Curve tilt of instanced curves to follow sampled geometry

Question

I have a node set up that requires curves to follow a plane. (the plane is manipulated outside of geonodes and the curves follow)

Is it possible to have the curve tilt follow the normals of the plane?

*Edit: Ended up following the solution in this thread: https://blender.stackexchange.com/a/275131/88095 *

Current set up, with the tilt of the curve unaffected by the normals of the plane: Desired result, note the normal/tilt of the curve adhering to the normals of the plane (i've used instance on points as demonstration, but not what I'm after):

Answer 1

This GN group will return the signed angle between two vectors A and B about the axis N:

So could be used to rotate an existing Curve Normal about its tangent to align with another vector. In this case, it's aligning to the normal of a surface, on to which the curve is projected:

With this kind of result.. (excuse poor sampling):

.. although I think, if you're instancing on the curve, rather than orienting a mesh-profile for it, you'll run into fewer problems with interpolation by aligning the instances themselves, than the tilt of the curve they're on.

Answer 2

(Using Blender 3.6.5)

NB: This post is to illustrate/to demonstrate/to complement Robin Betts's answer.

Objective

Let $\vec{N}$ (in blue), $\vec{A}$ (in magenta), $\vec{B}$ (in green) be three vectors such that $\vec{N}$ is not collinear to $\vec{A}$, neither to $\vec{B}$. The objective is to compute the rotation angle around the axis defined by $\vec{N}$, bringing $\vec{A}$ to be coplanar to $\vec{N}$ and $\vec{B}$ (in red).

Mathematics

Rotation around the axis defined by $\vec{N}$ is affecting only the vector components projected in a plane perpendicular to $\vec{N}$ (illustrated by the blue triangle). The vector component align with $\vec{N}$ is unchanged.
Let $\vec{A}^\perp$ and $\vec{B}^\perp$ be the restrictions of $\vec{A}$ and $\vec{B}$ perpendicular to $\vec{N}$ (illustrated as thinner vectors connected by triangles of same colour). These are defined by removing the projection along the axis, as: $$\left\{ {\begin{array}{rcl} \vec{A}^\perp & = & \vec{A} - (\vec{A} \cdot \vec{n}) \times \vec{n} \\ \vec{B}^\perp & = & \vec{B} - (\vec{B} \cdot \vec{n}) \times \vec{n} \end{array}} \right. $$ where $\vec{n} = \vec{N} / \| \vec{N}\|$ is the unitary vector align with the axis.
From the definitions of dot product and cross product, $$\left\{ {\begin{array}{rcl} \vec{A}^\perp \cdot \vec{B}^\perp & = & \| \vec{A}^\perp \| \times \| \vec{B}^\perp \| \times \cos{\alpha} \\ \vec{A}^\perp \wedge \vec{B}^\perp & = & \| \vec{A}^\perp \| \times \| \vec{B}^\perp \| \times \sin{\alpha} \times \vec{n} \end{array}} \right. $$ where $\alpha$ is the signed angle from $\vec{A}^\perp$ to $\vec{B}^\perp$, $\tan{\alpha}$ is computed as: $$ \tan{\alpha = \frac{\sin{\alpha}}{\cos{\alpha}} = \frac{(\vec{A}^\perp \wedge \vec{B}^\perp) \cdot \vec{n}}{\vec{A}^\perp \cdot \vec{B}^\perp}}$$ Taking advantage of $\vec{n} \cdot \vec{n} = 1$, $\vec{A}^\perp \cdot \vec{B}^\perp$ is written: $$\vec{A}^\perp \cdot \vec{B}^\perp = \vec{A} \cdot \vec{B} - (\vec{A} \cdot \vec{n}) \times (\vec{B} \cdot \vec{n})$$ Taking advantage of $\vec{n} \wedge \vec{n} = \vec{0}$ and of $\vec{U} \wedge \vec{n} \perp \vec{n}$, $(\vec{A}^\perp \wedge \vec{B}^\perp) \cdot \vec{n}$ is written: $$(\vec{A}^\perp \wedge \vec{B}^\perp) \cdot \vec{n} = (\vec{A} \wedge \vec{B}) \cdot \vec{n}$$ As a conclusion: $$\tan{\alpha} = \frac{(\vec{A} \wedge \vec{B}) \cdot \vec{n}}{\vec{A} \cdot \vec{B} - (\vec{A} \cdot \vec{n}) \times (\vec{B} \cdot \vec{n})}$$

Geometry Nodes graph

Without hypothesis for $\vec{N}$, $\vec{A}$ and $\vec{B}$, $\alpha$ can then be computed with the following Node Group:

Resources

Blender file:

Moving vertices A,B and N, different configurations can be dynamically simulated.