Rotating vector functions

It looks correct to me. One way to see this is to realize that $$ R^T(\theta) = R(\theta)^{-1} $$

So your formula says that to find the value at $(x, y, z)$, we should first rotate $(x, y, z)$ by $-\theta$, and then find the projection of the resulting vector onto the (original) $x-$, $y-$, and $z-$axes, i.e., find the coordinates in the original domain of the vector field. [Those are your three dot products].

We then find the vector that's "attached there" in the original vector field (that's the application of "E") and rotate that into the new coordinates (that's the $R(\theta)$). So yes, it's fine.

If you're willing to write $E$ as a function of a single 3-component vector rather than a function of three separate arguments, then this all simplifies a good deal: $$ E_\text{new}(\mathbf v) = R \cdot E( R^{-1}(\mathbf v)), $$ at which point my description above can be read much more easily.

In particular, this version (with no "thetas") works in any dimension, for any rotation matrix $R$.

If I may I would like to shed a little light on this for any students who might be struggling with this sort of thing.

There are two vectors at play here, the vector field E(v), and the position vector v(x,y).

It is clear from the question that we want to rotate the whole thing, as demonstrated by the graphics.

When we apply the rotation matrix R to the vector field we get $\mathbf{R}\cdot \mathbf{E}(\mathbf{v})$ and this rotates all of the vectors E by pi/2 CW, but the E vectors remain where they are so to speak, at (x,y).

To rotate the E vectors about the origin we need to change the argument v(x,y) of E, or put another way, rotate v by pi/2 CW. If we make a transformation $\mathbf{v}'(x',y') = \mathbf{R}\cdot\mathbf{v}(x,y)$ then the (x,y) coordinates of every E vector get rotated pi/2 CW about the origin to (x'y'), and so does the position of every vector E.

This has the effect of "moving" every E vector to the position we want it. A little manipulation gives

$$\begin{pmatrix} {x} \\ {y} \end{pmatrix}= \mathbf{R}^{-1} \cdot \begin{pmatrix} x' \\ y' \end{pmatrix} $$

so we get

$$\mathbf{E}_{\mbox{new}}(x',y') = \mathbf{R}\cdot\mathbf{E}\left(\mathbf{R}^{-1}\cdot \begin{pmatrix} x' \\ y' \end{pmatrix}\right)$$.

The notation above is a little misleading because (x',y') is often used to indicate the (x,y) coordinates when transformed to a new coordinate system. In this case we have not changed the coordinate system to (x',y') but we have changed the argument of E.

Let's look at the specific example above… E(x,y) = (x,1) (first graphic) and $\mathbf{E}_{\mbox{new}}(x',y') = \mathbf{R}\cdot\mathbf{E}\left(\mathbf{R}^{-1}\cdot \begin{pmatrix} x' \\ y' \end{pmatrix}\right) = (1,y')$ (second graphic).

I am only an amateur at this stuff and will be happy to accept corrections from those with greater knowledge.

PS Sorry about the poor formatting but I will fix it up when I learn how.