Torque direction meaning

While the notion of the 'meaning' is somewhat subjective so this may not provide any more meaning but, you should think about the origin of the torque. Namely as the rate of change of the angular momentum: $$textbf{N} = frac{dtextbf{L}}{dt},$$ Now the fact is orthogonal to the plane containing the position vector and force vector is a direct result of the definition of angular momentum $$textbf{L} = textbf{r} times textbf{p} $$ where $textbf{p}$ is the linear momentum. So one can see the torque as the vector acting on the angular momentum.

This is where I claim that it is not a convention, but a requirement that the angular momentum and hence torque lie perpendicular to the plane. Imagine you could define it in some other way, where it didn't lie perpendicular to this plane, then for a constant angular momentum (e.g no torque) the angular momentum vector would have to rotate with the system and would not be constant! For this reason the choice of orthogonality, is not really a choice or convention but it describes the system.

This is my own personal interpretation so I don't claim it as exactly true in any sense.

In electrical engg, itis very clear in that the current carrying conductor and the magnetic field produced by it are mutually perpendicular to each other.In mechanics it is very difficult to assign a physical meaning for directions of angular velocity,angular momentum and torque.

When we said a direction we mean the motion is just in that direction .So torque is a vector quantity appears in the normal direction of RXF plane and not appear in any other angle than 90 degree. That is my understanding

When I was taught about torque it seemed lifeless to me having no physical significance unlike other quantities- force, velocity etc. I developed my own intuition about torque after some analysis. I think the direction of torque is indicating the axis around which the object rotates. Also the direction in which we curl our fingers(in right hand thumb rule) indicates sense of rotation of the object around the axis.

The geometry of torque has nothing to do with motion and the equations of motion. You can have torques in statics and their magnitude and direction are important and insightful. You do need to consider a torque vector, as well as any force vector, applied together to extract the geometry of the situation.

Torque gives us the line of action of force.

A force vector $boldsymbol{F}$ gives us the magnitude of the force, as well as the direction it acts upon. What it does not give us, is the location in space where the force is applied. This line of action as it is called is only available from the torque this force produces $boldsymbol{tau} = boldsymbol{r} times boldsymbol{F}$.

A point on the line of action closest to the reference location where torque is measured is found by

$$ boldsymbol{r}_{rm action} = frac{ boldsymbol{F} times boldsymbol{tau}}{ | boldsymbol{F} |^2} tag{1}$$

The magnitude of torque is a measure of the perpendicular distance to the line of action, and the direction of torque is perpendicular to both the force direction and the location of the line of action. It is also the same direction velocity would be pointing of the line of action was a rotation axis. The math is identical, as shown in the linked answers below.

Just as the torque of a force $boldsymbol{F}$ located at $boldsymbol{r}$ is $boldsymbol{tau} = boldsymbol{r} times boldsymbol{F}$, the velocity of a rigid rotating about $boldsymbol{r}$ is also $boldsymbol{v} = boldsymbol{r} times boldsymbol{omega}$. If you can understand and visualize the velocity field of a rotating body, you can understand and visualize the torque field of a force vector.

---

Related Answers

1. https://physics.stackexchange.com/a/518467/392
2. https://physics.stackexchange.com/a/516069/392

---

Proof of expression (1)

Transfer the torque from the reference point to the line of action and show that the torque on the line of action is zero (use vector triple product identity).

$$require{cancel} begin{aligned} boldsymbol{tau}_{rm action} & = boldsymbol{tau} - boldsymbol{r}_{rm action} times boldsymbol{F} & = boldsymbol{tau} - frac{(boldsymbol{F} times boldsymbol{tau})}{| boldsymbol{F} |^2} times boldsymbol{F} & = boldsymbol{tau} - frac{ boldsymbol{tau}(boldsymbol{F}cdot boldsymbol{F}) - boldsymbol{F} (cancel{ boldsymbol{tau} cdot boldsymbol{F}}) }{| boldsymbol{F} |^2} & = boldsymbol{tau} - frac{ boldsymbol{tau} | boldsymbol{F}|^2}{| boldsymbol{F} |^2} = boldsymbol{0} end{aligned} $$