{\boldsymbol{q}}\newcommand{\vecp}{\boldsymbol{p}}\newcommand{\vecv}{\boldsymbol{v}}$ I'm going to write quaternions in italic letters with hats, scalars in italic letters without hats and 3D vectors with italic bold letters. This is just so it's easer to see what is what.

If you have a unit quaternion $\hat q$ then it consists of the scalar (or real) part $q_0$ and the vector (or imaginary) part $\vecq$: $$\hat q=\begin{bmatrix}q_0\\\vecq\end{bmatrix}$$

Suppose you have a point in space $\vecv=\textstyle\begin{bmatrix}x\\y\\z\end{bmatrix}$ then you can identify it with a quaternion $\hat v$: $$ \hat v = \begin{bmatrix}0\\\vecv\end{bmatrix}.$$

If you want to apply the rotation that is encoded in the unit quaternion $\hat q$ to the vector $\vecv$ you do, as you wrote $$ \hat v' = \hat q\circ \hat v\circ \hat q^{-1},$$ where $\circ$ denotes quaternion multiplication and then use the vector part of $\hat v'$ as the rotated vector $\vecv'$.

You already know $$ \begin{bmatrix}q_0\\\vecq\end{bmatrix}\circ\begin{bmatrix}p_0\\\vecp\end{bmatrix} = \begin{bmatrix}p_0 q_0 -\vecq\cdot\vecp\\p_0 \vecq + q_0\vecp + \vecq\times\vecp\end{bmatrix}, $$ where $\cdot$ is the dot product of two vectors and for a unit quaternion $$ \hat q^{-1} = \begin{bmatrix}q_0\\\vecq\end{bmatrix}^{-1}=\begin{bmatrix}q_0\\-\vecq\end{bmatrix}.$$ Now we try to simplify the rotation. For this, we will first apply our quaternion multiplication two times, then expand the terms and use that $(\vecq\times\vecv)\cdot\vecq = 0$, since $\vecq\times\vecv$ is perpendicular to $\vecq$ by definition, $\vecq\times\vecv=-\vecv\times\vecq$, the BAC-CAB identity $\veca\times(\vecb\times\vecc) = \vecb(\veca\cdot\vecc) - \vecc(\veca\cdot\vecb)$ and that $\hat q$ is a unit quaternion $q_0^2 + \vecq\cdot\vecq = 1$. \begin{align} \hat v' &= \hat q \circ \hat v\circ \hat q^{-1} =\begin{bmatrix}q_0\\\vecq\end{bmatrix}\circ\begin{bmatrix}0\\\vecv\end{bmatrix}\circ \hat q^{-1}\\ &=\begin{bmatrix}0 - \vecq\cdot\vecv\\q_0\vecv + \boldsymbol{0} +\vecq\times\vecv\end{bmatrix}\circ\begin{bmatrix}q_0\\-\vecq\end{bmatrix}\\ &=\begin{bmatrix}-(\vecq\cdot\vecv)q_0 - (q_0\vecv + \vecq\times\vecv)\cdot(-\vecq)\\ q_0(q_0\vecv + \vecq\times\vecv) + (-\vecq\cdot\vecv)(-\vecq) + (q_0\vecv + \vecq\times\vecv)\times(-\vecq)\end{bmatrix}\\ &=\begin{bmatrix}-(\vecq\cdot\vecv)q_0 + q_0(\vecv\cdot\vecq) - (\vecq\times\vecv)\cdot\vecq\\ q_0^2\vecv + q_0\vecq\times\vecv + (\vecq\cdot\vecv)\vecq - q_0\vecv\times\vecq - (\vecq\times\vecv)\times\vecq\end{bmatrix}\\ &=\begin{bmatrix}0\\ q_0^2\vecv + 2q_0\vecq\times\vecv + (\vecq\cdot\vecv)\vecq + \vecq\times(\vecq\times\vecv)\end{bmatrix}\\ &=\begin{bmatrix}0\\ (q_0^2+\vecq\cdot\vecq)\vecv - (\vecq\cdot\vecq)\vecv+ 2q_0\vecq\times\vecv + (\vecq\cdot\vecv)\vecq + \vecq\times(\vecq\times\vecv)\end{bmatrix}\\ &=\begin{bmatrix}0\\ \vecv + 2q_0\vecq\times\vecv + \vecq(\vecq\cdot\vecv) - \vecv(\vecq\cdot\vecq) + \vecq\times(\vecq\times\vecv)\end{bmatrix}\\ &=\begin{bmatrix}0\\ \vecv + 2q_0\vecq\times\vecv + 2\vecq\times(\vecq\times\vecv)\end{bmatrix}. \end{align} From this, we get the formula for the rotated vector $\vecv'$: $$ \vecv' = \vecv + 2q_0\vecq\times\vecv + 2\vecq\times(\vecq\times\vecv).$$

(Note that you have some errata in your question, like wrong signs etc.)

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