# Rotate vector by angles between a unit vector and the positive x axis [duplicate]

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If I know that a vector $\vec{y}$ resulted from shifting another vector by a vector $\vec{d}$, it is easy to get the original vector back as simple as $\vec{x}=\vec{y}-\vec{d}$.

Now suppose that $\vec{y}$ resulted from rotating $\vec{x}$ by the angles between a given unit vector $\vec{e}$ and the positive $x$ axis. How can I get the $\vec{x}$ back as something like

\begin{align} \vec{x}=\vec{y}\curvearrowright\vec{e} & \qquad\text{rotate "clockwise" by angles of }\vec{e}\\ \text{when it was created by}\\ \vec{y}=\vec{x}\curvearrowleft\vec{e} & \qquad\text{rotate "counter clockwise" by angles of }\vec{e} \end{align}

Obviously I made up the operations $\curvearrowleft$ and $\curvearrowright$ to symbolize counter- and clockwise rotation (if only for the fun of finding these symbols :-).

The only idea I have is to use a rotation matrix created from $\vec{e}$, but this looks daunting. Am I missing something simpler, like in the complex plane, where this would just be a multiplication?.

FOLLOWUP question: What would be $\frac{d}{dt} (\vec{x}\curvearrowleft\vec{e}(t))$?

I think I need this to answer this: Letting a function $f$ move through space along a path s while rotating it to keep it aligned with ds/dt

I tried that Vector operation: add angles, multiply absolue values. How is it called? but it did not work.

## marked as duplicate by Brahadeesh, Namaste, José Carlos Santos, Shailesh, LeucippusAug 27 '18 at 6:28

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• It’s “just a multiplication” with matrices, too. In fact, there’s a one-to-one correspondence between complex numbers with modulus $1$ and $2\times2$ rotation matrices. – amd Aug 25 '18 at 7:17