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April 18th, 2019, 11:10 AM   #1
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How to find the rotation vector by deriving the final vector with respect to the disp

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][1]][1]

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?


[1]: https://i.stack.imgur.com/hYDJn.png
MaharshiKintada is offline  
 
April 19th, 2019, 02:39 AM   #2
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How would $\frac{\partial v}{\partial d}$ be defined?
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