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Hi i know this is a really really simple question but it has me confused.

I want to calculate the cross product of two vectors $$ \vec a \times \vec r. $$ The vectors are given by $$ \vec a= a\hat z,\quad \vec r= x\hat x +y\hat y+z\hat z. $$ The vector $\vec r$ is the radius vector in cartesian coordinates.

My problem is: I want to calculate the cross product in cylindrical coordinates, so I need to write $\vec r$ in this coordinate system.

The cross product in cartesian coordinates is $$ \vec a \times \vec r=-a y\hat x+ax\hat y, $$ however how can we do this in cylindrical coordinates? Thank you

user1111
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The radius vector $\vec{r}$ in cylindrical coordinates is $\vec{r}=\rho\hat{\rho}+z\hat{z}$. Calculating the cross-product is then just a matter of vector algebra:

$$\vec{a}\times\vec{r} = a\hat{z}\times(\rho\hat{\rho}+z\hat{z})\\ =a(\rho(\hat{z}\times\hat{\rho})+z(\hat{z}\times\hat{z}))\\ =a\rho(\hat{z}\times\hat{\rho})\\ =a\rho\hat{\phi},$$

where in the last line we've used the orthonormality of the triad $\{\hat{\rho},\hat{\phi},\hat{z}\}$.

David H
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    @user1111 Because the unit vectors vary from point to point, as opposed to cartesian unit vectors. That means that $\hat{r}(1,0,0) \neq \hat{r}(0,1,0)$, for example. I wrote $\hat{r} \equiv \hat{r}(x,y,z)$ to mean that $\hat{r}$ is a function of position. – Mark Fantini May 19 '14 at 21:50
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    Do not feel stupid, that is one of the greatest difficulties in adapting to different coordinate systems. I suggest you don't think of it as a radius vector but as a position vector. Therefore position depends on the distance in the $xy$ plane and height. It should be easier to think that way. – Mark Fantini May 19 '14 at 21:54