Here is a 2N3055 connected as a common-emitter switch. The load R=E/I=12/5 = 2.4Ω is connected between the collector and the positive end of the 12V power supply.
I am showing that to turn on the 3055, your PIR will have to source more than 130mA at 10V to get the base current up to a level where the 3055 turns on. Look at the sim plots. The X-axis is the voltage of V2, the base source. The blue trace is the current into the base of the 3055. The green trace is the load current. Note that the input voltage V2 has to get to >9V (which corresponds to a base current of >130mA to turn on the 3055.
Note that by rights, when the 3055 is used as a saturated switch, the base current should be much higher, about Ic/10, or 0.5A to turn the 3055 on as well as it should be. It is doubtful that the PIR can source 0.5A, so you would do better to use an NFET, or possibly put a small NPN in front of the 3055 to make a Darlington pair.