ExampleOfQuaternionsAndDirectionCosineMatrices 2026-08-24 00:36:15 2026-08-24 00:36:15 Example quaternions and direction cosine matrices 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). % If you use \includegraphics{figure} with no file extension, PhysicsLibrary % will prefer figure.png before falling back to figure.eps. \usepackage{graphicx} \DeclareGraphicsExtensions{.png,.jpg,.jpeg,.pdf,.eps} % 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*{Quaternions and Direction Cosine Matrices: Examples, Exercises, and Solutions} This companion article belongs to the PhysicsLibrary entry \emph{Quaternions and Direction Cosine Matrices}. 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^* = R(q)\mathbf v \end{equation} for unit quaternions. Passive coordinate changes are represented by the associated DCM relation \begin{equation} {}^{A}\mathbf v = C^A_B\,{}^{B}\mathbf v, \qquad C^A_B = R({}^{A}q_B). \end{equation} \section{Exercises} \subsection*{Exercise 1} Starting from the vector rotation formula \begin{equation} \mathbf v'= (q_w^2-\mathbf q\cdot\mathbf q)\mathbf v +2\mathbf q(\mathbf q\cdot\mathbf v) +2q_w(\mathbf q\times\mathbf v), \end{equation} derive the compact matrix formula for $R(q)$. \subsection*{Exercise 2} Expand the compact formula to obtain the component matrix \begin{equation} R(q)= \begin{bmatrix} 1-2(q_y^2+q_z^2) & 2(q_xq_y-q_wq_z) & 2(q_xq_z+q_wq_y)\\ 2(q_xq_y+q_wq_z) & 1-2(q_x^2+q_z^2) & 2(q_yq_z-q_wq_x)\\ 2(q_xq_z-q_wq_y) & 2(q_yq_z+q_wq_x) & 1-2(q_x^2+q_y^2) \end{bmatrix}. \end{equation} \subsection*{Exercise 3} For \begin{equation} q=\frac{1+\mathbf k}{\sqrt2}, \end{equation} compute $R(q)$. \subsection*{Exercise 4} Using the matrix from Exercise 3, compute the images of $\mathbf e_1$, $\mathbf e_2$, and $\mathbf e_3$. \subsection*{Exercise 5} Show directly from the compact matrix formula that $R(-q)=R(q)$. \subsection*{Exercise 6} Show that for a unit quaternion, \begin{equation} R(q^*)=R(q)^T. \end{equation} \subsection*{Exercise 7} Let \begin{equation} q=\cos\frac{\pi}{3}+\mathbf i\sin\frac{\pi}{3} =\frac{1}{2}+\frac{\sqrt3}{2}\mathbf i. \end{equation} Find $R(q)$. \subsection*{Exercise 8} Let \begin{equation} q=\cos\frac{\pi}{3}+\frac{\mathbf i+\mathbf j+\mathbf k}{\sqrt3}\sin\frac{\pi}{3}. \end{equation} Compute $R(q)$ and interpret the result geometrically. \subsection*{Exercise 9} Show that the columns of $R(q)$ are the images of $\mathbf e_1$, $\mathbf e_2$, and $\mathbf e_3$. \subsection*{Exercise 10} Write the passive DCM relation for the frame chain $B \to A \to N$ and derive \begin{equation} C^N_B = C^N_A C^A_B. \end{equation} \subsection*{Exercise 11} Suppose \begin{equation} R= \begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}. \end{equation} Recover a quaternion representing this rotation. \subsection*{Exercise 12} Given the proper rotation matrix \begin{equation} R= \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{bmatrix}, \end{equation} find a quaternion representing it and explain why the trace formula using $q_w$ is numerically delicate here. \subsection*{Exercise 13} Show that if $q$ is unit, then $R(q)$ is orthogonal and has determinant $+1$. \subsection*{Exercise 14} Explain the difference between the active matrix relation $\mathbf v'=R(q)\mathbf v$ and the passive DCM relation ${}^{A}\mathbf v=C^A_B\,{}^{B}\mathbf v$. \subsection*{Exercise 15} A software package stores quaternions in scalar-last order. A student copies the scalar-first matrix formula without changing the component mapping. Describe the error and how to fix it. \section{Solutions} \subsection*{Solution 1} Write the three terms separately: \begin{equation} (q_w^2-\mathbf q\cdot\mathbf q)\mathbf v = (q_w^2-\mathbf q^T\mathbf q)I_3\mathbf v, \end{equation} \begin{equation} 2\mathbf q(\mathbf q\cdot\mathbf v)=2\mathbf q\mathbf q^T\mathbf v, \end{equation} and \begin{equation} 2q_w(\mathbf q\times\mathbf v)=2q_w[\mathbf q]_\times\mathbf v. \end{equation} Adding the three contributions gives \begin{equation} R(q)= (q_w^2-\mathbf q^T\mathbf q)I_3 + 2\mathbf q\mathbf q^T + 2q_w[\mathbf q]_\times. \end{equation} \subsection*{Solution 2} Insert \begin{equation} \mathbf q= \begin{bmatrix}q_x\\q_y\\q_z\end{bmatrix}, \qquad [\mathbf q]_\times= \begin{bmatrix} 0 & -q_z & q_y\\ q_z & 0 & -q_x\\ -q_y & q_x & 0 \end{bmatrix}, \end{equation} into the compact matrix formula and expand entry by entry. Using $q_w^2+q_x^2+q_y^2+q_z^2=1$ simplifies the diagonal terms, yielding the stated matrix. \subsection*{Solution 3} For $q_w=q_z=1/\sqrt2$ and $q_x=q_y=0$, so \begin{equation} R(q)= \begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}. \end{equation} \subsection*{Solution 4} The columns are the images of the basis vectors, so \begin{equation} R(q)\mathbf e_1= \begin{bmatrix}0\\1\\0\end{bmatrix}, \qquad R(q)\mathbf e_2= \begin{bmatrix}-1\\0\\0\end{bmatrix}, \qquad R(q)\mathbf e_3= \begin{bmatrix}0\\0\\1\end{bmatrix}. \end{equation} Thus the $x$ axis rotates into $y$, the $y$ axis rotates into $-x$, and the $z$ axis is unchanged. \subsection*{Solution 5} Replacing $q$ by $-q$ changes $q_w$ to $-q_w$ and $\mathbf q$ to $-\mathbf q$. The terms in the matrix become \begin{equation} (-q_w)^2 - (-\mathbf q)^T(-\mathbf q) = q_w^2-\mathbf q^T\mathbf q, \end{equation} \begin{equation} 2(-\mathbf q)(-\mathbf q)^T = 2\mathbf q\mathbf q^T, \end{equation} and \begin{equation} 2(-q_w)[-\mathbf q]_\times = 2q_w[\mathbf q]_\times. \end{equation} Hence $R(-q)=R(q)$. \subsection*{Solution 6} A unit quaternion satisfies $q^{-1}=q^*$. The inverse active rotation matrix is therefore the transpose of the forward matrix. More directly, quaternion conjugation corresponds to reversing the rotation, so \begin{equation} R(q^*)=R(q)^{-1}=R(q)^T. \end{equation} \subsection*{Solution 7} Here $q_w=1/2$, $q_x=\sqrt3/2$, and $q_y=q_z=0$. Therefore, \begin{equation} R(q)= \begin{bmatrix} 1 & 0 & 0\\ 0 & -\frac12 & -\frac{\sqrt3}{2}\\ 0 & \frac{\sqrt3}{2} & -\frac12 \end{bmatrix}. \end{equation} This is the matrix for a $120^\circ$ active rotation about the $x$ axis. \subsection*{Solution 8} Here \begin{equation} q_w=\frac12, \qquad q_x=q_y=q_z=\frac12. \end{equation} Substituting into the component formula gives \begin{equation} R(q)= \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}. \end{equation} This cyclically permutes the coordinate axes: $\mathbf e_1\mapsto\mathbf e_2$, $\mathbf e_2\mapsto\mathbf e_3$, and $\mathbf e_3\mapsto\mathbf e_1$. \subsection*{Solution 9} For any matrix $A$, \begin{equation} A= \begin{bmatrix} A\mathbf e_1 & A\mathbf e_2 & A\mathbf e_3 \end{bmatrix}. \end{equation} Applying this to $A=R(q)$ shows that its columns are exactly the rotated basis vectors. \subsection*{Solution 10} The passive chain is \begin{equation} {}^{N}\mathbf v = C^N_A\,{}^{A}\mathbf v, \qquad {}^{A}\mathbf v = C^A_B\,{}^{B}\mathbf v. \end{equation} Substituting the second relation into the first gives \begin{equation} {}^{N}\mathbf v = C^N_A C^A_B\,{}^{B}\mathbf v, \end{equation} so \begin{equation} C^N_B = C^N_A C^A_B. \end{equation} \subsection*{Solution 11} The matrix is the $+90^\circ$ active rotation about the $z$ axis, so one choice is \begin{equation} q=\frac{1+\mathbf k}{\sqrt2}. \end{equation} The equally valid choice $-q$ represents the same matrix. \subsection*{Solution 12} The matrix is a $180^\circ$ rotation about the $x$ axis, so one choice is \begin{equation} q=\mathbf i. \end{equation} The trace is $1-1-1=-1$, so the formula \begin{equation} q_w = \frac12\sqrt{1+\operatorname{tr}(R)} \end{equation} returns $q_w=0$. Dividing by $4q_w$ is therefore numerically dangerous. One must instead use a branch based on the large value of $q_x$. \subsection*{Solution 13} For a unit quaternion the corresponding map is a proper rotation, so it preserves dot products and orientation. Hence the matrix is orthogonal and has determinant $+1$. Algebraically one can verify \begin{equation} R(q)^T R(q)=I_3 \end{equation} from the compact formula or from the fact that the inverse rotation is given by $q^*$. \subsection*{Solution 14} The active formula $\mathbf v'=R(q)\mathbf v$ rotates the vector itself in a fixed basis. The passive formula ${}^{A}\mathbf v=C^A_B\,{}^{B}\mathbf v$ changes the coordinates used to describe the same geometric vector. The same orientation change may be described either way, but the physical interpretation is different. \subsection*{Solution 15} The student has confused storage layout with component meaning. In the scalar-first formula, the first symbol is the scalar component $q_w$. If a library stores $[q_x,q_y,q_z,q_w]^T$, the entries must be remapped before the formula is used. Otherwise the resulting matrix corresponds to a different quaternion. \section*{License} This article is an original synthesis prepared for PhysicsLibrary and intended for release under CC BY-SA 4.0.