ExampleOfCompositionOfRotationsAndQuaternionOrder 2026-08-23 23:15:13 2026-08-23 23:15:13 Example composition of rotations and quaternion order converted to passive rotation convention % 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*{Composition of Rotations and Quaternion Order: Examples, Exercises, and Solutions} This companion article belongs to the PhysicsLibrary entry \emph{Composition of Rotations and Quaternion Order}. All exercises are stated first. Complete solutions appear only after the exercise section so that the article is self-study friendly. \section{Convention declaration} We use right-handed frames, Hamilton multiplication, scalar-first display notation, and the active rotation rule \begin{equation} \mathbf v'=q\mathbf v q^* \end{equation} for unit quaternions. If $q_1$ acts first and $q_2$ acts second, then the net quaternion is \begin{equation} q_{\mathrm{net}}=q_2q_1. \end{equation} \section{Exercises} \subsection*{Exercise 1} Show directly from the active rotation formula that if $q_1$ acts first and $q_2$ acts second, then the net quaternion is $q_2q_1$. \subsection*{Exercise 2} Let \begin{equation} q_z=\frac{1+\mathbf k}{\sqrt2} \end{equation} represent a $+90^\circ$ rotation about the $z$ axis, and let \begin{equation} q_x=\frac{1+\mathbf i}{\sqrt2} \end{equation} represent a $+90^\circ$ rotation about the $x$ axis. Starting from $\mathbf v=\mathbf j$, compute the final vector when the order is first $q_z$ then $q_x$. \subsection*{Exercise 3} Repeat Exercise 2 but reverse the order: first $q_x$ then $q_z$. Compare the result with Exercise 2. \subsection*{Exercise 4} Using \begin{equation} q_x=\frac{1+\mathbf i}{\sqrt2}, \qquad q_y=\frac{1+\mathbf j}{\sqrt2}, \end{equation} start from $\mathbf v=\mathbf k$ and verify that first $q_x$ then $q_y$ gives $-\mathbf j$, while first $q_y$ then $q_x$ gives $\mathbf i$. \subsection*{Exercise 5} Let \begin{equation} q_1=\cos\frac{\alpha}{2}+\widehat{\mathbf u}\sin\frac{\alpha}{2}, \qquad q_2=\cos\frac{\beta}{2}+\widehat{\mathbf u}\sin\frac{\beta}{2} \end{equation} be rotations about the same unit axis $\widehat{\mathbf u}$. Show that \begin{equation} q_2q_1=\cos\frac{\alpha+\beta}{2}+\widehat{\mathbf u}\sin\frac{\alpha+\beta}{2}. \end{equation} \subsection*{Exercise 6} Suppose $q$ is unit. Show that applying $q$ and then $q^*$ gives the identity rotation. \subsection*{Exercise 7} A software library stores quaternions as $[q_x,q_y,q_z,q_w]^T$. Does that fact alone change the composition rule for active rotations? Explain briefly. \subsection*{Exercise 8} Write the frame-chain rule for ${}^{N}q_A$ and ${}^{A}q_B$, and derive ${}^{N}q_B$. \subsection*{Exercise 9} Explain the difference between an intrinsic $z$-then-$x$ sequence and an extrinsic $z$-then-$x$ sequence. \subsection*{Exercise 10} Show that if $R(q)$ is the active rotation matrix associated with $q$, then \begin{equation} R(q_2q_1)=R(q_2)R(q_1). \end{equation} \subsection*{Exercise 11} Let \begin{equation} q_1=\frac{1+\mathbf i}{\sqrt2}, \qquad q_2=\frac{1+\mathbf j}{\sqrt2}. \end{equation} Compute the explicit quaternion products $q_2q_1$ and $q_1q_2$ and verify that they differ. \subsection*{Exercise 12} Give a passive-rotation formula corresponding to the active sequence ``first $q_1$, then $q_2$''. \subsection*{Exercise 13} Why do small-angle rotations often appear to commute in first-order linearized attitude-error models even though finite rotations do not commute exactly? \subsection*{Exercise 14} Let \begin{equation} q_1=\frac{1+\mathbf k}{\sqrt2}, \qquad q_2=\frac{1-\mathbf k}{\sqrt2}. \end{equation} Find $q_2q_1$ and interpret the result physically. \subsection*{Exercise 15} A student says, ``Since multiplication is associative, the order of quaternion rotations does not matter.'' Identify the mistake in this statement. \section{Solutions} \subsection*{Solution 1} If $q_1$ acts first, then $\mathbf v_1=q_1\mathbf v q_1^*$. If $q_2$ acts next, then \begin{equation} \mathbf v_2=q_2\mathbf v_1 q_2^*=q_2(q_1\mathbf v q_1^*)q_2^*. \end{equation} By associativity, \begin{equation} \mathbf v_2=(q_2q_1)\mathbf v(q_1^*q_2^*). \end{equation} Since $(q_2q_1)^*=q_1^*q_2^*$, \begin{equation} \mathbf v_2=(q_2q_1)\mathbf v(q_2q_1)^*. \end{equation} Hence the net quaternion is $q_2q_1$. \subsection*{Solution 2} First rotate $\mathbf j$ about $z$ by $+90^\circ$: \begin{equation} q_z\mathbf j q_z^*=-\mathbf i. \end{equation} Then rotate $-\mathbf i$ about $x$ by $+90^\circ$. A vector on the $x$ axis is unchanged, so \begin{equation} q_x(-\mathbf i)q_x^*=-\mathbf