Unit Dual Quaternions, Line Transformations and Plücker Coordinates

BEWARE! THIS DRAFT HAS SOME SYMBOLS THAT DO NOT MATCH THOSE WE HAVE PREVIOUSLY USED. THE LETTER USED FOR TRANSLATION WAS $\vec{d}$ BUT HERE IT IS $\vec{t}$. THE SYMBOL FOR PARALLEL TRANSLATION TO THE AXIS WAS $\vec{d_{\Vert}}$ BUT HERE IT IS $\vec{d}$ AND THEREFORE $d = \Vert\vec{d}\Vert$ SO $\vec{d} = d\vec{l}$.

In Parallelograms, Plücker Coordinates and Dual Numbers we had a line $L$ with Plücker coordinates $\big(\vec{l}, \vec{m} \big)$ where $\vec{l}$ is a unit vector in the direction of line $L$ and $\vec{m} = \vec{p}\times\vec{l}$ where $\vec{p}$ can be chosen perpendicular to $L$ so $\vec{l}\cdot\vec{m} = 0$. Let the dual quaternion of these Plücker coordinates be:

$$\pmb{l} + \epsilon\pmb{m} = \vec{l} + \epsilon\vec{m}\tag{1}$$

The standard dual quaternion $\pmb{r_0} + \frac{\epsilon}{2}\vec{t}\pmb{r_0}$ for translation and rotation is not suited for Plücker Coordinates since it assumes a rotation axis through the origin and that means $\vec{p_{\bot}} = 0$ and $\vec{m} = 0$ which isn't what we want. What we want is this displacement to occur around axis $\mathrm{L}$ which does not pass through the origin but does pass through some point $\mathrm{P}$. We can obtain the dual quaternion for this axis offset by $\vec{p}$ by applying the translation dual quaternion $1 + \epsilon\frac{\vec{p}}{2}$ to the rotation dual quaternion $\pmb{\hat{r}}$ for the unit vector $\vec{l}$. However, from the post Unit Dual Quaternion For Translation And Rotation we can simply change this unit dual quaternion into the regular unit rotation quaternion $\pmb{r_0}$:

$$\pmb{\hat{r}} = \pmb{r_0} + \epsilon(0)$$ Recall that a regular rotation quaternion in terms of $\theta$ was derived in Extracting the Cross and Dot Products from the Quaternion Rotation Operator and for the unit vector $\vec{l}$ would be:

$$\pmb{r_0} = \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l}$$

Since this will be a transformation operation of the regular quaternion $\pmb{r_0}$, this quaternion has to be multiplied on both sides by the translation quaternion. We have seen this "quaternion sandwich operation" before with rotations but now we are doing it with a translation. The left side is multiplied by $1 + \epsilon\frac{\vec{p}}{2}$ and right is multiplied by the conjugate $1 - \epsilon\frac{\vec{p}}{2}$:

$$\begin{align} \Big(1 + &\epsilon\frac{\vec{p}}{2} \Big)\pmb{r_0}\Big(1 - \epsilon\frac{\vec{p}}{2} \Big) = \Big(1 + \epsilon\frac{\vec{p}}{2} \Big)\Big( \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l}\Big) \Big(1 - \epsilon\frac{\vec{p}}{2} \Big)\\ &= \bigg(\cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \frac{\epsilon}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{p} + \frac{\epsilon}{2}\sin{\Big(\frac{\theta}{2}\Big)}\vec{p}\vec{l}\bigg)\Big(1 - \epsilon\frac{\vec{p}}{2} \Big)\\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \frac{\epsilon}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{p} + \frac{\epsilon}{2}\sin{\Big(\frac{\theta}{2}\Big)}\vec{p}\vec{l} - \frac{\epsilon}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{p} - \frac{\epsilon}{2}\sin{\Big(\frac{\theta}{2}\Big)}\vec{l}\vec{p} \\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{p} \times \vec{l}\Big) \tag{2} \end{align}$$

However, to obtain a general (e.g. screw) displacement dual qauternion, we need to use another translation quaternion, $1 + \epsilon\frac{\vec{d}}{2}$, for the displacement $\vec{d}$ parallel to the rotation axis $L$ to equation $(2)$. Here we are concatenating/combine operations instead of transforming one operation, so there is only a simple one-side multiplication that needs to be done.

