Parallelograms, Plucker Coordinates and Dual Numbers

Figure $\mathrm{1}$ will be used to point out some properties of parallelograms that have bearing on Plucker coordinates. The two parallelograms $\mathrm{1}$ and $\mathrm{2}$ shown in the figure have the same base lengths $\mathrm{b}$ and the same length heights $\mathrm{h}$.

Figure 1

The area $\mathrm{A}$ of a parallelogram is just:

$$\mathrm{A} = \mathrm{b}\mathrm{h}\tag{1}$$

So both parallelograms have the same area:

$$\mathrm{A_1} = \big(\mathrm{a} + \mathrm{b}\big)\mathrm{h} - \frac{1}{2}\big(\mathrm{a}\mathrm{h} + \mathrm{a}\mathrm{h}\big) = \mathrm{b}\mathrm{h}$$ $$\mathrm{A_2} = \big(\mathrm{c} + \mathrm{b}\big)\mathrm{h} - \frac{1}{2}\big(\mathrm{c}\mathrm{h} + \mathrm{c}\mathrm{h}\big) = \mathrm{b}\mathrm{h}$$

Therefore

$$\mathrm{A_1} = \mathrm{A_2}$$

This will be the case for all parallelograms of same length bases and heights independent of the lengths like $\mathrm{a}$ and $\mathrm{c}$ and therefore independent of the length of their diagonals like $\mathrm{d_1}$ and $\mathrm{d_2}$.

Since the acute angle between a diagonal $\mathrm{d}$ and it's base $\mathrm{b}$ is $\theta$ then the area of a parallelogram can be written as:

$$\mathrm{A} = \mathrm{b}\mathrm{d}\sin{(\theta)}\tag{2}$$

If we associate the vector $\vec{b}$ with the base $\mathrm{b}$ and vector $\vec{d}$ with the diagonal $\mathrm{d}$ then the area of the parallelogram is the same as the absolute value of $\vec{b} \times \vec{d}$:

$$\Vert \vec{b} \times \vec{d} \Vert = \Vert \vec{b} \Vert\Vert \vec{d} \Vert \sin{(\theta)} = \mathrm{A}\tag{3}$$

Which also means from our two parallelograms that:

$$\begin{align} \Vert \vec{b} \Vert\Vert \vec{d_1} \Vert \sin{(\theta_1)} &= \Vert \vec{b} \Vert\Vert \vec{d_2} \Vert \sin{(\theta_2)} \\ \Vert \vec{b} \times \vec{d_1} \Vert &= \Vert \vec{b} \times \vec{d_2} \Vert \end{align}$$

Let the direction of the cross products be given by the unit vector $\vec{u_m}$ then:

$$\vec{b} \times \vec{d} = \Vert \vec{b} \Vert\Vert \vec{d} \Vert \sin{(\theta)} \vec{u_m}$$

If both cross products have the same direction $\vec{u_m}$ we then have:

$$\vec{b} \times \vec{d_1} = \vec{b} \times \vec{d_2}$$

Figure $\mathrm{2}$ shows a line $\mathrm{L}$ in $\mathrm{3D}$ space with vectors $\vec{p}$ and $\vec{p_{\perp}}$ to points on $\mathrm{L}$, a vector $\vec{l}$ parallel with line $\mathrm{L}$ and the vector $\vec{m} = \vec{p} \times \vec{l}$.

