This is a clarification to a series of emails I am having with an individual that seems to be confused with Phasors, Rotation and Reference Points.
First let me discuss phase rotation.
Imagine I was looking at a rotating wheel that was moving counterclockwise as in the picture below.

What would I see? I would see A first followed by B and then C. This means my phase rotation is ABC. Now I could also say the phase rotation is BCA and CAB. They all mean the same thing.
Now imagine I rearranged the wheel as below.

What would I see? I would see A first followed by C and then B. This means my phase rotation is ACB. Now I could also say the phase rotation is CBA and BAC. They all mean the same thing.
So here is the important point:
ALWAYS ASSUME THE PHASORS ARE ROTATING COUNTERCLOCKWISE!
There is no such thing as “counterclockwise” or “clockwise” rotation with regards to phase rotation.

(Side note:Yes the motor shaft or generator shaft can be called counterclockwise and clockwise and Yes the phase rotation effects which direction the shaft physically spins. But this has nothing to do with the problems we are solving. Also consider this, when you say a motor shaft is rotating “clockwise” are you looking directly at the shaft or are you sitting on the motor looking out watching it turn? Even here you can see that saying a motor rotates “clockwise”, a reference point has to be given for you to truly understand what you mean.)

The rotation also says nothing about weather current is lagging or leading! Consider the diagram below:

Now what would I see? I would see Va first, than after 30° I would see Ia. Therefore Ia lags Va by 30°. I would than see Vb followed by Ib (30° later). This is ABC rotation with a lagging current.
Now consider the diagram below.

I would see Va first, than after 90° I would see Ib. After 30° I would see Vb.  Than Ic than Vc. Finally, I would see Ia followed by Va 30° later. This is the current leading the voltage by 30°.
When solving these problems let me repeat what I stated above:
ALWAYS ASSUME THE PHASORS ARE ROTATING COUNTERCLOCKWISE!

Now in the above examples I always chose Va as my reference point. In other words, Va is considered 0°. Once I choose a reference point, when I speak of angles it is always with regards to the original reference point chosen. For load calculation it is preferable to use the voltage at 0° so that the angle of the current I represents its actual lag or lead value.
I can choose any reference point as long as I am keeping track of what I am doing. Consider the problem below:

A single phase synchronous generator is connected to an infinite bus through a series inductive reactance of 1.00 per unit. The terminal voltage of the generator is taken as the reference and is 1.1∠0° per unit. The infinite bus voltage is 1∠-30° per unit. What is the transmitted power, in per unit, to the infinite bus from the generator?

Now I am using Vt as my reference. So any angle solved will be in reference to Vt. So let me solve for I:

Now here would be an incorrect answer:
P=VIcos(θ)=1*0.552*cos(-25°)=0.5002
Look at the vector diagram to understand why the above is incorrect:

Remember I choose Vt as my reference. Now the angle I is solved with reference to Vt. The angle used in the power equation should be the difference between Vr and I.
Now here would be the correct answer:
P=VIcos(θ)=1*0.552*cos(5°)=0.5498.
I leads Vr by 5° (30-25).

Let me repeat what I stated above:
For load calculation it is preferable to use the voltage at 0° so that the angle of the current I represents its actual lag or lead value.

The above statement is very important. To use the above statement in the problem above I will use Vr as my reference and resolve it.

Now this -175° might look strange, but it can be fixed by moving the vector to point in the actual direction the current is flowing. cos(-175) is a negative number. First lets draw it with the -175° and solve it:

P=VIcos(θ)=1*0.552*cos(-175°)=  -0.5498.
The fact that I got “negative” power just shows I choose the wrong direction for I.
Now let us fix the vector notation. (Remember if you have a vector as Vac=480 @ -60° than Vca=480 @ 120°. Another way of stating its 180° out.) So the current is at 175°. if I was to rotate it 180°, I would wind up with 5°.

P=VIcos(θ)=1*0.552*cos(5°)=  0.5498.

I also could have just done I= (-Vt+Vr)/X. and gotten the same thing:

P=VIcos(θ)=1*0.552*cos(-85-(-90)°)=0.552*cos(5°)=  0.5498.