I don't require to draw it, but explain why (if) my current phasor diagram is incorrect. I have explained on the picture why im unsure given the signage of the equation in consideration. Here's the image:


I would assume the -\sqrt{6}A would be related to the axis, though if i force that to obey i.e. by having it in the third quadrant (as it must obey real and imaginary), then surely that would contridict the \omega t - \frac{\pi}{4} as a  - \frac{\pi}{4} doesn't just flip over to the third quadrant.
Of course i could define wt to be large enough, though given that it can take a negative or postive value: -\sqrt{6}A and \sqrt{6}A
we'd only be interested in the first and third quadrants (see image + equations in it)