ExampleOfRotatingVectorsWithQuaternions 2026-08-23 20:49:30 2026-08-23 20:49:30 Example rotating vectors with quaternions converted to passive rotation convention quaternion vector rotation unit quaternion Rodrigues formula pure quaternion active rotation passive rotation exercises worked solutions % this is the default PhysicsLibrary preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \section*{Rotating Vectors with Quaternions: Examples, Exercises, and Solutions} This entry is the self-study companion to the PhysicsLibrary article \emph{Rotating Vectors with Quaternions}. The convention throughout is Hamilton multiplication in a right-handed frame, with scalar-first quaternion display and active vector rotation \begin{equation} \boxed{\mathbf v'=q\mathbf vq^*.} \end{equation} The quaternion is assumed to have unit norm unless a problem states otherwise. \section{Exercises} \begin{enumerate} \item \textbf{Pure-quaternion representation.} Write the vector \[ \mathbf v=2\hat{\mathbf x}-3\hat{\mathbf y}+4\hat{\mathbf z} \] as a scalar-first pure quaternion. \item \textbf{$+90^\circ$ about $+z$.} Use quaternion multiplication to rotate \[ \mathbf v=\hat{\mathbf x} \] through $+90^\circ$ about $+z$. Show every quaternion multiplication step. \item \textbf{$+90^\circ$ about $+x$.} Rotate \[ \mathbf v=\hat{\mathbf y} \] through $+90^\circ$ about $+x$ and determine the resulting Cartesian vector. \item \textbf{A vector parallel to the axis.} Let \[ q=\cos 35^\circ+\mathbf k\sin35^\circ. \] What physical rotation does $q$ represent? Apply it to $\mathbf v=5\mathbf k$ and explain the result geometrically. \item \textbf{Rotation through $180^\circ$.} Rotate \[ \mathbf v=2\mathbf i-3\mathbf j+4\mathbf k \] through $180^\circ$ about $+z$. \item \textbf{A symmetric $120^\circ$ rotation.} Let \[ \widehat{\mathbf u}=\frac{\mathbf i+\mathbf j+\mathbf k}{\sqrt3}, \qquad \theta=120^\circ. \] Construct $q$ and show that the active rotation sends \[ \mathbf i\longrightarrow\mathbf j. \] \item \textbf{Rodrigues formula.} Using \[ \widehat{\mathbf u}=\mathbf k, \qquad \theta=60^\circ, \qquad \mathbf v=2\mathbf i+\mathbf j+3\mathbf k, \] evaluate the rotated vector using Rodrigues' formula. \item \textbf{Quaternion sign ambiguity.} Prove directly that $q$ and $-q$ rotate every vector identically under the sandwich action. \item \textbf{Inverse rotation.} If \[ \mathbf v'=q\mathbf vq^*, \] prove that \[ \mathbf v=q^*\mathbf v'q. \] \item \textbf{Norm preservation.} Use quaternion norm multiplicativity to prove that a unit quaternion preserves the Euclidean magnitude of every rotated vector. \item \textbf{Deriving Rodrigues' formula.} Starting from \[ q=c+s\widehat{\mathbf u}, \qquad q^*=c-s\widehat{\mathbf u}, \] derive \[ \mathbf v' =\mathbf v\cos\theta +(\widehat{\mathbf u}\times\mathbf v)\sin\theta +\widehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v)(1-\cos\theta). \] \item \textbf{Vector-only implementation.} Show that, for a unit quaternion $q=(q_w,\mathbf q)$, \[ \mathbf v' =\mathbf v +2q_w(\mathbf q\times\mathbf v) +2\mathbf q\times(\mathbf q\times\mathbf v). \] Then rewrite it using \[ \mathbf t=2(\mathbf q\times\mathbf v). \] \item \textbf{Active versus passive.} A fixed physical vector has coordinates \[ \mathbf v_A=\hat{\mathbf x} \] in frame $A$. Frame $B$ is obtained by rotating the coordinate axes $+90^\circ$ about $+z$ relative to $A$. Using the same unit quaternion that would actively rotate vectors by $+90^\circ$ about $+z$, determine the coordinates $\mathbf v_B$. \item \textbf{Scalar-first and scalar-last software arrays.} The quaternion for a $+90^\circ$ rotation about $+z$ is displayed in PhysicsLibrary scalar-first form as \[ \left[\frac{\sqrt2}{2},0,0,\frac{\sqrt2}{2}\right]^T. \] Write the same quaternion in scalar-last storage. Explain whether the Hamilton multiplication law changes. \item \textbf{Debugging the wrong sandwich order.} A program intends to actively rotate $\hat{\mathbf x}$ by $+90^\circ$ about $+z$, but it computes \[ q^*\mathbf i q. \] Predict the output and explain the error. \item \textbf{Nonunit quaternion pitfall.