In 3D, torque is defined as $\vec{r} \times \vec{F}$ which is a vector, therefore having both a direction perpendicular to the plane of $\vec{F}$ and $\vec{r}$ and a magnitude of $\text{F}\cdot\text{r}\cdot\text{sin} \theta$.

However, while in 2D torque still has a magnitude of $\text{F}\cdot\text{r}\cdot\text{sin} \theta$ it does not seem to have a direction.

So in 2D is torque a vector or just a directionless number? If it is directionless, how come the non-angular equivalent, force, is a vector in 2D though.

  • $\begingroup$ In any case torque is a vector. $\endgroup$
    – user82794
    Aug 2, 2015 at 6:46

2 Answers 2


Torque is a mathematical object called a bivector, produced by taking the wedge product of $\mathbf{r}$ and $\mathbf{F}$. Bivectors can be thought of as area elements of planes; the magnitude is equal to $rF \sin \theta$ and the plane itself contains $\mathbf{r}$ and $\mathbf{F}$.

By a great coincidence, in three dimensions, there are three distinct planes (xy, yz, and zx), and also three distinct axes (z, x, and y). As such, we can associate any bivector with a vector, which all textbooks do to avoid having to introduce extra math. (Cross product actually means "take the wedge product, then convert the resulting bivector into a vector".) The problem is that this doesn't work in any dimension other than three.

In two dimensions, there's only one plane, so you can associate bivectors with scalars. That is, you can say "the torque is +3 Newton-meters".

The other answer addressed two dimensions by adding a third dimension, but that doesn't work for four dimensions. In four dimensions, there are six planes, so you can't associate the wedge product with either a scalar or vector.


Torque is a vector whose direction is always out of plane. The same with angular velocity

$$ \vec{\tau} = \vec{r} \times \vec{F}$$ $$ (0,0,\tau) = (x,y,0) \times (F_x,F_y,0) = (0,0,x F_y - y F_x)$$ $$ \vec{v} = \vec{\omega} \times \vec{r}$$ $$ (v_x,v_y,0) = (0,0,\omega) \times (x,y,0) = (-y\, \omega,x \,\omega,0)$$

I always think of planar quantities as a 2D projection of their 3D equivalents.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.