$$\begin{align} \bigg(1 + &\epsilon\frac{\vec{d}}{2}\bigg)\bigg( \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{p} \times \vec{l}\Big) \bigg) \\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\bigg(\sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{p} \times \vec{l}\Big) + \frac{1}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{d} + \frac{1}{2}\sin{\Big(\frac{\theta}{2}\Big)}\vec{d}\vec{l}\bigg)\tag{3} \end{align}$$ and since $\vec{d}$ is parrallel to $\vec{l}$ then $\vec{d}\vec{l} = \big( \vec{d}\times\vec{l}\big) - \vec{d}\cdot\vec{l} = -d$ and equation $(3)$ becomes

$$\begin{align} & \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\bigg(\sin{\Big(\frac{\theta}{2}\Big)}\vec{m} + \frac{\vec{d}}{2}\cos{\Big(\frac{\theta}{2}\Big)} - \frac{d}{2}\sin{\Big(\frac{\theta}{2}\Big)}\bigg)\\ &= \cos{\Big(\frac{\theta}{2}\Big)} - \epsilon\frac{d}{2}\sin{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\Big(\sin{\Big(\frac{\theta}{2}\Big)}\vec{m} + \frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{l} \Big)\tag{4} \end{align}$$

AN ALTERNATIVE GEOMETRIC DERIVATION OF EQUATION $(4)$

In Figure 1, the screw axis $\mathrm{L}$ has unit direction vector $\vec{l}$ and is represented by Plücker coordinates $(\vec{m}, \vec{l})$ where $\vec{m} = \vec{p}\times\vec{l}$ and $\vec{p}$ is from the origin and can be to any point $\mathrm{P}$ on line $\mathrm{L}$. The following formulas are general for any point $\mathrm{P}$ but we will choose $\vec{p}$ perpendicular to $\mathrm{L}$ so that $\vec{p} = \vec{p}_{\bot}$. Furthermore, $\vec{t}$ is the translation vector of any point by the screw displacement so the distance traveled parallel to $\mathrm{L}$ is $d = \vec{t}\cdot\vec{l}$.

FIGURE 1

To find $\vec{m}$ in terms of the screw displacement, start by screw displacing the origin point of the coordinate system at the tail of $\vec{p}_{\bot}$ as shown in Figure 1. This establishes the geometry we need to work with in finding the equation for the location of point $\mathrm{P}$ at the tip of $\vec{p}_{\bot}$ only in terms of $\vec{t}$, $\vec{l}$, and the rotation angle $\theta$. Next, proceed in a similar manner as the derivation of equations $(2)$ to $(5)$ in post Mozzi-Chasles' Screw Theorem, Half-Turn Equivalence and Combining Successive Screw Displacements.

For $\vec{p_2}+\vec{p_1}$, begin by using the relation

$$\tan{\Big(\frac{\theta}{2}\Big)} = \frac{\Vert \vec{p_2} - \vec{p_1} \Vert}{\Vert \vec{p_2} + \vec{p_1} \Vert}$$

then cross-multiplying $\vec{p_2}-\vec{p_1}$ with the unit vector $\vec{l}$ to get:

$$\vec{p_2}+\vec{p_1} = \cot{\Big(\frac{\theta}{2}\Big)} \Big(\vec{p_2}-\vec{p_1}\Big) \times \vec{l}$$

The vector that is half the length of $\vec{p_2}+\vec{p_1}$ but in the opposite direction from the bisecting point $\mathrm{B}$ to point $\mathrm{P}$ at the tip of $\vec{p}_{\bot}$ is:

$$-\frac{1}{2}\Big(\vec{p_2}+\vec{p_1}\Big) = \frac{1}{2} \cot{\Big(\frac{\theta}{2}\Big)}\vec{l}\times\Big(\vec{p_2}-\vec{p_1}\Big)$$

and because

$$\Big(\vec{p_2}-\vec{p_1}\Big) = \vec{t} - \Big(\vec{t}\cdot\vec{l} \Big) \vec{l}$$

then we get

$$-\frac{1}{2}\Big(\vec{p_2}+\vec{p_1}\Big) = \frac{1}{2} \cot{\Big(\frac{\theta}{2}\Big)}\vec{l}\times\vec{t}$$

