Calculate angle between two vectors, given their rotation w.r.t. a third vector.

Question

I have three vectors, $\vec{a},\vec{b},$ and $\vec{c}$ in $n$-dimensional space. I know the coordinates of all three vectors and their dot products. Both $\vec{a}$ and $\vec{b}$ are rotated away from $\vec{c}$ by an angle $\alpha$, in their own respective directions, obtaining $\vec{a}'$ and $\vec{b}'$. What is the angle between $\vec{a}'$ and $\vec{b}'$?

I have worked on finding the generalized rotation matrices in $n$-dimensions, calculating $\vec{a}'$ and $\vec{b}'$ using Clifford Algebra using the answers from the posts below. However, these methods require many matrix operations and are too slow. I am wondering whether there is a neater solution without requiring the calculation of the two rotated $\vec{a}'$ and $\vec{b}'$ vectors first.

Generalized rotation matrix in N dimensional space around N-2 unit vector Finding the rotation matrix in n-dimensions How can I calculate a $4\times 4$ rotation matrix to match a 4d direction vector?

Answer 1

Assuming that $a, b, c$ have norm 1, we have \begin{align} a' &= 2(a\cdot c)a - c\\ b' &= 2(b\cdot c)b - c\\ \end{align} Indeed, this implies $a'\cdot a = a\cdot c$ and $a'\cdot c = 2(a\cdot c)^2-1$ which is the cosine of the double angle. Alternatively, it is obvious geometrically that $\frac{a'+c}{2}$ is the orthogonal projection of $c$ on $a$, that is to say $(a\cdot c) a$.

Hence \begin{equation} a'\cdot b' = 4 (a\cdot c)(b\cdot c)(a\cdot b) - 2(a\cdot c)^2 - 2(b\cdot c)^2 +1\end{equation}