Figure 2

The vector $\vec{p_{\perp}}$ happens to be perpendicular to $\mathrm{L}$ but from our previous discussion, a vector $\vec{q}$ to any point on $\mathrm{L}$ will satisfy the following equality:

$$\vec{q} \times \vec{l} = \vec{p} \times \vec{l} = \vec{m}$$

Now if you have $\vec{l}$ and $\vec{m}$ then how do you find $\vec{q}$ given that $\vec{q}$ can be to any point on the line $\mathrm{L}$? We can make this well-defined by choosing $\vec{q} = \vec{p_{\perp}}$. With this choice, $\vec{l}$, $\vec{p_{\perp}}$ and $\vec{m}$ are orthogonal thus $\vec{l} \cdot \vec{p_{\perp}} = 0$ and vector $\vec{l} \times \vec{m}$ is in the same direction as $\vec{p_{\perp}}$ with magnitude

$$\begin{align} \Vert \vec{l} \times \vec{m} \Vert &= \Vert \vec{l} \Vert\Vert \vec{m} \Vert \\ &= \Vert \vec{l} \Vert\Vert \vec{p_{\perp}} \times \vec{l} \Vert \\ &= {\Vert \vec{l} \Vert}^{2}\Vert \vec{p_{\perp}}\Vert \end{align}$$

and therefore

$$\Vert \vec{p_{\perp}} \Vert = \frac{\Vert \vec{l} \times \vec{m} \Vert}{{\Vert \vec{l} \Vert}^{2}} = \frac{\Vert \vec{m} \Vert}{\Vert \vec{l} \Vert}\tag{4}$$ $$\vec{p_{\perp}} = \frac{\vec{l} \times \vec{m}}{{\Vert \vec{l} \Vert}^{2}}\tag{5}$$

By making $\vec{l}$ a unit vector, $\Vert \vec{l} \Vert = 1$, we get:

$$\Vert \vec{p_{\perp}} \Vert = \Vert \vec{m} \Vert\tag{6}$$ $$\vec{p_{\perp}} = \vec{l} \times \vec{m}\tag{7}$$

Another way to get equation $(7)$ is with a vector triple product expansion of $\vec{l} \times \vec{m} = \vec{l} \times \big(\vec{p_{\perp}} \times \vec{l}\big)$.

$$\begin{align} \vec{l} \times \big(\vec{p_{\perp}} \times \vec{l}\big) &= \big(\vec{l} \cdot \vec{l} \big)\vec{p_{\perp}} - \big(\vec{l} \cdot \vec{p_{\perp}} \big) \vec{l} \\ &= {\Vert \vec{l} \Vert}^{2} \vec{p_{\perp}} \\ {\Vert \vec{l} \Vert}^{2} \vec{p_{\perp}} &= \vec{l} \times \vec{m} \\ \vec{p_{\perp}} &= \frac{\vec{l} \times \vec{m}}{{\Vert \vec{l} \Vert}^{2}} = \vec{l} \times \vec{m} \tag{8} \end{align}$$

Unit vector $\vec{l}$ and vector $\vec{m}$ will be the Plucker coordinates $\big(\vec{l}, \vec{m}\big)$ of line $\mathrm{L}$. The dual number representation $\pmb{\hat{l}}$ of these coordinates is:

$$\begin{align} \pmb{\hat{l}} &= \pmb{l} + \epsilon \pmb{m} \\ &= \big(0 + \vec{l}\big) + \epsilon \big(0 + \vec{m}\big) \\ &= \vec{l} + \epsilon \vec{m} \end{align}$$

where $\vec{l} = \mathrm{x_l}\vec{i} + \mathrm{y_l}\vec{j} + \mathrm{z_l}\vec{k}$ and $\vec{m} = \mathrm{x_m}\vec{i} + \mathrm{y_m}\vec{j} + \mathrm{z_m}\vec{k}$.

Unit Dual Quaternion For Translation And Rotation

Like in regular unit quaternion product or "sandwich" for a rotation, we will use the unit dual quaternion $\pmb{\hat{q}}$ and one of it's conjugates $\pmb{\bar{\hat{q}}^{*}}$ on a dual quaternion of a vector $\vec{v} = a\vec{i} + b\vec{j} + c\vec{k}$ to first perform product ("sandwich") for a translation of $\vec{v}$.

