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 $\displaystyle -\sqrt{6}A$ 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 $\displaystyle \omega t - \frac{\pi}{4}$ as a $\displaystyle - \frac{\pi}{4}$ 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: $\displaystyle -\sqrt{6}A$ and $\displaystyle \sqrt{6}A$
we'd only be interested in the first and third quadrants (see image + equations in it)