The TIP120 does not need 120mA at the base for normal operation, that's the absolute maximum rating, above which you don't want to go.
The spec you are mostly interested in is the hFE (current gain), which for a darlington is very high, since it's two transistors connected in a way so the current gains multiply.
For the TIP120 it's given as minimum 1,000 (compare with a typical 200 for a single bipolar transistor)
Also important are the max collector current (5A) and the collector emitter voltage (60V)
The main disadvantages are that the base-emitter voltage is doubled compared to a single transistor (~1.4V), and the saturation voltage is higher (typically ~0.8V compared to ~0.2V at low currents)
These points are rarely a problem for a simple switch driven from a micro pin. At higher collector-emitter currents though, the Vsat rises and can interfere with desired operation and cause problems with dissipation.
For example, in the TIP120 datasheet note that at 3A Ice, Vsat is given as 2V, but at 5A it has risen to 4V. That's 20W of dissipation, a lot to heat to try and get rid of to keep the temperature down. So when switching a large current you need to take these factors into account, and maybe decide to look at a more suitable part (e.g. logic level, low Rsdon power MOSFET)
Since we have a gain of 1000, we hardly have to draw anything from the micro pin. Let's say we want to switch 1 Amp:
1A / 1000 = 1mA into the base needed.
If we have a drive voltage of 5V, then we subtract the Vbe from the drive voltage and divide by the current:
(5V - 1.4V) / 1mA = 3.6k resistor. To give it a bit of leeway select something a bit smaller like 2.2k. This still only draws ~1.6mA.
I wouldn't read too much into the different prices - the price of components is often dictated by how popular they are, the more they sell the less they cost. If you see better specs at a cheaper price, go for it ;-)
You can some pretty odd prices when the component is scarce/new/obsolete - I saw a 10uF ceramic capacitor priced at £7.50 (qty 1) on Farnell the other week...
There are two situations where a voltage divider can provide a relatively constant voltage to a load. First, if the load draws much less current than the current that flows through the voltage divider when no load is connected. Second, if the current drawn by the load is constant at the desired voltage. Motors and relays usually draw relatively large currents and so they don't fit in the first category. The current drawn by a motor can vary greatly depending on its load torque so a motor doesn't fit well in the second category either. The current drawn by a relay is also variable as it is switched on and off. Therefore, using a voltage divider is usually not a good choice for things like motors and relays.
The important question, which you haven't told us, how much difference there is between the available voltage and the desired load voltage. If the available voltage is much higher than the specified load voltage there is the possibility that the full supply voltage will be supplied for a short time and destroy your relay or motor. (For an inductive load the initial current when you switch it on must be zero, so the full supply voltage could appear on the load until current starts flowing.) In this situation you can't use PWM either, because the PWM pulses apply the full voltage.
If the difference between your supply voltage and the voltage needed for the motor or relay is small then you might get away with adding a series resistance but it's not a very good design. The best thing to do is provide a regulated voltage supply at the appropriate voltage for the load. Your choice of voltage regulator will depend on current and voltage requirements for the loads.
Best Answer
The 1k0 resistor is simply a legacy from a previously used circuit. The author has left this component in to demonstrate what happens if you add additional load to the circuit but has not explained the intermediate circuit.
Consider this version instead:
The first circuit shows that adding a load to a potential divider circuit will pull down the voltage at the junction.
Look at the equations in blue:
The author is saying that:
(i) if Rload is much greater than R2 (a factor of at least 10) it doesn't change the voltage too much and we can ignore it.
(ii) If R2 is similar or less than Rload then we need to re-calculate the voltage. (the author doesn't explain this very well and concentrates on the much less)
How the emitter follower work using this information?
(I've drawn the intermediate circuit to make this process clearer)
Q1 only takes a very small base current (because it has current amplification of at least 100). As long as this base current is at least 10 times smaller than the current going through R1, R2 it doesn't reduce the voltage at the junction too much. (remember that gold band resistors have a 5% tolerance so we expect some variation in any case)
The transistor's output is across a 1k0 resistor load (much lower than R2). It is supplied by current through the transistor and not the potential divider.
The base-emitter junction will drop about 0.6V. The output voltage at the emitter will be 0.6V less than the potential divider. (middle diagram)
The author then adds an even larger load (smaller resistance) across the emitter resistor (righthand diagram) to demonstrate this has no effect on the output voltage.
In other words the load can vary but the voltage across it stays the same. (which is the teaching point).
To answer your question - Yes, the 1k0 (R3) resistor could be removed, in practice it has no function and is simply part of the load but you need to take this circuit diagram in the context of the whole article which starts with a common emitter amplifier designed with R3 as a 1k0 resistor. Context here is important.