TITLE: How to calculate cross product of $\hat{x}$ and $-\hat{x}$ in cylindrical coordinates
QUESTION [0 upvotes]: I'm sorry if this seems like a stupid question. I have never had to calculate cross products of vectors that weren't in cartesian coordinates before.
If I try to write the $\hat{x}$ vector in cylindrical form, the result is $\hat{r}$, whereas $-\hat{x}$ would be $\hat{r} + \pi \hat{\theta}$. The cross product between them should be $\vec{0}$, however that isn't the case when I try to calculate it with the determinant, and what I end up getting is that the cross product is $\pi \hat{z}$ which doesn't make sense since the vectors are parallel.

REPLY [3 votes]: In cylindrical coordinates, you have the unit vectors
$$\begin{cases}\hat r = \frac{x\,\hat\imath+y\,\hat \jmath}{r} = \frac{r\cos(\theta)\,\hat \imath + r\sin(\theta)\,\hat \jmath}{r} = \cos(\theta)\,\hat \imath + \sin(\theta)\,\hat \jmath \\ \hat z = \hat k \\ \hat\theta = \hat z \times \hat r = -\sin(\theta)\,\hat\imath + \cos(\theta)\,\hat \jmath\end{cases}$$
Solving for the Cartesian unit vectors gives
$$\begin{cases}\hat\imath = \cos(\theta)\,\hat r - \sin(\theta)\,\hat\theta \\ \hat\jmath = \sin(\theta)\,\hat r + \cos(\theta)\,\hat\theta \\ \hat k = \hat z\end{cases}$$
So if $\hat x = x_1\,\hat\imath + x_2\,\hat\jmath + x_3\,\hat k$, where $x_1=r\cos(\theta)$ and $x_2=r\sin(\theta)$, its cylindrical coordinate representation would be
$$\hat x = (x_1\cos(\theta)+x_2\sin(\theta))\,\hat r + (x_2\cos(\theta)-x_1\sin(\theta))\,\hat\theta + x_3\,\hat z$$
$-\hat x$ is the same vector, but $\theta$ is replaced with $\theta\pm\pi$, which gives $\cos(\theta\pm\pi)=-\cos(\theta)$ and $\sin(\theta\pm\pi)=-\sin(\theta)$; ultimately the same effect as multiplying $\hat x$ by $-1$.
Then the cross product of $\hat x$ and $-\hat x$ is
$$\hat x\times(-\hat x) = 
\begin{vmatrix}
\hat r & \hat \theta & \hat z \\
x_1\cos(\theta)+x_2\sin(\theta) & x_2\cos(\theta)-x_1\sin(\theta) & x_3 \\
-x_1\cos(\theta)-x_2\sin(\theta) & -x_2\cos(\theta)+x_1\sin(\theta) & -x_3
\end{vmatrix} = \hat 0$$
since the second and third rows are linearly dependent.
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