# Question about torque reaction of a helicopter

This image below about the torque reaction of a helicopter has confused me a bit. I would like to ask whether i understand the principles here correctly. In the picture we see a helicopter whose main propeller is spinning Counter Clock Wise (CCW), as is indicated by the up most arrow. We know that the torque is $\vec{T} = \vec{r}\times\vec{F}$, so according to the right hand rule i want to calculate the direction of the torque of the main propeller here, or better yet to understand why it is in the direction illustrated.

We know that the main propeller is producing a force downwards to the air. And then the air is countering this force, by producing a lift force that lifts the helicopter upwards. Now, my problem is which one of the two forces should i consider when i calculate the Torque with the above equation? I believe i should take into account the force-action (from the helicopter to the air) and not the force-reaction (from the air to the helicopter). If i consider the force-action-dowwards using the right hand rule (i know how to do that correctly, i'm sure) then the direction of the torque should be that of the image below. However the direction that the main rotor is spinning confuses me whether i'm doing the right thing.

So finally in order to compensate for this torque, the tail rotor must produce the torque reaction in the opposite direction of the main torque.

My main question is whether my reasoning about the direction of the main torque is correct. Thanks in advance.