$$\begin{align} \big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)&\big( 1 + \epsilon\vec{v}\big)\big( \pmb{q_{0}^{*}} - \epsilon\pmb{q_{\epsilon}^{*}} \big) \\ &= \big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)\big( \pmb{q_{0}^{*}} - \epsilon\pmb{q_{\epsilon}^{*}} \big) + \big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)\epsilon\vec{v}\big( \pmb{q_{0}^{*}} - \epsilon\pmb{q_{\epsilon}^{*}} \big) \\ &= \pmb{q_0}\pmb{q_{0}^{*}} - \epsilon\big(\pmb{q_0}\pmb{q_{\epsilon}^{*}} - \pmb{q_{0}^{*}}\pmb{q_{\epsilon}}\big) - {\epsilon}^2\pmb{q_{\epsilon}}\pmb{q_{\epsilon}^{*}} + \epsilon\pmb{q_0}\vec{v}\pmb{q_{0}^{*}} - {\epsilon}^2\big(\pmb{q_0}\vec{v}\pmb{q_{\epsilon}^{*}} - \pmb{q_{\epsilon}}\vec{v}\pmb{q_{0}^{*}}\big) - {\epsilon}^3\pmb{q_{\epsilon}}\vec{v}\pmb{q_{\epsilon}^{*}} \\ &= 1 - \epsilon\big(\pmb{q_0}\pmb{q_{\epsilon}^{*}} - \pmb{q_{0}^{*}}\pmb{q_{\epsilon}}\big) + \epsilon\pmb{q_0}\vec{v}\pmb{q_{0}^{*}}\tag{1} \end{align}$$

Expanding the following terms from equation $\big(1\big)$:

$$\begin{align} \pmb{q_0}\pmb{q_{\epsilon}^{*}} - \pmb{q_{0}^{*}}\pmb{q_{\epsilon}} &= \big(w_0 + \vec{q}_{0}\big)\big( w_{\epsilon} - \vec{q}_{\epsilon}\big) - \big(w_0 - \vec{q}_{0}\big)\big( w_{\epsilon} + \vec{q}_{\epsilon}\big) \\ &= -2w_0\vec{q}_{\epsilon} + 2\vec{q}_{0}w_{\epsilon} \end{align}$$

Since $\pmb{\hat{q}}$ is a unit dual quaternion, we can choose $w_0 = 1$, $\vec{q}_{0} = 0$ and $w_{\epsilon} = 0$ then

$$w_0^2 + x_0^2 + y_0^2 + z_0^2 = (1)^2 + (0)^2 + (0)^2 + (0)^2 = 1$$ $$w_0 w_{\epsilon} + x_0 x_{\epsilon} + y_0 y_{\epsilon} + z_0 z_{\epsilon} = (1)(0) + (0)x_{\epsilon} + (0)y_{\epsilon} + (0)z_{\epsilon} = 0$$

These values satisfy the unit condition and we get:

$$-2w_0\vec{q}_{\epsilon} + 2\vec{q}_{0}w_{\epsilon} = -2\vec{q}_{\epsilon}$$

Likewise for the remaining term of equation $\big(1\big)$ we have

$$\epsilon\pmb{q_0}\vec{v}\pmb{q_{0}^{*}} = \epsilon\vec{v}$$

Thus this quaternion product has no rotational effect on vector $\vec{v}$ and equation $\big(1\big)$ becomes:

$$1 + 2\epsilon\vec{q}_{\epsilon} + \epsilon\vec{v}\tag{2}$$

which is a dual quaternion representation of the resultant of $\vec{v}$ added to the vector associated with $2\vec{q}_{\epsilon}$. In other words, a translation of the point $\vec{v}$ in direction $\vec{q}_{\epsilon}$ by distance $\Vert 2\vec{q}_{\epsilon} \Vert$. If the translation vector is $\vec{d} = x_{d}\vec{i} + y_{d}\vec{j} + z_{d}\vec{k}$ then we can correct the translation distance doubling in our dual quaternion term $\big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)$ by using half the length of $\vec{d}$:

$$\begin{align} \big(1 + \frac{\epsilon}{2}\vec{d}\big)\big(1 + \epsilon\vec{v}\big)\big(1 - \frac{\epsilon}{2}\vec{d^{*}}\big) &= \big(1 + \epsilon\vec{v} + \frac{\epsilon}{2}\vec{d}\big)\big(1 - \epsilon\vec{d^{*}}\big) \\ &= 1 - \frac{\epsilon}{2}\vec{d^{*}} + \epsilon\vec{v} + \frac{\epsilon}{2}\vec{d} \\ &= 1 + \epsilon\vec{d} + \epsilon\vec{v} \end{align}$$

