# Transformation of unit vectors from cartesian coordinate to cylindrical coordinate

Let $(\hat i, \hat j, \hat k)$ be unit vectors in Cartesian coordinate and $(\hat e_\rho, \hat e_\theta, \hat e_z)$ be on spherical coordinate. Using the relation, $$\hat e_\rho = \frac{\frac{\partial \vec r}{\partial \rho}}{ \left | \frac{\partial \vec r}{\partial \rho} \right |}, \hat e_\theta = \frac{\frac{\partial \vec r}{\partial \theta}}{ \left | \frac{\partial \vec r}{\partial \theta} \right |}, \;\; \hat e_z = \frac{\frac{\partial \vec r}{\partial z}}{ \left | \frac{\partial \vec r}{\partial z} \right |}$$ We have the relation $$\begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix}$$ $$\text { Let } A = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}$$ To express unit vectors of Cartesian coordinate in Spherical coordinates, the author uses, $$\hat i = \frac { \begin{vmatrix} \hat e_{\rho} & \sin \phi & 0\\ \hat e_{\theta} & \cos \phi & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\ \hat j = \frac { \begin{vmatrix} \cos \phi & \hat e_{\rho} & 0\\ -\sin \phi & \hat e_{\theta} & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\\hat k = \frac { \begin{vmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & \hat e_z \end{vmatrix} }{|A|}$$ Which I cannot understand! Can anyone help me to understand it? $$\begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix}$$ Looks intuitive but certainly the previous way is faster. I would like to know above relation works if it works. Thank you!!

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That is just Cramer's Rule applied to the linear system given by the equation following: "We have the relation..." see: en.wikipedia.org/wiki/Cramer's_rule – David Wheeler Jun 15 '12 at 19:39
No ... i meant how could write put that unit vectors in column of matrix A?? – Santosh Linkha Jun 15 '12 at 19:41
The $\hat{z}$ at the bottom right of the last numerator is probably a typoe and should be $1$. – Arturo Magidin Jun 15 '12 at 19:45
Yes that's an error, it was $\hat e_z$ – Santosh Linkha Jun 15 '12 at 19:48
@DavidWheeler thanks!! – Santosh Linkha Jun 15 '12 at 21:27

In fact unit vectors are components of the determinant, not the matrix $A$. There is nothing wrong with it. Determinant is really an antisymmetric linear form, so you still have vector quantities on both sides of the relation. EDIT After a closer look the formulae do not seem entirely correct. According to Cramer's rule the numerator must be look like the system determinant with one column replaced by the RHS:
$$\hat i = \frac { \begin{vmatrix} \hat e_{\rho} & \sin \theta & 0\\ \hat e_{\theta} & \cos \theta & 0\\ \hat e_z & 0 & 1 \end{vmatrix} }{|A|} \\ \hat j = \frac { \begin{vmatrix} \cos \theta & \hat e_{\rho} & 0\\ -\sin \theta & \hat e_{\theta} & 0\\ 0 & \hat e_z & 1 \end{vmatrix} }{|A|} \\\hat k = \frac { \begin{vmatrix} \cos \theta & \sin \theta & \hat e_{\rho}\\ -\sin \theta & \cos \theta & \hat e_{\theta}\\ 0 & 0 & \hat e_z \end{vmatrix} }{|A|}$$ Which computationally lead to the same result. The symbol for polar angle should be consistent throughout.