} Let \[ r=2q, \] where $q$ is unit. Show that \[ r\mathbf v r^*=4q\mathbf vq^*. \] Explain why a nonunit quaternion should not be used with the conjugate sandwich formula when a pure rotation is intended. \end{enumerate} \section{Solutions} \subsection*{1. Pure-quaternion representation} The scalar part is zero, so \begin{equation} \boxed{ \mathbf v \longleftrightarrow \begin{bmatrix} 0&2&-3&4 \end{bmatrix}^{T}. } \end{equation} \subsection*{2. $+90^\circ$ about $+z$} The rotation quaternion is \[ q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf k, \qquad q^*=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\mathbf k. \] First, \begin{equation} \begin{aligned} q\mathbf i &=\frac{\sqrt2}{2}\mathbf i +\frac{\sqrt2}{2}\mathbf k\mathbf i \\ &=\frac{\sqrt2}{2}(\mathbf i+\mathbf j). \end{aligned} \end{equation} Then \begin{equation} \begin{aligned} q\mathbf i q^* &=\frac12(\mathbf i+\mathbf j)(1-\mathbf k) \\ &=\frac12(\mathbf i+\mathbf j-\mathbf i\mathbf k-\mathbf j\mathbf k) \\ &=\frac12(\mathbf i+\mathbf j+\mathbf j-\mathbf i) \\ &=\mathbf j. \end{aligned} \end{equation} Therefore \begin{equation} \boxed{\hat{\mathbf x}\rightarrow\hat{\mathbf y}.} \end{equation} \subsection*{3. $+90^\circ$ about $+x$} By the right-hand rule, a positive quarter-turn about $+x$ carries $+y$ toward $+z$. Thus \begin{equation} \boxed{\hat{\mathbf y}\rightarrow\hat{\mathbf z}.} \end{equation} The quaternion is \[ q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf i. \] Direct multiplication gives the same result. \subsection*{4. Vector parallel to the axis} The quaternion phase is $35^\circ$, so the physical rotation angle is \[ \theta=70^\circ. \] The axis is $+z$. Since $\mathbf v=5\mathbf k$ is parallel to the axis, \begin{equation} \boxed{q(5\mathbf k)q^*=5\mathbf k.} \end{equation} A rotation leaves every vector on its own axis unchanged. \subsection*{5. Rotation through $180^\circ$} A $180^\circ$ rotation about $+z$ reverses the $x$ and $y$ components and leaves the $z$ component unchanged. Therefore \begin{equation} \boxed{ \mathbf v'=-2\mathbf i+3\mathbf j+4\mathbf k. } \end{equation} \subsection*{6. Symmetric $120^\circ$ rotation} Since \[ \cos60^\circ=\frac12, \qquad \sin60^\circ=\frac{\sqrt3}{2}, \] we obtain \begin{equation} \boxed{ q=\frac12(1+\mathbf i+\mathbf j+\mathbf k). } \end{equation} Using Rodrigues' formula with $\widehat{\mathbf u}=(1,1,1)/\sqrt3$ and $\mathbf v=\mathbf i$ gives \begin{equation} \boxed{\mathbf v'=\mathbf j.} \end{equation} Thus this $120^\circ$ rotation cyclically carries the coordinate axes $x\rightarrow y\rightarrow z\rightarrow x$. \subsection*{7. Rodrigues formula} For $\widehat{\mathbf u}=\mathbf k$, \[ \widehat{\mathbf u}\times\mathbf v =\mathbf k\times(2\mathbf i+\mathbf j+3\mathbf k) =2\mathbf j-\mathbf i, \] and \[ \widehat{\mathbf u}\cdot\mathbf v=3. \] With $\cos60^\circ=1/2$ and $\sin60^\circ=\sqrt3/2$, \begin{equation} \begin{aligned} \mathbf v' ={}&\frac12(2\mathbf i+\mathbf j+3\mathbf k) +\frac{\sqrt3}{2}(2\mathbf j-\mathbf i) +\frac32\mathbf k \\ ={}&\left(1-\frac{\sqrt3}{2}\right)\mathbf i +\left(\frac12+\sqrt3\right)\mathbf j +3\mathbf k. \end{aligned} \end{equation} Hence \begin{equation} \boxed{ \mathbf v' =\left(1-\frac{\sqrt3}{2}\right)\mathbf i +\left(\frac12+\sqrt3\right)\mathbf j +3\mathbf k. } \end{equation} \subsection*{8. Quaternion sign ambiguity} Because $(-q)^*=-q^*$, \begin{equation} (-q)\mathbf v(-q)^* =(-q)\mathbf v(-q^*) =q\mathbf vq^*. \end{equation} Therefore \begin{equation} \boxed{q\text{ and }-q\text{ generate the same vector rotation}.} \end{equation} \subsection*{9. Inverse rotation} Start with \[ \mathbf v'=q\mathbf vq^*. \] Multiply on the left by $q^*$ and on the right by $q$: \begin{equation} \begin{aligned} q^*\mathbf v'q &=q^*(q\mathbf vq^*)q \\ &=(q^*q)\mathbf v(q^*q) \\ &=\mathbf v. \end{aligned} \end{equation} Thus \begin{equation} \boxed{\mathbf v=q^*\mathbf v'q.