Therefore unit dual quaternion for a translation $\pmb{\hat{t}}$ is:

$$\pmb{\hat{t}} = 1 + \frac{\epsilon}{2}\vec{d}\tag{3}$$

Now we perform a second dual quaternion product on $1 + \epsilon\vec{v}$ for the rotation on $\vec{v}$. Using the unit dual quaternion $\pmb{\hat{r}} = \pmb{r_0} + \epsilon(0)$, which means $\pmb{r_0}$ is a unit quaterion, gives:

$$\begin{align} \pmb{\hat{r}}\big( 1 + \epsilon\vec{v} \big) \pmb{\bar{\hat{r}}^{*}} &= \pmb{r_0}\big( 1 + \epsilon\vec{v} \big) \pmb{{r_0}^{*}} \\ &= 1 + \epsilon \pmb{r_0}\vec{v}\pmb{{r_0}^{*}} \end{align}$$

Combining the two products for translation then rotation gives the dual quaterion $\pmb{\hat{s}}$:

$$\begin{align} \pmb{\hat{s}} &= \pmb{\hat{r}}\pmb{\hat{t}} \\ &= \pmb{r_0}\big(1 + \frac{\epsilon}{2}\vec{d}\big) \\ &= \pmb{r_0} + \frac{\epsilon}{2}\pmb{r_0}\vec{d}\tag{4} \end{align}$$

and it's conjugate

$$\begin{align} \pmb{\bar{\hat{s}}^{*}} &= \pmb{\bar{\hat{t}}}^{*}\pmb{\bar{\hat{r}}}^{*} \\ &=\pmb{{r_0}^{*}} - \frac{\epsilon}{2}\vec{d}^{*}\pmb{{r_0}^{*}}\tag{5} \end{align}$$

Applying the quaternion product with $\pmb{\hat{s}}$ to vector $\vec{v}$:

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

However, $\vec{d}$ is a pure regular quaternion, so

$$-\vec{d}^{*} = \vec{d}$$

Therefore

$$\pmb{\hat{s}}\big( 1 + \epsilon\vec{v}\big)\pmb{\bar{\hat{s}}^{*}} = 1 + \epsilon\pmb{r_0}\big(\vec{v} + \vec{d}\big)\pmb{{r_0}^{*}}\tag{6}$$

If we reverse the order of the operations and perform a rotation then translation, we get:

$$\begin{align} \pmb{\hat{s}}\big( 1 + \epsilon\vec{v}\big)\pmb{\bar{\hat{s}}^{*}} &= \big(\pmb{r_0} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\big)\big( 1 + \epsilon\vec{v}\big)\big( \pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{{r_0}^{*}}\vec{d}^{*}\big) \\ &= \big(\pmb{r_0} + \epsilon\pmb{r_0}\vec{v} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\big)\big( \pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{{r_0}^{*}}\vec{d}^{*}\big)\\ &= \pmb{r_0}\pmb{{r_0}^{*}} - \frac{\epsilon}{2}\pmb{r_0}\pmb{{r_0}^{*}}\vec{d}^{*} + \epsilon\pmb{r_0}\vec{v}\pmb{{r_0}^{*}} + \frac{\epsilon}{2}\vec{d}\pmb{r_0}\pmb{{r_0}^{*}}\\ &= 1 + \epsilon\big(\pmb{r_0}\vec{v}\pmb{{r_0}^{*}} + \vec{d}\big)\tag{7} \end{align}$$

