Question: If I were wanting to drive i.e. industrial relays which require 24v from a Microchip PIC what would I need to look at in terms of a general transistor circuit (and it's pitfalls when driving such a component).

Answer: 
I personally would follow stevenvh's answer only I would move R2 to the pic side of R1. The reason for this is R1 and R2 are acting as a voltage divider and therefore you are not getting the full pic voltage on to the gate pin of the MOSFET. by moving R2 to the other side of R1 you'll get the full voltage at the gate pin. 
Admittedly the voltage loss will be very small as the resistor values chosen here will only loose 0.003v @ 3.3v so it wouldn't really make a difference in this example but with different resistor values it could make a difference so out of principle I would always put R2 on the other side of R1
I recently did something very similar to this as a beginner, and found MOSFETS very useful. You need a 24V source from somewhere to switch your relay, which should be connected across the drain-source pins of the MOSFET. You apply a voltage to the Gate pin (using the pinout from your PIC), and this essentially switches the relay-driving circuit on.

I made the mistake of taking the gate-threshold voltage in the datasheets as the voltage required to fully turn it, so rather than worrying about this it's much easier to just get a logic-level MOSFET, such as the IRL520. You need to make sure that it's capable of switching 24V and whatever current your relay will draw.

You also need to connect a diode in parallel with the relay to prevent damage to the transistor when it switches.