# How to find the rotation vector by deriving the final vector with respect to the displacement?

My understanding of a rotation of a vector can be done by using a 2D rotation matrix as shown below,

$$R(\theta )=\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}$$.

This rotates column vectors by means of the following matrix multiplication,

$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix}$$

For example, if you rotate the vector x=$$\begin{bmatrix}1\\1 \end{bmatrix}$$ by 45 degrees (clockwise), then the new vector is $$\begin{bmatrix} \sqrt2 \\ 0 \end{bmatrix}$$.

Other Method:

If I have only initial and final coordinates of the vectors

The initial vector is, V = $$\begin{bmatrix}1\\1 \end{bmatrix}$$ and the final vector is, v = V+d = $$\begin{bmatrix} \sqrt2 \\ 0 \end{bmatrix}$$.

The displacement between these vectors is d = $$\begin{bmatrix} \sqrt2-1 \\ -1 \end{bmatrix}$$.

Can I derive the final vector v with respect to displacement $$\frac{\partial{v}}{\partial{d}}$$ to get the rotation vector? [but returns a identity matrix]

If so, does $$\frac{\partial{v}}{\partial{d}} * d$$ can be used to cross-check?

• I have asked for the last part of the question (cross-check) in here math.stackexchange.com/questions/3177310/… – maharshi kintada Apr 16 at 3:19
• A minor point that I initially found a bit confusing, and which other people might as well, is that your drawing of the initial vector is of $(-1,-1)$, not $(1,1)$. The arrow should be on the upper right part, not the lower left. – John Omielan Apr 16 at 3:19
• Thanks @JohnOmielan. I have corrected the drawing. – maharshi kintada Apr 16 at 3:22