Time *is* orthogonal to space. Check that $(1, 0, 0, 0) \cdot (0, 1, 0, 0) = 0$ and likewise for all other spacelike unit vectors, where $\cdot$ represents the Lorentz invariant scalar product and I put time in the 0th coordinate.

The inclination originates in the Lorentz transformation that you use to go from one observer's point of view to another one's. Perfoming a boost creates an $x$ dependence in $t'$ and a $t$ dependence in $x'$:
$$ t' = \gamma \left(t - \frac{v}{c^2} x\right), x' = \gamma (x - v t)$$

This also happens in regular euclidean space, e.g. when perfoming a rotation about the $z$ axis, giving
$$ x' = \cos\theta \ x - \sin\theta\ y,\quad y' = \sin\theta\ x + \cos\theta\ y$$
So after a rotation a previousely vertical line will look inclined as well.

Edit: Observe that there is no t′ dependence in x′, even after the transformation. So, while time is *always* orthogonal to space, an object at rest for observer $O$ (say the observer's nose) will not be at rest for an observer $O'$ that moves away from observer $O$ at a speed $v \neq 0$.

The time that the light takes to travel to the observer is not included. The time an event happens in a reference frame is defined as the time that a clock at the location of the event would show. To tell time in a reference frame, Special Relativity imagines clocks at all relevant locations, and all those clocks properly synchronised within the reference frame. It is important to keep that definition in mind, because it also leads to Relativity of Simultaneity, which many novices don't count with.

Which $\Delta t$ is to be used to calculate the time the journey takes for Bob?

In reference frame A, Bob reaches $d_A$ at time $t_{0,A}$ and Alice at time $t_{0,A} + \Delta t_A$, so the journey would take 11 years in reference frame A.
In reference frame B, the journey takes $\Delta t_A /\gamma$ years.

Would Alice see Bobs time run slower or faster? (suppose there is a
clock on board)

As perceived from reference frame A, the clocks in reference frame B run slower. If with "see", you mean what Alice sees through the telescope, then you need to take the Doppler effect into account, and Alice would see the clocks of Bob run faster.

The Doppler shift is caused because each tick of the moving clock occurs at a point a bit closer to Alice than the previous tick. The light from the previous tick had to travel a bit of time to get to that same point, and therefore the two ticks appear closer together to Alice when viewed through the telescope. That is, the clock appears through the telescope to run faster than it actually does.

In an opposite scenario where Bob travels from the earth to the planet, would Alice see Bobs time run slower or faster?

Perceived from reference frame A, Bobs time runs just as slow as on the way towards the Earth. Through the telescope, Alice would see an additional slow-down due to the receding Doppler effect.

## Best Answer

You shouldn't use the "subjective/objective" distinction for a place where "relative/absolute" is much more appropriate, because they mean different things. For something to be subjective, it must be dependent on the knowledge or state of mind of an observer.

As an example, suppose we define "depth" as "length along the direction an observer is facing". This direction can be used to define a "depth axis", and measurements of depth would depend on projections to that axis. Then how deep an object is depends on the circumstances of the observers, and so is relative. But it is not subjective: it doesn't matter is the observer is delusional about they're facing, etc.

In special relativity, time is rather analogous to this. In spacetime, your

four-velocityis kind of like 'direction you're facing' would be in ordinary space. It defines the temporal axis of your inertial frame, and time measurements would be projections to that axis, similarly to the previous case.