# Representing displacement vectors in cylindrical coordinates and finding the distance in cylindrical coordinates? In cartesian coordinates, we can derive the vector $\vec v_3$ by vector subtraction $\vec v_2-\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\vec v_3$ which then is: $$\lvert \vec v_3\rvert = \lvert \vec v_2-\vec v_1 \rvert= \lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \rvert = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

But how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\rho_1$ and $\theta_1$ for $\vec v_1$ as well as $\rho_2$ and $\theta_2$ for $\vec v_2$ in cylindrical coordinates $(\rho,\theta,z)$, where $z=0$ in this example.

Your problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\vec v_3|^2=(\vec v_2-\vec v_1)\cdot(\vec v_2-\vec v_1)=|\vec v_1|^2+|\vec v_2|^2-2\vec v_2\cdot \vec v_1$$ Now using cylindrical coordinates, $|\vec v_i|^2=r_i^2+z_i^2$ and $\vec v_2\cdot \vec v_1=r_1r_2\cos(\theta_2-\theta_1)+z_1z_2$. Then the final answer will be $$|\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)+(z_2-z_1)^2$$
• Yes, I've seen this solution, but isn't it based in terms of cartesian coordinates/base vectors, $\hat e_x$, $\hat e_y$ and $\hat e_z$ after all? This since, I guess, you must express a distance in constant base vectors? I'm a bit confused about how to interpret the problem I have to admit. How would it look if I want to express the solution completely in cylindrical coordinates with $\vec v_1=\rho_1 \hat e_\rho (\theta_1)$ and base vectors $\hat e_\rho$, $\hat e_\theta$, and $\hat e_z$ etc? Would that be possible? Sep 6, 2018 at 5:46
• I did not use Cartesian coordinates. The $z$ coordinate is there in cylindrical system. The only other reference to the Cartesian system is the angle $\theta$. The issue that you have is that the basis of the cylindrical coordinate system changes with the vector, therefore equations will be more complicated. Sep 6, 2018 at 6:38
• If you choose one of the vectors as the reference, the cylindrical coordinate system is just a Cartesian system rotated with respect to the original. $\hat e_\rho$ is the new $\hat e_x$, and $\hat e_\theta$ is the new $\hat e_y$. Sep 6, 2018 at 7:19