How do points on a grid satisfy the grid equation? Currently, I'm working on a presentation regarding Penrose tilings. During my research, I've become interested in the Pentagrid method of construction, that was introduced by N.G. de Bruijn in 1981 (original paper). However, since I won't talk exclusively about this construction method, I kept searching for lower-level sources. A difficult task, it appeared, but I found a thesis of Laura Effinger-Dean called "The Empire Problem in Penrose Tilings".
I read it through, but I'm only really interested in chapter 4 (Pentagrids) and I can't get a grasp of the grid equation. On page 33, we can read:

A grid is an infinite collection of regularly spaced parallell lines. The points of the grid satisfy the grid equation:
$$\vec{x} \cdot \vec{\epsilon} + \gamma = k$$
for some integer $k$. Every line in the grid is perpendicular to a grid vector $\vec{\epsilon}$. The distance between consecutive lines is $\frac{1}{\vec{\epsilon}}$. The grid is shifted from the origin by a distance $-\gamma$ in the direction of $\vec{\epsilon}$.

I guess we can choose whatever $k$ we want, but it must remain fixed for all points on the grid. Furthermore, I suppose $\vec{x}$ is a point on the grid (that must satisfy the equation). What I don't get is how we can construct a grid, with multiple (infinitely to be precise) parallell lines, using this equation.
How should I interpret the $\vec{x}\cdot\vec{\epsilon}$? Since we're working with vectors, my initial thougt was a dot product, but I'm not sure. Then the first term would always be zero (every point must be orthogonal to the grid vector), but if $\gamma$ and $k$ are fixed, how can we get more than one line?
Am I missing something or making wrong assumptions?
Thanks in advance!
 A: Since there seems to be some confusion, rather than continuing in the comments I'll try to write up a thorough and concrete answer.
The classic way folks are taught to write lines is through the relation $y =mx + b$, where $m$ is the slope, and $b$ is the $y$-intercept. Hence all points on the line are of the form:
$$\left(x,mx+b\right)$$
Then it always holds that:
$$(x,mx+b) \cdot (1,-1/m) = -b/m$$
The dot product is bilinear (it's linear in each component), so we can divide this equation through by the right-hand side scalar to obtain:
$$(x,mx+b) \cdot (-m/b, b) = 1$$
So by an appropriate choice of normal vector $v$, every line in the plane (more generally, any $(n-1)$-dimensional affine subspace of $\mathbb{R}^n$, called a hyperplane) can be written in the form:
$$\{x : x \cdot v = 1\}$$
(More generally $\{x : x \cdot \epsilon = b\}$ for some $b \in \mathbb{R}$)
Now, notice that if $b=0$ the derivation I just wrote doesn't work (why?). In this case, the formulation will be:
$$\{x : x \cdot v = 0\}$$
In fact, this is better than the $y= mx+b$ relation! For example, there is no way to express the points on the $y$-axis as a function, but we can easily describe them as $\{x : x \cdot (1,0) = 0\}$.

Now to your problem. Fix $\epsilon \in \mathbb{R}^2$, and consider the lines:
$$L_1 = \{x : x \cdot \epsilon = 0\}$$
$$L_2 = \{x : x \cdot \epsilon = 1\}$$
The notion of "distance apart" has a little more subtlety than is immediate. Give two lines, I can always pick one point on one line, and another on the other so that the distance between those two points are arbitrarily far apart. What we want to find is the "distance between corresponding points," whatever that means.
Without loss of generality, since we can always rotate the line arrangement without changing distances, we can assume $\epsilon = (0,r)$ for some $r \in \mathbb{R}$. Thus $L_1$ is just the $x$-axis. This seems like a good way to check the distance--- all we need to do is examine the $y$-coordinate of the points on $L_2$.
Indeed, suppose $l = (l_1,l_2) \in L_2$. Then $(l_1,l_2) \cdot (0,r) = 1$, so $l_2r = 1$, so $l_2 = 1/r$. Thus $L_1$ and $L_2$ are (in this notion of distance) a distance of $1/r = 1/\|\epsilon\|$ apart.
By linearity of the expression, this holds for all $k$, and by rotation and translation this holds for all $\epsilon$ and $\gamma$.
There's also a way to formulate this abstractly via the general expression (things cancel nicely), but I believe this concrete example illustrates it most clearly.
