Transformation of two vectors in high dimensional space, which makes the angle unchanged

Question

In the $n$-dimensional Euclidean space, given two vectors $\vec{OA}$ and $\vec{OB}$ (not collinear), their angle AOB is $\theta$. Now there is a vector $\vec{OC}$, the angle AOC is also equal to $\theta$. I want the general transformation for all the possible $\vec{OC}$s.

In case I describe it incorrectly, I give a example in $3D$. Imagine a circular cone, $\vec{OA}$ is like the rotation axis, $\vec{OB}$ is like one of the generatrixes, then all the possible $\vec{OC}$s also form the set of generatrixes.

(ps: I don't think it's a simple pure rotation problem...)

Answer 1

I dont know if this is what you want, but maybe it helps:

$ M = \{ V \in \mathbb{R}^n| \angle(A-O,V-O) = \angle(A-O,C-O) \} = \{V \in \mathbb{R}^n | \frac{<A-O,V-O>}{||A-O|| \ ||V-O||} =\frac{<A-O,C-O>}{||A-O|| \ ||C-O||} \} = \{V \in \mathbb{R}^n | \frac{<A-O,V-O>}{ \ ||V-O||} =\frac{<A-O,C-O>}{ \ ||C-O||} =: c \} = \{V \in \mathbb{R}^n | {<A-O,\frac{V-O}{||V-O||}>}{ \ } = <A-O,\frac{C-O}{ \ ||C-O||}> \} = \{V \in \mathbb{R}^n | {<A-O,\frac{V-O}{||V-O||}-\frac{C-O}{ \ ||C-O||}>}{ \ } = 0 \} \ i.e. \ \frac{V-O}{||V-O||}-\frac{C-O}{ \ ||C-O||} \in Ker( <A-O, *>) = (A-O)^\bot $ Hence, all the Vectors in the orthogonal complement of $ A-O$. Now to get back to the $V$, take any vector $v$ in $ (A-O)^\bot , $add $\ c, $ multiply by any nonzero number $a$, (pointing in the same direction, i.e. don't switch signs) and add O.

Answer 2

Based on @a_student 's answer, I found a candidate solution. Let $\vec{CV} = \lambda \vec{v}$, and $\vec{OV} = \vec{OC} + \vec{CV}$. $ \because \angle VOA = \angle COA,$ $\therefore \frac{<\vec{OV},\ \vec{OA}>}{||\vec{OV}|| \ ||\vec{OA}||} =\frac{<\vec{OC},\ \vec{OA}>}{||\vec{OC}|| \ ||\vec{OA}||}, $ $\therefore \frac{<\vec{OC}+\lambda \vec{v},\ \vec{OA}>}{||\vec{OC}+\lambda \vec{v}|| \ ||\vec{OA}||} =\frac{<\vec{OC},\ \vec{OA}>}{||\vec{OC}|| \ ||\vec{OA}||}, $ For now, only one unknown variable $\lambda $ is left, we can build a quadratic equation with one unknown. It has two solutions for $\lambda $, one of which is 0 (means V and C is overlapped). The other one is $\frac{-2(<\vec{v},\ \vec{OA}> \cdot ||\vec{OC}||^2 - <\vec{OC},\ \vec{OA}>^2 \cdot <\vec{v},\ \vec{OC}>)}{<\vec{v},\ \vec{OA}>^2 \cdot ||\vec{OC}||^2 - <\vec{OC},\ \vec{OA}>^2 \cdot ||\vec{v}||^2 }.$

I guess there are other types of solutions.