From Figure 1 we see that three successive vectors, the first starting from the coordinate system origin to the mid-point of $\vec{t}$, then the second from the mid-point of $\vec{t}$ to point $\mathrm{B}$ and finally the third from point $\mathrm{B}$ towards point $\mathrm{P}$, give the same result as $\vec{p}_{\bot}$. So we have:

$$\begin{align} \vec{p}_{\bot} &= \frac{1}{2}\vec{t} - \frac{1}{2}\Big(\vec{t}\cdot\vec{l} \Big) \vec{l} - \frac{1}{2}\Big(\vec{p_2}+\vec{p_1}\Big) \\ &= \frac{1}{2}\vec{t} - \frac{1}{2}\Big(\vec{t}\cdot\vec{l} \Big) \vec{l} + \frac{1}{2} \cot{\Big(\frac{\theta}{2}\Big)}\vec{l}\times\vec{t}\\ &= \frac{1}{2} \bigg(\vec{t} - \Big(\vec{t}\cdot\vec{l} \Big) \vec{l} + \cot{\Big(\frac{\theta}{2}\Big)}\vec{l}\times\vec{t} \bigg) \end{align}$$

Finally, for $\vec{m}$ we have:

$$\vec{m} = \vec{p}_{\bot} \times \vec{l} = \frac{1}{2} \bigg(\vec{t}\times \vec{l} + \cot{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\times\vec{t}\Big)\times \vec{l} \bigg)\tag{5}$$

We know from post Unit Dual Quaternion For Translation And Rotation that we use the dual qauternion rotation and translation operator:

$$\pmb{r_0} + \frac{\epsilon}{2}\vec{t}\pmb{r_0}\tag{6}$$

We recall from equation $(23)$ of the post Extracting the Cross and Dot Products from the Quaternion Rotation Operator that:

$$\pmb{r_0} = \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l}\tag{7}$$

Substituting equation $(7)$ into $(6)$ gives:

$$\begin{align} \pmb{r_0} + \frac{\epsilon}{2}\vec{t}\pmb{r_0} &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)}\vec{t} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{t}\vec{l}\bigg)\\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)}\vec{t} - \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{t}\cdot\vec{l}\Big) + \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{t}\times\vec{l}\Big)\bigg)\tag{8}\\ \end{align}$$

Now let's multiply both sides of equation $(5)$ by $\sin{\big(\frac{\theta}{2} \big)}$:

$$\sin{\Big(\frac{\theta}{2} \Big)}\vec{m} = \frac{1}{2} \bigg(\sin{\Big(\frac{\theta}{2} \Big)}\Big(\vec{t}\times \vec{l}\Big) + \cos{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\times\vec{t}\Big)\times \vec{l} \bigg)\tag{9}$$

rearranging $(9)$:

$$\sin{\Big(\frac{\theta}{2} \Big)}\Big(\vec{t}\times \vec{l}\Big) = 2\sin{\Big(\frac{\theta}{2} \Big)}\vec{m} - \cos{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\times\vec{t}\Big)\times \vec{l}\tag{10}$$

Substituting $(10)$ into $(8)$ gives:

$$\begin{align} \pmb{r_0} + \frac{\epsilon}{2}&\vec{t}\pmb{r_0}\\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l}\\ &+ \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)}\vec{t} - \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{t}\cdot\vec{l}\Big) + 2\sin{\Big(\frac{\theta}{2} \Big)}\vec{m} - \cos{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\times\vec{t}\Big)\times \vec{l}\bigg)\\\\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l}\\ &+ \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)}\vec{t} - \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{t}\cdot\vec{l}\Big) + 2\sin{\Big(\frac{\theta}{2} \Big)}\vec{m} + \cos{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\cdot\vec{t}\Big)\vec{l} - \cos{\Big(\frac{\theta}{2}\Big)}\vec{t}\bigg)\\\\ &= \cos{\Big(\frac{\theta}{2}\Big)} - \frac{\epsilon}{2}\sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{t}\cdot\vec{l}\Big)\\ &+ \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\bigg( \sin{\Big(\frac{\theta}{2} \Big)}\vec{m} + \frac{1}{2}\cos{\Big(\frac{\theta}{2}\Big)}\Big(\vec{l}\cdot\vec{t}\Big)\vec{l}\bigg)\\ \\ &= \cos{\Big(\frac{\theta}{2}\Big)} - \epsilon\frac{d}{2}\sin{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\bigg(\sin{\Big(\frac{\theta}{2}\Big)}\vec{m} + \frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)}\vec{l} \bigg)\tag{11} \end{align}$$

