Question: Consider **2D spacetime** with two inertial reference frame S and $S'$, where $S'$ is moving in the $S$ positive spatial direction at velocity $vt$, along with the usual graphical representation with $t$ on the vertical axis and $x$ on the horizontal axis. Use units such that $c = 1$.

Consider the space-time displacement vector which points forward in time in the coordinates of $S'$ at $(0,0)$. I.e., it is displacement vector that connects the clock at point $(0,0)$ in $S'$ to the clock at point $(1,0)$ in $S'$.

Is it true that it follows that the clock at $(1,0)$ in $S'$ is at $(t,vt)$ in $S$, for some value of $t$?

If so, does it follow that viewing the time-axis of $S'$ in the $S$ coordinates, it is rotated to the right from vertical by an angle $\theta$ such that $v = \tan(\theta)$?

Is there any analogous statement that can be made about the direction in $S$ of the $S'$ space-displacement vector, i.e. the vector with components $(0,1)$ in $S'$? I can't come up with one and it seems that the answer has to be "no" because I haven't yet used anything about Lorentz invariance of the interval and the rotation of the space and time axes is equivalent to Lorentz invariance. But I would not have expected an asymmetry here between time and space.

Answer: 
If you choose to draw the frames so that the $S$ axes are horizontal and vertical on your paper, then they look like this (with the $S$ frame in black and the $S'$ frame in blue):

[![enter image description here](https://i.stack.imgur.com/aCBv8.jpg)](https://i.stack.imgur.com/aCBv8.jpg)

The marked points are $(1,0)$ and $(0,1)$ in the $S'$ frame.
[![enter image description here](https://i.stack.imgur.com/VhdIa.png)](https://i.stack.imgur.com/VhdIa.png)

Here You can easily visualise the scenario, the worldline of $S'$ is represented by $ct'$ and transformation rules are as follows :

$ \displaystyle 
\left[ {\begin{array}{c}
 ct'\\
 x' \\
 \end{array} } \right] = 
$
$ \displaystyle 
\left[ {\begin{array}{cc}
 \gamma & -\beta \gamma \\
 -\beta \gamma & \gamma \\
 \end{array} } \right]
$
$ \displaystyle 
\left[ {\begin{array}{c}
 ct\\
 x \\
 \end{array} } \right] 
$

The vector with components $(0,1)$ in $S'$ would lie on the new space basis as defined by the transformation.