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A general approach would be to construct a so called axis-angle representation and convert that to a rotation matrix representation. On order to do so, one could start with a normal vector of the triangle. Let's take the $P_1P_2$ line and construct the perpendicular of the triangle from $P_3$ to $P_1P_2$. Let's call $P_4$ the point where the ...

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Your 6d vector can also be considered a frame. Lets call it the frame of the object. Convert your 6d vector to something easier to work with like Homogeneous Transforms. Another alternative is Quaternion + Translation, but homogeneous matrices are easier. Homogeneous Transform: A 4x4 matrix that looks like: $T=\begin{bmatrix} R & t \\ 0_3 & 1 \end{... 2 I am just going to explain from the basics. So feel free to skip through the first part and scroll to the bottom if you want the answer. Basics: The 3 parameters of your pose are$x,y,\theta$. These can be stored as homogeneous matrix which is the combination of the translation($x,y$) and the rotation($\theta$). It looks like so$$\begin{bmatrix} cos(\theta) ... 1 If$T^{ee}_{root}$is the homogeneous transformation that relates coordinates expressed in the end-effector frame into coordinates expressed in the root frame, then it holds that$T^{root}_{ee} = \left( T^{ee}_{root} \right)^{-1}\$. Interestingly, the computation of the matrix inverse is straightforward and can be done by applying the rules reported in https:/...

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You can use the rotation matrix. linear_v= R*sensor_linear_v + p X sensor_angular_v p is the translation vector and the X is the cross operator angular_v=R*sensor_angular_v

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Yes it is. The rotation matrix of the coordinates system B relative to the coordinate system A or its inverse backwards, A relative to B.

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