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 April 18th, 2019, 11:10 AM #1 Newbie   Joined: Apr 2019 From: Pune Posts: 1 Thanks: 0 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
 April 19th, 2019, 02:39 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,613 Thanks: 2071 How would $\frac{\partial v}{\partial d}$ be defined?

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