Equations $(4)$ and $(11)$ are the same and can be simplified further since a Taylor series expansion of every function $f$ of dual numbers yeilds the rule that:

$$f\big(a + \epsilon b \big) = f\big(a\big) + \epsilon b f^{'}\big(a\big)$$

hence

$$\begin{align} \cos{\Big(\frac{\theta + \epsilon d}{2}\Big)} &= \cos{\Big(\frac{\theta}{2}\Big)} - \epsilon\frac{d}{2}\sin{\Big(\frac{\theta}{2}\Big)}\\ \sin{\Big(\frac{\theta + \epsilon d}{2}\Big)} &= \sin{\Big(\frac{\theta}{2}\Big)} + \epsilon\frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)} \end{align}$$

Letting $\sigma = \pmb{r_0} + \frac{\epsilon}{2}\vec{t}\pmb{r_0}$ and substituting the two equations above into $(4)$ or $(11)$ gives us:

$$\begin{align} \sigma &= \pmb{r_0} + \frac{\epsilon}{2}\vec{t}\pmb{r_0}\\ &=\cos{\Big(\frac{\theta + \epsilon d}{2}\Big)} + \bigg(\sin{\Big(\frac{\theta}{2}\Big)} + \epsilon\frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)}\bigg)\vec{l} + \sin{\Big(\frac{\theta}{2}\Big)}\epsilon\vec{m}\\ &=\cos{\Big(\frac{\theta + \epsilon d}{2}\Big)} + \bigg(\sin{\Big(\frac{\theta}{2}\Big)} + \epsilon\frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)}\bigg)\vec{l} + \bigg(\sin{\Big(\frac{\theta}{2}\Big)} + \epsilon\frac{d}{2}\cos{\Big(\frac{\theta}{2}\Big)}\bigg)\epsilon\vec{m}\\ &= \cos{\Big(\frac{\theta + \epsilon d}{2}\Big)} + \sin{\Big(\frac{\theta + \epsilon d}{2}\Big)}\Big(\vec{l} + \epsilon\vec{m} \Big)\tag{12} \end{align}$$

By introducing the following dual angle and dual vector

$$\begin{align} \bar{\theta} &= \theta + \epsilon d \\ \bar{\vec{l}} &= \vec{l} + \epsilon\vec{m} \end{align}$$

then equation $(12)$ can be rewritten into a very succinct form:

$$\sigma = \cos{\Big(\frac{\bar{\theta}}{2}\Big)} + \bar{\vec{l}}\sin{\Big(\frac{\bar{\theta}}{2}\Big)}\tag{13}$$

IGNORE BELOW.

Applying the dual quaternion displacement operator to the Plücker Coordinates

$$\begin{align} \pmb{\hat{s}}\big(\vec{l} + \epsilon\vec{m}\big)\pmb{\bar{\hat{s}}^{*}} &= \big(\pmb{r_0} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\big)\big( \vec{l} + \epsilon\vec{m}\big)\big( \pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{{r_0}^{*}}\vec{d}^{*}\big) \\ &= \big(\pmb{r_0}\vec{l} + \epsilon\pmb{r_0}\vec{m} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\vec{l}\big)\big( \pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{{r_0}^{*}}\vec{d}^{*}\big)\\ &= \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\vec{d}^{*} + \epsilon\pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\\ &= \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} + \epsilon\bigg(\pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \frac{1}{2}\Big(\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\vec{d}^{*} + \vec{d}\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\Big) \bigg)\tag{2} \\ \end{align}$$

