# $\nabla \times \underline{v}$ - Results in a vector perpendicular to these two vectors?

Say $v = -y\hat{i} + x\hat{j}$

If we take the cross product of $\underline{v}$ with $\nabla$ we get

$\left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{d}{dx} & \frac{d}{dy} & \frac{d}{dz} \\ -y & x & 0 \end{array} \right|$

$= 2k$

So as the cross product of two vectors gives a vector that is perpendicular to those two vectors, the results of $2k$ seems to be saying that $\nabla$ is some vector with non-zero $\hat{i}$ and $\hat{j}$ components and a zero $\hat{k}$ component.

But $\nabla = \frac{d}{dx}\hat{i} + \frac{d}{dy}\hat{j} + \frac{d}{dz}\hat{k}$ doesn't give any indication of the magnitude of the $\hat{i}, \hat{j}$, and $\hat{k}$ components. So I don't understand how $\nabla \times \underline{v}$ resulted in vector of $2k$?

It's a useful mnemonic to remember that the curl is given by a rule formally similar to the formal determinant rule giving the cross product of two vectors, enough that we often denote $\nabla \times X := \mathrm{curl} X$ but that's all it is, a mnemonic; in particular $\nabla$ is neither a vector nor a vector field. When you "evaluate the determinant" in this mnemonic, you apply the (partial) derivatives in the middle row to the corresponding functions in the last row. As you noted, the $\hat{k}$-component of the curl is
$$\frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y) = 2$$
$$\mathrm{curl} (-y \hat{i} + x \hat{j}) = 2 \hat{k}.$$