Having to points $A = (-1,2)$ and $B = (1,0)$ and their respective homogenouse coordinates $(1:-1:2)$ and $(1:1:0)$ the line $f$ connecting both points is given by $f = A \lor B$.
In $\mathbb{R}^3$ one can simple calculate the cross product $A \times B$:
$ A \times B = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \Rightarrow (-1:1:1) $
Plotting these vectors:
Okay that's that, but it's kind of hard to me to understand why this is working. Could anybody explain to me the reason that this is working?
I know having two points $A, B \in \mathbb{R}^2$ a line $f$ connecting those points can be said to be the 1-dimensional hyperplane H or subspace of $\mathbb{R}^2$ but I'm lacking intuition why one can calculate this line like that.
If a vector $v$ represents a given point $p$, then any multiple $\lambda v$ of $v$ represents the same point (as long as $\lambda\neq0$). So you can think of a point in $\mathbb{RP}^2$ as a line through the origin in $\mathbb R^3$. The set of all points on a line in $\mathbb{RP}^2$ likewise corresponds to the union of many such lines, i.e. to a plane through the origin in $\mathbb R^3$. The normal vector of that plane in $\mathbb R^3$ corresponds to the coordinate vector of a line in $\mathbb{RP}^2$.
So to sum things up, a point in the projective plane is modeled by a one-dimensional linear subspace of $\mathbb R^3$, which can be characterized as the span of any non-zero element of that space. Similarly a line of the projective plane is modeled by a two-dimensional linear subspace of $\mathbb R^3$, which can be characterized as the space orthogonal to its normal vector.
This orthogonality is where the cross product comes in: if you have $a$ and $b$, then the cross-product $a\times b$ will be orthogonal to both $a$ and $b$. So if $a$ and $b$ describe points, then $a\times b$ is a vector orthogonal to both, so the plane orthogonal to that will contain $a$ and $b$.
This is an explanation aimed at geometric intuition. There are algebraic ways to look at things, but as you included a geometric picture in your question, I guess that this approach here might be better suited to help you understand.