We can modify the regular unit quaternion $\pmb{r_0} = r_0 + \vec{r}$ into one that uses angles of rotation around the axis $\vec{r}$. See Extracting the Cross and Dot Products from the Quaternion Rotation Operator. The scalar part is $\cos{\Big(\frac{\theta}{2}\Big)}$ and the vector part is the unit vector $\vec{u}$ of $\vec{r}$ scaled by $\sin{\Big(\frac{\theta}{2}\Big)}$.

$$\pmb{r_0} = \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{u}\tag{8}$$

Substituting $(8)$ into $(4)$:

$$\begin{align} \pmb{\hat{s}} &= \pmb{r_0} + \frac{\epsilon}{2}\pmb{r_0}\vec{d} \\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{u} + \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{u} \bigg)\vec{d} \\ &= \cos{\Big(\frac{\theta}{2}\Big)} + \sin{\Big(\frac{\theta}{2}\Big)}\vec{u} + \frac{\epsilon}{2}\bigg( \cos{\Big(\frac{\theta}{2}\Big)}\vec{d} + \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{u}\times\vec{d}\Big) - \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{u}\cdot\vec{d}\Big) \bigg)\tag{9} \end{align}$$

Likewise, substituting $(8)$ into $(5)$:

$$\begin{align} \pmb{\hat{s}} &= \pmb{{r_0}^{*}} - \frac{\epsilon}{2}\vec{d}^{*}\pmb{{r_0}^{*}} \\ &= \cos{\Big(\frac{\theta}{2}\Big)} - \sin{\Big(\frac{\theta}{2}\Big)}\vec{u} - \frac{\epsilon}{2} \bigg(\cos{\Big(\frac{\theta}{2}\Big)}\vec{d}^{*} - \sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{d}^{*} \times \vec{u}\Big) +\sin{\Big(\frac{\theta}{2}\Big)}\Big(\vec{d}^{*}\cdot\vec{u}\Big)\bigg)\tag{10} \\ \end{align}$$

How To Represent A 3D Vector As A Unit Dual Qauternion

To see how any $3D$ vector $\vec{v} = \big(a, b, c\big)$ can be represented as a unit dual quaternion, let's start with the dual quaternion $\pmb{\hat{q}} = \pmb{q_0} + \epsilon\pmb{q_{\epsilon}}$ with quaternions components $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$. The real and imaginary parts of $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$ are given as follows with the imaginary part italicized and an arrow above:

$$\begin{align} \pmb{q_0} &= w_0 + \vec{q}_{0} = w_0 + \big(x_0\vec{i} + y_0\vec{j} + z_0\vec{k}\big)\\ \pmb{q_{\epsilon}} &= w_{\epsilon} + \vec{q}_{\epsilon} = w_{\epsilon} + \big(x_{\epsilon}\vec{i} + y_{\epsilon}\vec{j} + z_{\epsilon}\vec{k} \big) \end{align}$$

Their respective conjugates are:

$$\begin{align} \pmb{q_{0}^{*}} &= w_0 - \vec{q}_{0} = w_0 - \big(x_0\vec{i} + y_0\vec{j} + z_0\vec{k}\big)\\ \pmb{q_{\epsilon}^{*}} &= w_{\epsilon} - \vec{q}_{\epsilon} = w_{\epsilon} - \big(x_{\epsilon}\vec{i} + y_{\epsilon}\vec{j} + z_{\epsilon}\vec{k} \big) \end{align}$$

Unlike regular quaternions, there are three different conjugations of dual quaternions which depend on whether:

1. The dual number is conjugated

   $$\pmb{\bar{\hat{q}}} = \pmb{q_0} - \epsilon\pmb{q_{\epsilon}}$$

   This finds little use except in deriving the third type of conjugation $\pmb{\bar{\hat{q}}^{*}}$.