i. \end{equation} Therefore the final vector is $-\mathbf i$. \subsection*{Solution 3} First rotate $\mathbf j$ about $x$ by $+90^\circ$: \begin{equation} q_x\mathbf j q_x^*=\mathbf k. \end{equation} Then rotate $\mathbf k$ about $z$ by $+90^\circ$. A vector on the $z$ axis is unchanged, so the final vector remains $\mathbf k$. This differs from the result of Exercise 2, so the order matters. \subsection*{Solution 4} Using the standard axis rotations, \begin{equation} q_x\mathbf k q_x^*=-\mathbf j. \end{equation} Rotating $-\mathbf j$ about $y$ leaves it unchanged, so first $q_x$ then $q_y$ gives $-\mathbf j$. Also, \begin{equation} q_y\mathbf k q_y^*=\mathbf i, \end{equation} and rotating $\mathbf i$ about $x$ leaves it unchanged. So first $q_y$ then $q_x$ gives $\mathbf i$. \subsection*{Solution 5} Because both quaternions use the same axis $\widehat{\mathbf u}$, \begin{equation} \begin{aligned} q_2q_1 &=\left(\cos\frac{\beta}{2}+\widehat{\mathbf u}\sin\frac{\beta}{2}\right) \left(\cos\frac{\alpha}{2}+\widehat{\mathbf u}\sin\frac{\alpha}{2}\right)\\ &=\cos\frac{\beta}{2}\cos\frac{\alpha}{2} -\sin\frac{\beta}{2}\sin\frac{\alpha}{2}\\ &\qquad +\widehat{\mathbf u}\left( \sin\frac{\beta}{2}\cos\frac{\alpha}{2} +\cos\frac{\beta}{2}\sin\frac{\alpha}{2} \right), \end{aligned} \end{equation} where $\widehat{\mathbf u}^{\,2}=-1$. The angle-addition formulas then give \begin{equation} q_2q_1=\cos\frac{\alpha+\beta}{2}+\widehat{\mathbf u}\sin\frac{\alpha+\beta}{2}. \end{equation} \subsection*{Solution 6} The net quaternion is \begin{equation} q^*q=qq^*=1, \end{equation} so the second rotation exactly undoes the first. Therefore the composition is the identity rotation. \subsection*{Solution 7} No. Scalar-last is only a storage convention. The composition rule depends on the algebra and the rotation interpretation, not on memory layout. For active rotations with Hamilton multiplication, if $q_1$ acts first and $q_2$ acts second, the net quaternion is still $q_2q_1$. \subsection*{Solution 8} The frame-chain rule is \begin{equation} \boxed{{}^{N}q_B={}^{N}q_A{}^{A}q_B.} \end{equation} It is derived by substituting the map from $B$ to $A$ into the map from $A$ to $N$ and regrouping, exactly as for successive active rotations. \subsection*{Solution 9} In an extrinsic $z$-then-$x$ sequence, both rotations are taken about axes fixed in the reference frame. In an intrinsic $z$-then-$x$ sequence, the second rotation is about the body's current $x$ axis after the first rotation has already occurred. The axis labels can therefore refer to different physical axes in the two descriptions. \subsection*{Solution 10} For any vector $\mathbf v$, \begin{equation} R(q_2q_1)\mathbf v=(q_2q_1)\mathbf v(q_2q_1)^*=q_2(q_1\mathbf v q_1^*)q_2^*=R(q_2)R(q_1)\mathbf v. \end{equation} Since this holds for every $\mathbf v$, the matrices are equal: \begin{equation} R(q_2q_1)=R(q_2)R(q_1). \end{equation} \subsection*{Solution 11} Compute \begin{equation} \begin{aligned} q_2q_1 &=\frac{1}{2}(1+\mathbf j)(1+\mathbf i)\\ &=\frac{1}{2}(1+\mathbf i+\mathbf j+\mathbf j\mathbf i)\\ &=\frac{1}{2}(1+\mathbf i+\mathbf j-\mathbf k), \end{aligned} \end{equation} because $\mathbf j\mathbf i=-\mathbf k$. Similarly, \begin{equation} q_1q_2=\frac{1}{2}(1+\mathbf i+\mathbf j+\mathbf k). \end{equation} The products differ, so the corresponding rotations differ. \subsection*{Solution 12} The passive formula is obtained by using the inverse action. If $q_1$ acts first and $q_2$ acts second in the active sense, then the passive coordinate change uses \begin{equation} \mathbf v_{\mathrm{passive}}=(q_2q_1)^*\mathbf v(q_2q_1)=q_1^*q_2^*\mathbf v q_2q_1. \end{equation} \subsection*{Solution 13} In a first-order small-angle model, products of two small rotation increments are second-order quantities and are neglected. The noncommutativity is then invisible at first order. For finite rotations, those higher-order terms are not negligible, so the order matters. \subsection*{Solution 14} Since $q_2=q_1^*$, \begin{equation} q_2q_1=q_1^*q_1=1. \end{equation} The net effect is no rotation. The second quarter-turn exactly cancels the first. \subsection*{Solution 15} Associativity means that one may regroup factors without changing their order, for example $(q_2q_1)q_0=q_2(q_1q_0)$. It does \emph{not} mean one may swap the order of factors. Commutativity would be the property $q_2q_1=q_1q_2$, and that property fails in general for finite rotations. \section*{License} This article is an original synthesis prepared for PhysicsLibrary and intended for release under CC BY-SA 4.0.