Letting $\vec{l^{'}} + \epsilon\vec{m^{'}} = \pmb{\hat{s}}\big(\vec{l} + \epsilon\vec{m}\big)\pmb{\bar{\hat{s}}^{*}}$ then:

$$\begin{align} \vec{l^{'}} &= \pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\tag{3} \\ \vec{m^{'}} &= \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \frac{1}{2}\Big( \pmb{r_0} \vec{l} \pmb{{r_0}^{*}} \vec{d}^{*} + \vec{d}\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\Big) \\ &= \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \frac{1}{2}\bigg( \Big(\pmb{r_0} \vec{l} \pmb{{r_0}^{*}}\Big) \cdot \vec{d} - \Big(\pmb{r_0} \vec{l} \pmb{{r_0}^{*}} \times \vec{d}\Big) - \vec{d} \cdot \Big(\pmb{r_0} \vec{l} \pmb{{r_0}^{*}}\Big) + \Big(\vec{d} \times \pmb{r_0} \vec{l} \pmb{{r_0}^{*}} \Big)\bigg) \\ &= \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \vec{d} \times \pmb{r_0} \vec{l} \pmb{{r_0}^{*}}\tag{4} \\ \end{align}$$

Therfore

$$\vec{l^{'}} + \epsilon\vec{m^{'}} = \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} + \epsilon\Big( \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \vec{d} \times \pmb{r_0} \vec{l} \pmb{{r_0}^{*}} \Big)\tag{5}$$

In terms of $\theta$ we use equation $(23)$ from Extracting the Cross and Dot Products from the Quaternion Rotation Operator and have

$$\begin{align} \vec{l^{'}} &= \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} = \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} \\ \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{m} \end{align}$$ which are substituted into $(4)$ above to give:

$$\begin{align} \vec{m^{'}} &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{m} + \cos{\Big(\frac{\theta}{2}\Big)}\vec{d} + \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{d}\times\vec{l}\Big)\tag{6} \\ \end{align}$$

Therefore:

$$\vec{l^{'}} + \epsilon\vec{m^{'}} = \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{l} + \epsilon\bigg(\cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{m} + \cos{\Big(\frac{\theta}{2}\Big)}\vec{d} + \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{d}\times\vec{l}\Big)\bigg)\tag{7}$$

Since $\vec{m} = \vec{p}\times\vec{l}$ in equation $(6)$, let's first look at what happens to $\vec{p}$ with transformation. From equation $(7)$ in Unit Dual Quaternion For Translation And Rotation, we have for $1 + \epsilon\vec{p}$:

$$1 + \epsilon\vec{p^{'}} = 1 + \epsilon\big(\pmb{r_0}\vec{p}\pmb{{r_0}^{*}} + \vec{d}\big)$$

and therefore:

$$\vec{p^{'}} = \pmb{r_0}\vec{p}\pmb{{r_0}^{*}} + \vec{d}$$

Now since vectors $\vec{l}$ and $\vec{p}$ are rotated around the same axis by the same amount then their cross-product $\vec{m}$ is as well so $\pmb{r_0}\vec{m}\pmb{{r_0}^{*}} = \pmb{r_0}\big(\vec{p}\times\vec{l}\big)\pmb{{r_0}^{*}} = \pmb{r_0}\vec{p}\pmb{{r_0}^{*}} \times \pmb{r_0}\vec{l}\pmb{{r_0}^{*}}$ and:

$$\begin{align} \vec{m^{'}} &= \vec{p^{'}}\times\vec{l^{'}} \\ &= \Big( \pmb{r_0}\vec{p}\pmb{{r_0}^{*}} + \vec{d} \Big) \times \Big( \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} \Big) \\ &= \pmb{r_0}\vec{p}\pmb{{r_0}^{*}} \times \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} + \vec{d} \times \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} \\ &= \pmb{r_0}\vec{m}\pmb{{r_0}^{*}} + \vec{d} \times \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} \end{align}$$

Since the cross-product of two pure quaternions $\vec{a}$ and $\vec{b}$ is $\vec{a}\times\vec{b} = \frac{1}{2}\Big(\vec{b}\vec{a}^{*} + \vec{a}\vec{b} \Big)$ then:

$$\vec{d} \times \pmb{r_0}\vec{l}\pmb{{r_0}^{*}} = \frac{1}{2}\Big(\pmb{r_0}\vec{l}\pmb{{r_0}^{*}}\vec{d}^{*} + \vec{d}^{*}\pmb{r_0}\vec{l}\pmb{{r_0}^{*}} \Big)$$