# How does one explicitly compute the result of a rotation on a higher dimensional vector?

So a while ago I asked if one could use the octonions to rotate vectors in $\mathbb{R}^n$ for n<8 the same way the quaternions are used for $\mathbb{R}^3$. I've since learned that one must use elements of the even subalgebra of the clifford algebra of $\mathbb{R}^n$ to rotate vectors in $\mathbb{R}^n$. The quotient spaces of the tensor algebra, and the tensor algebra itself, are still things I'm trying to wrap my head around. But on to what I've thought of so far: given $v\in\mathbb{R}^n$ as well as the 2D subspace I want to rotate it in and the angle $\theta$ my initial thought is to write $$v=|v|n_1e^{I\phi}$$ where $\phi$ is an angle depending on the choice of $n_1$ and $n_2$ which are orthonormal and span the aforementioned 2D subspace and $I=n_1\wedge n_2$. It then seems straightforward to multiply the above quantity by $e^{I\theta}$. So my questions are: is this analysis correct? and does this mean that in $\mathbb{R}^n$ for n>3 that to compute a rotation you have to have a scalar ($\theta$) and two orthonormal vectors as data as opposed to a single vector denoting the axis rotation? Because, unless I'm misinformed, in dimension greater than three every plane has more than 1 direction for its normal vectors.

• In general rotation is defined in a plane (not wrt a normal vector... that's an idiosyncrisy of 3 dimensions). So two orthonormal vectors and an angle sounds about right. – spaceisdarkgreen Dec 30 '16 at 5:07
• Your formula will only work if the vector you want to rotate is in the plane of rotation. In general you will need to use the sandwich product. – Andrey Sokolov May 29 '17 at 23:06