Consider the 2-D $EG(2,2^2)$ geometry. Let $\alpha$ be a primitive element of $GF(2^{2\times 2})$. The incident vector for the line $\mathcal{L} = \{\alpha^7, \alpha^8, \alpha^{10}, \alpha^{14}\}$ is $(0 0 0 0 0 0 0 1 1 0 1 0 0 0 1)$.
I know that since $m=2$ and $s=2$, for each point in $EG(m,2^s)$, there are $\frac{2^{ms} - 1}{2^s - 1}$ lines intersecting at this point.
So in this case, I can generate $\frac{15}{3} = 5$ lines $\mathcal{L}_1,\mathcal{L}_2,\mathcal{L}_3,\mathcal{L}_4,\mathcal{L}_5$ that intersecting at, say point $\alpha^{14}$, where $\mathcal{L}_1=\mathcal{L}$ as given above. These five lines are:
$\mathcal{L}_1 = \mathcal{L}, \mathcal{L}_2 = \{\alpha, \alpha^5, \alpha^{13},\alpha^{14}\}$, $\mathcal{L}_3 = \{1, \alpha^2,\alpha^6,\alpha^{14}\}$, $\mathcal{L}_4 = \{0, \alpha^4,\alpha^9,\alpha^{14}\}$, $\mathcal{L}_5=\{\alpha^3,\alpha^{11},\alpha^{12},\alpha^{14}\}$.
I did manage to find all the incident vectors according to the example given for the incident vector $\mathcal{L}$. But I wasn't sure about the incident vector for $\mathcal{L}_4$, which contains element $0$. How exactly do I generate the incident vector for $\mathcal{L}_4$? Could anyone guide me please.
Okay, from what I understand, you are working with the affine plane $\mathrm{AG}(2,4)$, thinking of this as the vector space $\mathbb{F}_{4}^{2}$, associated through a field reduction map as $\mathbb{F}_{16}$ (considered as a $2$-dimensional vector space over its subfield of order $4$). There are indeed $\frac{q^s -1}{q-1}$ lines through a point in $\mathrm{AG}(s,q)$, and you can associate the points of the space with $\{0\} \cup \{ \alpha^{i} : 1 \leq i \leq q^s -1\}$, where $\alpha$ is a primitive element of $\mathbb{F}_{q^{s}}$. It makes sense to consider them in this order.
This means your incidence vector for $\mathcal{L}$ is wrong; it should have 16 entries (since there are 16 points). It should have an extra $0$ at the beginning, if you are ordering in the way I put above. This should also solve your problem with giving the vector for $\mathcal{L}_{4}$.