\subsection{Positioning}
\shortdefinition[Position Vector]
$_{\color{blue}\fbox{W}}\,\vec{t}\,_{\color{red}\fbox{B}} = \, _{\color{blue}\fbox{W}}\,\vec{t}\,_{\color{ForestGreen}\fbox{W}}\,_{\color{red}\fbox{B}}$,
{\color{blue} Original Frame}, {\color{red} End point}, {\color{ForestGreen} Target Frame},
\hl{$\sin = s$, $\cos = c$}

\shortdefinition[State vector] $x_R$: $x$, $v$ of rob in $W$, pos of sensors

\shortdefinition[Rot. Mat.] $\mat{R}_{z}(\psi) = \begin{bmatrix}
        c(\psi) & -s(\psi) & 0 \\
        s(\psi) & c(\psi)  & 0 \\
        0       & 0        & 1
    \end{bmatrix}$\\
$\mat{R}_y(\theta) = \begin{bmatrix}
        c(\theta)  & 0 & s(\theta) \\
        0        & 1 & 0       \\
        -s(\theta) & 0 & c(\theta) \\
    \end{bmatrix};
    \mat{R}_x(\varphi)
    \begin{bmatrix}
        1 & 0       & 0        \\
        0 & c(\varphi) & -s(\varphi) \\
        0 & s(\varphi) & c(\varphi)
    \end{bmatrix}$

\shortremark Application: ${_W} \vec{a} = \mat{R}_{WB} {_B} \vec{a}$

\shortlemma $\mat{R}_{BW} = \mat{R}_{WB}^{-1} = \mat{R}_{WB}^\top$, $\det(\mat{R}_{WB}) = 1$ (orth.)

\shortremark Cols of $\mat{R}_{WB}$ are basis vec. of Frame $\underset{\rightarrow}{\cF}{_B}$ in $\underset{\rightarrow}{\cF}{_W}$

\shortdefinition[Euler Angles] Yaw ($z$), Pitch ($y$), Roll ($x$), mult. rotation matrices, e.g.
$\mat{R}_{EB} = \mat{R}_z(\psi) \cdot \mat{R}_y(\theta) \cdot \mat{R}_x(\varphi)$, \hl{bound.}.
$\qquad [\vec{n}]^\times = \vec{n} \vec{x}^\top$ (matrix from vec + arg $\vec{x}$)

\shortdefinition[Rot. Vec]
$\vec{\alpha} = \alpha \vec{n}$ ($\vec{n}$ normal)\\
$\mat{R}(\alpha, \vec{n}) = \mat{I}_3 + \sin(\alpha)[\vec{n}]^\times + (1 - \cos(\alpha))([\vec{n}]^\times)^2$

\shortdefinition[Quaternions] $q = q_w + q_x i + q_y j + q_z k$ with\\
$i^2 = j^2 = k^2 = -1$, ($ij = -ji = k$, same for $jk$ and $ki$)
% TODO: Finish this

\shortdefinition[Transf. M] $\mat{T}_{AB} = \begin{bmatrix}
        \mat{R}_{AB}        & {_A}\vec{t}_B \\
        \mat{0}_{1\times 3} & 1
    \end{bmatrix}$\\
$\mat{T}_{BA} = \mat{T}_{AB}^{-1} =
    \begin{bmatrix}
        \mat{R}_{AB}^\top    & -\mat{R}_{AB}^\top {_A}\vec{t}_B \\
        \mat{0}_{1 \times 3} & 1
    \end{bmatrix}$
$\mat{T}_{AC} = \mat{T}_{AB} \mat{T}_{BC}$
