why does $|r'|^2 = r'^2 + r^2\theta^{\prime2}$?

Question

For a particle moving in 3 dimensions. I know Angular momentum is $mr^2θ$ in the z dimension

edit: not sure how to do absolute value in math terms

here is some more info Consider a particle moving in three dimensions with no force. I think its motion is $ r = rcosθ x + rsinθ y + 0z $

Answer 1

Under a radial force, motion is confined to a plane orthogonal to the conserved angular momentum. Polar coordinates in that plane satisfy$$\vec{r}=r\hat{r}\implies\vec{r}^\prime=r^\prime\vec{r}+r\hat{r}^\prime.$$Since $\hat{r}=\hat{i}\cos\theta+\hat{j}\sin\theta$, $\frac{d\hat{r}}{d\theta}=-\hat{i}\sin\theta+\hat{j}\cos\theta$ is orthogoanl to $\hat{r}$, and$$\hat{r}^\prime=\theta^\prime\frac{d\hat{r}}{d\theta}\implies\vec{r}^\prime\cdot\vec{r}^\prime=r^{\prime2}+r^2\theta^{\prime2}.$$