${ \left[ {\bf{s}} \right]_ \times }$ is the skew-symmetric matrix of $\bf{s}$.

Rational Formulation of Residual
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We will show that the residuals from point-to-point, point-to-line and point-to-plane correspondences have a general form $$\label{equ:6}
{{r}_{g}}={{\mathbf{a}}^{T}}\mathbf{Rb}+{{\mathbf{a}}^{T}}\mathbf{t}+c$$ where $\mathbf{a}$ and $\mathbf{b}$ are two $3$-dimensional vectors and $c$ is a scalar. We first consider the point-to-plane residual ${{r}_{\pi }}$ in (\[equ:1\]). It is obvious that $\mathbf{a}=\mathbf{n},\ \mathbf{b}={{\mathbf{y}}_{\pi }}$ and $c=-{{\mathbf{n}}^{T}}{{\mathbf{y}}_{\pi }}$. Let us then consider the point-to-line residual. We define ${{\mathbf{r}}_{l}}=\left[ r_{l}^{1},r_{l}^{2},r_{l}^{3} \right]^T$ and ${{\mathbf{A}}_{l}}=\left( {{\mathbf{I}}_{3}}-\mathbf{d}{{\mathbf{d}}^{T}} \right)=\left[ \mathbf{a}_{l}^{1};\mathbf{a}_{l}^{2};\mathbf{a}_{l}^{3} \right]$, where $\mathbf{a}_{l}^{1},\mathbf{a}_{l}^{2}$ and $\mathbf{a}_{l}^{3}$ are the three rows of ${{\mathbf{A}}_{l}}$. Using this notation, equation (\[equ:2\]) can be written as $$\label{equ:7}
r_{l}^{i}=\mathbf{a}_{l}^{i}\left( \mathbf{R}{{\mathbf{x}}_{l}}+\mathbf{t}-{{\mathbf{y}}_{l}} \right),\ i=1,2,3$$ It is easy to find that $\mathbf{a}={{\left( \mathbf{a}_{l}^{i} \right)}^{T}},\ \mathbf{b}={{\mathbf{x}}_{l}}$ and $c=-\mathbf{a}_{l}^{i}{{\mathbf{y}}_{\pi }}$. Lastly, we consider the point-to-point correspondence. Similar to the point-to-line correspondence, we define ${{\mathbf{r}}_{p}}={{\left[ r_{p}^{1},r_{p}^{2},r_{p}^{3} \right]}^{T}}$ and ${{\mathbf{I}}_{3}}=\left[ \mathbf{e}_{l}^{1};\mathbf{e}_{l}^{2};\mathbf{e}_{l}^{3} \right]$. Substituting this notation into (\[equ:3\]), it is easy to find $$\label{equ:8}
r_{p}^{i}=\mathbf{e}_{p}^{i}\left( \mathbf{R}{{\mathbf{x}}_{p}}+\mathbf{t}-{{\mathbf{y}}_{p}} \right),\ i=1,2,3$$ The similarity between (\[equ:8\]) and (\[equ:7\]) is obvious. Therefore, $r_{p}^{i}$ also has the general form (\[equ:6\]).

Let $\mathbf{t}=\left[ {{t}_{1}};{{t}_{2}};{{t}_{3}} \right]$. Substituting (\[equ:5\]) into the general residual ${{r}_{g}}$ in (\[equ:6\]) and adding the tree terms together yields $$\label{equ:9}
{r_g} = \frac{{{{\bf{k}}^T}{\bf{v}}}}{{1 + {{\bf{s}}^T}{\bf{s}}}}$$ 