2. The quaternion components are conjugated

   $$\pmb{\hat{q}^{*}} = \pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}$$

   The result of $\pmb{\hat{q}}\pmb{\hat{q}^{*}}$ is a dual scalar but it can be just a regular scalar when the dot-product $\pmb{q_{0}} \cdot \pmb{q_{\epsilon}} = 0$ (i.e. when $\pmb{q_{0}}$ and $\pmb{q_{\epsilon}}$ are orthogonal 4-tuples).

3. The dual number and quaternion components are both conjugated.

   $$\pmb{\bar{\hat{q}}^{*}} = \pmb{q_{0}^{*}} - \epsilon\pmb{q_{\epsilon}^{*}}$$

   The product $\pmb{\hat{q}}\pmb{\bar{\hat{q}}^{*}}$ is a dual quaternion whose real part is a scalar and dual part is a vector.

The conjugate $\pmb{\hat{q}^{*}} = \pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}$ is used to define the norm of $\pmb{\hat{q}}$:

$$\begin{align} \Vert \pmb{\hat{q}} \Vert &= \sqrt{\pmb{\hat{q}\hat{q}^{*}}} \\ &= \sqrt{\big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)\big(\pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}\big)} \\ &= \sqrt{\pmb{q_0q_{0}^{*}} + \epsilon\big(\pmb{q_0q_{\epsilon}^{*}} + \pmb{q_{\epsilon}q_0^{*}}\big)} \\ &= \sqrt{\pmb{q_0q_{0}^{*}} + 2\epsilon\big(\pmb{q_0} \cdot \pmb{q_{\epsilon}}\big)} \end{align}$$

where

$$\begin{align} \pmb{q_0q_{0}^{*}} &= \big(w_0 + x_0\pmb{i} + y_0\pmb{j} + z_0\pmb{k}\big)\big( w_0 - x_0\pmb{i} - y_0\pmb{j} - z_0\pmb{k}\big) \\ &= w_0^2 + x_0^2 + y_0^2 + z_0^2 \end{align}$$

and

$$\begin{align} \pmb{q_0q_{\epsilon}^{*}} + \pmb{q_{\epsilon}q_0^{*}} &= 2\big(w_0 w_{\epsilon} + x_0 x_{\epsilon} + y_0 y_{\epsilon} + z_0 z_{\epsilon} \big) \\ &= 2\big(\pmb{q_0} \cdot \pmb{q_{\epsilon}}\big) \end{align}$$

A unit dual quaternion has a norm $\Vert \pmb{\hat{q}} \Vert = 1$. Choosing $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$ to be orthogonal makes the dot-product $\pmb{q_0} \cdot \pmb{q_{\epsilon}} = 0$ which then leaves us with $\sqrt{\pmb{q_0q_{0}^{*}}} = 1$ and therefore $w_0^2 + x_0^2 + y_0^2 + z_0^2 = 1$. Choosing $\vec{q}_{0} = 0$, $\mathrm{q_{\epsilon}} = 0$ and the coefficients of $\vec{q}_{\epsilon}$ to be $\big(a, b, c\big)$ of vector $\pmb{v}$ satisfies the conditions of the $\pmb{q_0} \cdot \pmb{q_{\epsilon}} = 0$ and $w_0^2 + x_0^2 + y_0^2 + z_0^2 = 1$:

$$w_0 w_{\epsilon} + x_0 x_{\epsilon} + y_0 y_{\epsilon} + z_0 z_{\epsilon} = w_0(0) + (0)a + (0)b + (0)c = 0$$
$$w_0^2 + x_0^2 + y_0^2 + z_0^2 = w_0^2 + (0)^2 + (0)^2 + (0)^2 = w_0^2$$ $$w_0^2 = 1$$

Thus $w_0 = 1$ and the vector $\vec{v} = \big(a, b, c\big)$ is encoded in a dual quaternion as:

$$\begin{align} &1 + \epsilon\big( a\vec{i} + b\vec{j} + c\vec{k} \big) \\ &1 + \epsilon\vec{v} \end{align}$$

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