Rotating Vectors with Quaternions: Examples, Exercises, and Solutions

This entry is the self-study companion to the PhysicsLibrary article rotating vectors with quaternions. The convention throughout is Hamilton multiplication in a right-handed frame, with scalar-first quaternion display and active vector rotation

$\displaystyle \boxed{\mathbf v'=q\mathbf vq^*.}$

(1)

The quaternion is assumed to have unit norm unless a problem states otherwise.

Exercises

Pure-quaternion representation. Write the vector

$\displaystyle \mathbf v=2\hat{\mathbf x}-3\hat{\mathbf y}+4\hat{\mathbf z}$
as a scalar-first pure quaternion.
$+90^\circ$ about . Use quaternion multiplication to rotate

$\displaystyle \mathbf v=\hat{\mathbf x}$
through $+90^\circ$ about . Show every quaternion multiplication step.
$+90^\circ$ about . Rotate

$\displaystyle \mathbf v=\hat{\mathbf y}$
through $+90^\circ$ about and determine the resulting Cartesian vector.
A vector parallel to the axis. Let

$\displaystyle q=\cos 35^\circ+\mathbf k\sin35^\circ.$
What physical rotation does represent? Apply it to $\mathbf v=5\mathbf k$ and explain the result geometrically.
Rotation through $180^\circ$ . Rotate

$\displaystyle \mathbf v=2\mathbf i-3\mathbf j+4\mathbf k$
through $180^\circ$ about .
A symmetric $120^\circ$ rotation. Let

$\displaystyle \widehat{\mathbf u}=\frac{\mathbf i+\mathbf j+\mathbf k}{\sqrt3}, \qquad \theta=120^\circ.$
Construct and show that the active rotation sends

$\displaystyle \mathbf i\longrightarrow\mathbf j.$
Rodrigues formula. Using

$\displaystyle \widehat{\mathbf u}=\mathbf k, \qquad \theta=60^\circ, \qquad \mathbf v=2\mathbf i+\mathbf j+3\mathbf k,$
evaluate the rotated vector using Rodrigues' formula.
Quaternion sign ambiguity. Prove directly that and rotate every vector identically under the sandwich action.
Inverse rotation. If

$\displaystyle \mathbf v'=q\mathbf vq^*,$
prove that

$\displaystyle \mathbf v=q^*\mathbf v'q.$
Norm preservation. Use quaternion norm multiplicativity to prove that a unit quaternion preserves the Euclidean magnitude of every rotated vector.
Deriving Rodrigues' formula. Starting from

$\displaystyle q=c+s\widehat{\mathbf u}, \qquad q^*=c-s\widehat{\mathbf u},$
derive

$\displaystyle \mathbf v' =\mathbf v\cos\theta +(\widehat{\mathbf u}\times\mathb... ...n\theta +\widehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v)(1-\cos\theta).$
Vector-only implementation. Show that, for a unit quaternion $q=(q_w,\mathbf q)$ ,

$\displaystyle \mathbf v' =\mathbf v +2q_w(\mathbf q\times\mathbf v) +2\mathbf q\times(\mathbf q\times\mathbf v).$
Then rewrite it using

$\displaystyle \mathbf t=2(\mathbf q\times\mathbf v).$
Active versus passive. A fixed physical vector has coordinates

$\displaystyle \mathbf v_A=\hat{\mathbf x}$
in frame . Frame is obtained by rotating the coordinate axes $+90^\circ$ about relative to . Using the same unit quaternion that would actively rotate vectors by $+90^\circ$ about , determine the coordinates $\mathbf v_B$ .
Scalar-first and scalar-last software arrays. The quaternion for a $+90^\circ$ rotation about is displayed in PhysicsLibrary scalar-first form as

$\displaystyle \left[\frac{\sqrt2}{2},0,0,\frac{\sqrt2}{2}\right]^T.$
Write the same quaternion in scalar-last storage. Explain whether the Hamilton multiplication law changes.
Debugging the wrong sandwich order. A program intends to actively rotate $\hat{\mathbf x}$ by $+90^\circ$ about , but it computes

$\displaystyle q^*\mathbf i q.$
Predict the output and explain the error.
Nonunit quaternion pitfall. Let

$\displaystyle r=2q,$
where is unit. Show that

$\displaystyle r\mathbf v r^*=4q\mathbf vq^*.$
Explain why a nonunit quaternion should not be used with the conjugate sandwich formula when a pure rotation is intended.

