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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

enter image description here

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?

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  • $\begingroup$ I have asked for the last part of the question (cross-check) in here math.stackexchange.com/questions/3177310/… $\endgroup$ – maharshi kintada Apr 16 at 3:19
  • $\begingroup$ 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. $\endgroup$ – John Omielan Apr 16 at 3:19
  • $\begingroup$ Thanks @JohnOmielan. I have corrected the drawing. $\endgroup$ – maharshi kintada Apr 16 at 3:22

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