How do I "change" the angle between three vectors in 3D space?

Question

In this question the answer shows how to get the angle between three points in 3D space. However I would like to know how to change the angle.

For example, if I have points $A,B$ and $C$ and I want to change the $ABC$ angle (where $B$ is the middle point), I should be able to get new points $C'$ and $A'$ given points $A,B,C$, the change in angle $\frac{d\theta}{dt}$ (or simply the desired angle if I know it) and the axes of the plane described by the 3 points.

I'm not sure if I described that 100% correctly so I'll also try and explain in an intuitive way. Imagine the three points at a certain angle. If I want the angle to get smaller or bigger I want to kind of rotate points $A$ and $C$ around point $B$ towards or away from each other. They'd rotate around the (y?) axis of the plane described by the three points. In other words the three points cannot rotate in such a way that causes the plane described by these three points to also rotate. I hope that kind of makes sense.

I'm not that great at maths so a kind of step by step explanation would be appreciated so I have a chance at understanding the answer rather than just being able to use it.

Answer 1

So, you have three points, $$\begin{cases} A = ( x_A ,\, y_A ,\, z_A ) \\ B = ( x_B ,\, y_B ,\, z_B ) \\ C = ( x_C ,\, y_C ,\, z_C ) \end{cases}$$ The angle $\theta = \angle ABC$ fulfills $$\cos(\theta) = \frac{ \overline{BA} \cdot \overline{BC} }{\left\lVert \overline{BA} \right\rVert \left\lVert \overline{BC} \right\rVert }$$ and $$\sin(\theta) = \frac{ \left\lVert \overline{BA} \times \overline{BC} \right\rVert}{\left\lVert \overline{BA} \right\rVert \left\lVert \overline{BC} \right\rVert}$$ where $\cdot$ denotes vector dot product, and $\times$ denotes vector cross product.

If we want to rotate $C$ around $B$ so that the angle $\angle ABC$ becomes $\theta + \varphi$, we first find the normal of the plane the three points form: $$N = ( x_N ,\, y_N ,\, z_N ) = \frac{\overline{BA} \times \overline{BC}}{\left\lVert\overline{BA} \times \overline{BC}\right\rVert}$$ If we first calculate the cross product components, $$\begin{cases} x = ( y_A - y_B ) ( z_C - z_B ) - ( z_A - z_B ) ( y_C - y_B ) \\ y = ( z_A - z_B ) ( x_C - x_B ) - ( x_A - x_B ) ( z_C - z_B ) \\ z = ( x_A - x_B ) ( y_C - y_B ) - ( y_A - y_B ) ( x_C - x_B ) \end{cases}$$ we get the components of the normal vector $N$ by "normalizing" the length of the resulting vector to unit length, or 1: $$\begin{cases} x_N = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \\ y_N = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \\ z_N = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \end{cases}$$ We now simply need to rotate $\overline{BC}$ around the vector $N$ by angle $\varphi$.

We can do that by applying Rodrigues' rotation formula to $\overline{BC} = ( x_C - x_B , y_C - y_B , z_C - z_B )$, to get the new position $C'$ relative to $B$. Essentially, $$C' = B + \overline{BC} \cos(\varphi) + \left( N \times \overline{BC} \right) \sin(\varphi) + N \left( N \cdot \overline{BC} \right) \left(1 - \cos(\varphi) \right)$$ Note that the definition of $\theta$ (and the rotation angle $\varphi$) above define it as positive counterclockwise. If $\angle ABC$ is clockwise — that is, $A$ is to the right of $C$, when viewing from $B$ —, then $\theta$ is negative, and a positive $\varphi$ will then decrease the angle.

If you want to rotate both $A$ and $C$ by the same amount, so that the angle $\angle ABC$ changes from $\theta$ to $\theta + \varphi$, you rotate $A$ by $-\varphi/2$ and $C$ by $\varphi/2$: $$A' = B + \overline{BA} \cos\left(\frac{\varphi}{2}\right) - \left( N \times \overline{BA} \right) \sin\left(\frac{\varphi}{2}\right) + N \left( N \cdot \overline{BA} \right) \left(1 - \cos\left(\frac{\varphi}{2}\right) \right)$$ because $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$; and $$C' = B + \overline{BC} \cos\left(\frac{\varphi}{2}\right) + \left( N \times \overline{BC} \right) \sin\left(\frac{\varphi}{2}\right) + N \left( N \cdot \overline{BC} \right) \left(1 - \cos\left(\frac{\varphi}{2}\right) \right)$$

Answer 2

With three points, there is always a unique plane passing through them. So considering this problem in 3D space is identical to doing it on a flat sheet of paper.

If your coordinates make this difficult, we can transform to better coordinates. We'll set $B$ as the origin. Letting $d_{AB}$ be the distance between $A$ and $B$, we'll put $A$ on the new $x$-axis, at the point $d_{AB}$ from the origin. Now draw a line from $C$ perpendicular to the new $x$-axis. This will be the direction of the new $y$-axis, so that all three points will be on this plane.

Once you're back in 2D, the problem should be easily visualizable.