Solutions

1. Pure-quaternion representation

The scalar part is zero, so

$\displaystyle \boxed{ \mathbf v \longleftrightarrow \begin{bmatrix} 0&2&-3&4 \end{bmatrix}^{T}. }$

(2)

2. $+90^\circ$ about

The rotation quaternion is

$\displaystyle q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf k, \qquad q^*=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\mathbf k.$

First,

$\begin{equation*}\begin{aligned} q\mathbf i &=\frac{\sqrt2}{2}\mathbf i +\frac{\... ...thbf i \ &=\frac{\sqrt2}{2}(\mathbf i+\mathbf j). \end{aligned}\end{equation*}$

Then

$\begin{equation*}\begin{aligned} q\mathbf i q^* &=\frac12(\mathbf i+\mathbf j)(1... ...f i+\mathbf j+\mathbf j-\mathbf i) \ &=\mathbf j. \end{aligned}\end{equation*}$

Therefore

$\displaystyle \boxed{\hat{\mathbf x}\rightarrow\hat{\mathbf y}.}$

(5)

3. $+90^\circ$ about

By the right-hand rule, a positive quarter-turn about $+x$ carries $+y$ toward $+z$ . Thus

$\displaystyle \boxed{\hat{\mathbf y}\rightarrow\hat{\mathbf z}.}$

(6)

The quaternion is

$\displaystyle q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf i.$

Direct multiplication gives the same result.

4. Vector parallel to the axis

The quaternion phase is $35^\circ$ , so the physical rotation angle is

$\displaystyle \theta=70^\circ.$

The axis is $+z$

. Since $\mathbf v=5\mathbf k$ is parallel to the axis,

$\displaystyle \boxed{q(5\mathbf k)q^*=5\mathbf k.}$

(7)

A rotation leaves every vector on its own axis unchanged.

5. Rotation through $180^\circ$

A $180^\circ$ rotation about $+z$ reverses the $x$ and $y$ components and leaves the $z$ component unchanged. Therefore

$\displaystyle \boxed{ \mathbf v'=-2\mathbf i+3\mathbf j+4\mathbf k. }$

(8)

6. Symmetric $120^\circ$ rotation

Since

$\displaystyle \cos60^\circ=\frac12, \qquad \sin60^\circ=\frac{\sqrt3}{2},$

we obtain

$\displaystyle \boxed{ q=\frac12(1+\mathbf i+\mathbf j+\mathbf k). }$

(9)

Using Rodrigues' formula with $\widehat{\mathbf u}=(1,1,1)/\sqrt3$ and $\mathbf v=\mathbf i$ gives

$\displaystyle \boxed{\mathbf v'=\mathbf j.}$

(10)

Thus this $120^\circ$ rotation cyclically carries the coordinate axes $x\rightarrow y\rightarrow z\rightarrow x$ .

7. Rodrigues formula

For $\widehat{\mathbf u}=\mathbf k$ ,

$\displaystyle \widehat{\mathbf u}\times\mathbf v =\mathbf k\times(2\mathbf i+\mathbf j+3\mathbf k) =2\mathbf j-\mathbf i,$

and

$\displaystyle \widehat{\mathbf u}\cdot\mathbf v=3.$

With $\cos60^\circ=1/2$ and $\sin60^\circ=\sqrt3/2$ ,

$\begin{equation*}\begin{aligned} \mathbf v' ={}&\frac12(2\mathbf i+\mathbf j+3\m... ... +\left(\frac12+\sqrt3\right)\mathbf j +3\mathbf k. \end{aligned}\end{equation*}$

Hence

$\displaystyle \boxed{ \mathbf v' =\left(1-\frac{\sqrt3}{2}\right)\mathbf i +\left(\frac12+\sqrt3\right)\mathbf j +3\mathbf k. }$

(12)

8. Quaternion sign ambiguity

Because $(-q)^*=-q^*$ ,

$\displaystyle (-q)\mathbf v(-q)^* =(-q)\mathbf v(-q^*) =q\mathbf vq^*.$

(13)

