From modern robotics chapter 3.2 and figure 3.7, how do we obtain $p_{b}=\begin{bmatrix}1 & -1 & 0\end{bmatrix}^{T}$ from $R_{b}$ and $p_{a}$? Because $R_{b}*p_{a} =\begin{bmatrix}-1 & 1 & 0\end{bmatrix}^{T}$ which is not $p_{b}$...

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  • $\begingroup$ AH okay, I see it now. So, $p_{a} = R_{b}^{a}p_{b}$ then $(R_{b}^{a})^{-1}p_{a} = p_{b}$ $\endgroup$ Sep 12, 2022 at 15:08

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One of the usage of rotation matrices is to represent the orientation of one frame relative to another. If we would like to represent the orientation of frame {$b$} relative to frame {a}, we need $R^a_b$, so if we have a vector in frame {$b$} (i.e. $p_b$) then the same vector expressed in frame {$a$} is $p_a=(R^a_b) p_b \implies p_b=(R^a_b)^{-1} p_a$. Notice $(R^a_b)^{-1} = (R^a_b)^T$ which is more efficient than the inverse.


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