# How is the displacement vector of a point object performing circular motion calculated in terms of the radius vector and the angular displacement?

How is the displacement vector of a point object performing circular motion given by $d\vec{s}=d\vec{\theta}\times\vec{r}$? This equation was stated on Wikipidia as a part of the proof for the equation for calculating the work done by a torque. I understand why the magnitude of d$\vec{s}$ is the same as that of $d\vec{\theta}\times\vec{r}$, but I'm struggling to understand how the directions of the two vectors would be the same.I'd really appreciate it if someone could help me understand this.

## 1 Answer

Angular displacement is change in angular position, so it's direction is not tangential but instead is perpendicular to the plane of instantaneous rotation. For instance, if an object is rotating, at some instant, in the x-y plane, $d\vec{\theta}$ points in the z-direction.

Therefore, $d\vec{\theta}\times\vec{r}$ lies in the plane the particle is (instantaneously) travelling in, but perpendicular to the radial direction, exactly as you would expect for a rotation (so that the distance of any given point from the origin is fixed).

Usually, with questions like this, the easiest way to gain intuition is to think about the plane in which the object is (instantaneously) rotating, because rotations in 2D are much simpler than rotations in 3D