Therefore

$\displaystyle \boxed{q\text{ and }-q\text{ generate the same vector rotation}.}$

(14)

9. Inverse rotation

Start with

$\displaystyle \mathbf v'=q\mathbf vq^*.$

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

$\displaystyle \boxed{\mathbf v=q^*\mathbf v'q.}$

(16)

10. Norm preservation

For unit $q$ ,

$\displaystyle \lVert q\rVert=\lVert q^*\rVert=1.$

Hence

$\begin{equation*}\begin{aligned} \lVert\mathbf v'\rVert &=\lVert q\mathbf vq^*\r... ...Vert\,\lVert q^*\rVert \ &=\lVert\mathbf v\rVert. \end{aligned}\end{equation*}$

So quaternion conjugation by a unit quaternion preserves Euclidean vector length.

11. Deriving Rodrigues' formula

Let

$\displaystyle q=c+s\widehat{\mathbf u}.$

The first product is

$\displaystyle q\mathbf v = \left( -s\widehat{\mathbf u}\cdot\mathbf v, \;c\mathbf v+s\widehat{\mathbf u}\times\mathbf v \right).$

(18)

Multiplication by $q^*=(c,-s\widehat{\mathbf u})$ cancels the scalar part and gives

$\begin{equation*}\begin{aligned} \mathbf v' ={}&(c^2-s^2)\mathbf v +2cs(\widehat... ...ehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v). \end{aligned}\end{equation*}$

Using

$\displaystyle c^2-s^2=\cos\theta, \qquad 2cs=\sin\theta, \qquad 2s^2=1-\cos\theta,$

produces

$\displaystyle \boxed{ \mathbf v' =\mathbf v\cos\theta +(\widehat{\mathbf u}\tim... ...\theta +\widehat{\mathbf u}(\widehat{\mathbf u}\cdot\mathbf v)(1-\cos\theta). }$

(20)

12. Vector-only implementation

Write $q=(q_w,\mathbf q)$ . Expanding the sandwich product and collecting cross-product terms gives

$\displaystyle \boxed{ \mathbf v' =\mathbf v +2q_w(\mathbf q\times\mathbf v) +2\mathbf q\times(\mathbf q\times\mathbf v). }$

(21)

$\displaystyle \mathbf t=2(\mathbf q\times\mathbf v),$

(22)

then

$\displaystyle \boxed{ \mathbf v'=\mathbf v+q_w\mathbf t+\mathbf q\times\mathbf t. }$

(23)

13. Active versus passive

The active quaternion for $+90^\circ$ about $+z$ is

$\displaystyle q=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathbf k.$

A passive coordinate change uses the inverse action,

$\displaystyle \mathbf v_B=q^*\mathbf v_Aq.$

For $\mathbf v_A=\mathbf i$ ,

$\displaystyle \boxed{\mathbf v_B=-\mathbf j.}$

(24)

Therefore the fixed vector has coordinates

$\displaystyle \boxed{ [0,-1,0]^T }$

(25)

in the rotated frame.

14. Scalar-first and scalar-last arrays

Scalar-first storage is

$\displaystyle \left[\frac{\sqrt2}{2},0,0,\frac{\sqrt2}{2}\right]^T.$

Scalar-last storage places the scalar at the end:

$\displaystyle \boxed{ \left[0,0,\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right]^T. }$

(26)

The Hamilton multiplication law does not change. Only the array indexing changes.

15. Debugging the wrong sandwich order

The expression

$\displaystyle q^*\mathbf i q$

applies the inverse rotation. Therefore the output is

$\displaystyle \boxed{-\mathbf j.}$

(27)

The intended active formula was $q\mathbf i q^*$ .

16. Nonunit quaternion pitfall

With $r=2q$ ,

$\displaystyle r^*=2q^*.$

Therefore

$\displaystyle r\mathbf vr^* =(2q)\mathbf v(2q^*) =4q\mathbf vq^*.$

(28)

Thus

$\displaystyle \boxed{ \lVert r\mathbf vr^*\rVert=4\lVert\mathbf v\rVert. }$

(29)

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.

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.

Bibliography

1: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public-domain historical source.
2: A. S. Hathaway, A Primer of Quaternions, 1896. Public-domain text; Project Gutenberg edition and L^ATEX source available. Project Gutenberg edition

License

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