} \end{equation} \subsection*{10. Norm preservation} For unit $q$, \[ \lVert q\rVert=\lVert q^*\rVert=1. \] Hence \begin{equation} \begin{aligned} \lVert\mathbf v'\rVert &=\lVert q\mathbf vq^*\rVert \\ &=\lVert q\rVert\,\lVert\mathbf v\rVert\,\lVert q^*\rVert \\ &=\lVert\mathbf v\rVert. \end{aligned} \end{equation} So quaternion conjugation by a unit quaternion preserves Euclidean vector length. \subsection*{11. Deriving Rodrigues' formula} Let \[ q=c+s\widehat{\mathbf u}. \] The first product is \begin{equation} q\mathbf v = \left( -s\widehat{\mathbf u}\cdot\mathbf v, \;c\mathbf v+s\widehat{\mathbf u}\times\mathbf v \right). \end{equation} Multiplication by $q^*=(c,-s\widehat{\mathbf u})$ cancels the scalar part and gives \begin{equation} \begin{aligned} \mathbf v' ={}&(c^2-s^2)\mathbf v +2cs(\widehat{\mathbf u}\times\mathbf v) \\ &+2s^2\widehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v). \end{aligned} \end{equation} Using \[ c^2-s^2=\cos\theta, \qquad 2cs=\sin\theta, \qquad 2s^2=1-\cos\theta, \] produces \begin{equation} \boxed{ \mathbf v' =\mathbf v\cos\theta +(\widehat{\mathbf u}\times\mathbf v)\sin\theta +\widehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v)(1-\cos\theta). } \end{equation} \subsection*{12. Vector-only implementation} Write $q=(q_w,\mathbf q)$. Expanding the sandwich product and collecting cross-product terms gives \begin{equation} \boxed{ \mathbf v' =\mathbf v +2q_w(\mathbf q\times\mathbf v) +2\mathbf q\times(\mathbf q\times\mathbf v). } \end{equation} If \begin{equation} \mathbf t=2(\mathbf q\times\mathbf v), \end{equation} then \begin{equation} \boxed{ \mathbf v'=\mathbf v+q_w\mathbf t+\mathbf q\times\mathbf t. } \end{equation} \subsection*{13. Active versus passive} The active quaternion for $+90^\circ$ about $+z$ is \[ q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf k. \] A passive coordinate change uses the inverse action, \[ \mathbf v_B=q^*\mathbf v_Aq. \] For $\mathbf v_A=\mathbf i$, \begin{equation} \boxed{\mathbf v_B=-\mathbf j.} \end{equation} Therefore the fixed vector has coordinates \begin{equation} \boxed{ [0,-1,0]^T } \end{equation} in the rotated frame. \subsection*{14. Scalar-first and scalar-last arrays} Scalar-first storage is \[ \left[\frac{\sqrt2}{2},0,0,\frac{\sqrt2}{2}\right]^T. \] Scalar-last storage places the scalar at the end: \begin{equation} \boxed{ \left[0,0,\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right]^T. } \end{equation} The Hamilton multiplication law does not change. Only the array indexing changes. \subsection*{15. Debugging the wrong sandwich order} The expression \[ q^*\mathbf i q \] applies the inverse rotation. Therefore the output is \begin{equation} \boxed{-\mathbf j.} \end{equation} The intended active formula was $q\mathbf i q^*$. \subsection*{16. Nonunit quaternion pitfall} With $r=2q$, \[ r^*=2q^*. \] Therefore \begin{equation} r\mathbf vr^* =(2q)\mathbf v(2q^*) =4q\mathbf vq^*. \end{equation} Thus \begin{equation} \boxed{ \lVert r\mathbf vr^*\rVert=4\lVert\mathbf v\rVert. } \end{equation} The conjugate sandwich formula represents a pure rotation only when the orientation quaternion is unit. A nonunit quaternion should be normalized, or the true inverse should be used if a general quaternion similarity transform is intended. \section{Sources} The rotation law is cross-checked against the public-domain treatments of Joly and Hathaway, both of which use the classical rotator $q(\cdot)q^{-1}$. The problems and worked solutions here are newly written for PhysicsLibrary. \begin{thebibliography}{9} \bibitem{Joly1905} C. J. Joly, \emph{A Manual of Quaternions}, Macmillan and Co., London, 1905. Public-domain historical source. \bibitem{Hathaway1896} A. S. Hathaway, \emph{A Primer of Quaternions}, 1896. Public-domain text; Project Gutenberg edition and \LaTeX{} source available. \PMlinkexternal{Project Gutenberg edition}{https://www.gutenberg.org/ebooks/9934} \end{thebibliography} \section*{License} Unless otherwise noted, this PhysicsLibrary entry is intended for release under the Creative Commons Attribution--ShareAlike 4.0 International license.