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 $\vec{m} = \mathrm{x_m}\vec{i} + \mathrm{y_m}\vec{j} + \mathrm{z_m}\vec{k}$.