I have a rectangular plane in 3D space, whose slope (height / length) along the X axis is say 0.3 and along the Y axis say 0.1. Now I rotate the plane around the Z axis by say 45°. How do I get the new slopes along X and Y axes respectively ? I've tried multiplying the slopes with sin(45) and / or cos(45), but both gave completely different results than I get by recreating that plane in a 3D design application (Blender), rotating it by 45° and calculating the slope from the vertex (point) positions.
EDIT: Here is a screenshot of a 3D rendering of the grid explained in the chat, on the left with both max-slope and max-bank = 0, at the right with both = 1 (raw terrain elevations without clamping). The problem is that for max-slope = 1 and max-bank = 0, applying the formula in the answer below, I get new max-slope = 0.328899 and max-bank = 0.944365, which results in no clamping at all in this example, and would result in insufficient clamping for steeper terrain.
If needed I can add more screenshots from different angles.
If we represent positions (points) and displacements (vectors) by $3 \times 1$ columns of numbers, then rotating through angle $\theta$ about the $z$-axis is accomplished by multiplying on the left by the rotation matrix $$ R_{\theta} = \left[ \begin{array}{@{}ccc@{}} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \phantom{-}\cos\theta & 0 \\ 0 & 0 & 1 \end{array} \right]. $$ If the plane has normal vector $n = (A, B, C)$, the rotated plane has normal vector $$ R_{\theta} n = \left[ \begin{array}{@{}c@{}} A\cos\theta - B\sin\theta \\ A\sin\theta + B\cos\theta \\ C \end{array} \right]. $$