fix: katex newline
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README.md
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README.md
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@ -222,11 +222,11 @@ $$
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随后我们需要将$\vec{d}$转化到`World`坐标系(简称为`W`系)中,已知两个坐标系有两个公共点,在`W`系中表示为原点$O$和发射点$\vec{r}=(x_0,y_0,z_0)$,在`P`系中则为$(0,0,\rho)$和原点$O^\prime$,则有:
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随后我们需要将$\vec{d}$转化到`World`坐标系(简称为`W`系)中,已知两个坐标系有两个公共点,在`W`系中表示为原点$O$和发射点$\vec{r}=(x_0,y_0,z_0)$,在`P`系中则为$(0,0,\rho)$和原点$O^\prime$,则有:
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$$
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$$
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\begin{bmatrix}
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\begin{bmatrix}
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x\\ y\\ z
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x\\\ y\\\ z
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\end{bmatrix} = \lambda R \begin{bmatrix}
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\end{bmatrix} = \lambda R \begin{bmatrix}
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u\\ v\\ w
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u\\\ v\\\ w
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\end{bmatrix} + \begin{bmatrix}
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\end{bmatrix} + \begin{bmatrix}
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x_0\\ y_0\\ z_0
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x_0\\\ y_0\\\ z_0
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\end{bmatrix}
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\end{bmatrix}
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\cdots(1)
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\cdots(1)
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$$
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$$
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@ -234,8 +234,8 @@ $$
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其中$\lambda$为尺度比例因子,设置为1;$R$被称为罗德里格矩阵,定义反对称矩阵:
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其中$\lambda$为尺度比例因子,设置为1;$R$被称为罗德里格矩阵,定义反对称矩阵:
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$$
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$$
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S = \begin{bmatrix}
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S = \begin{bmatrix}
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0,& -c,& -b \\
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0,& -c,& -b \\\
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c,& 0,& -a \\
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c,& 0,& -a \\\
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b,& a,& 0
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b,& a,& 0
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\end{bmatrix}
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\end{bmatrix}
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$$
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$$
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@ -243,8 +243,8 @@ $$
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则$R$可表示为$R = (I+S)(I-S)^{-1}$,展开后则有:
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则$R$可表示为$R = (I+S)(I-S)^{-1}$,展开后则有:
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$$
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$$
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R = \frac{1}{1+a^2+b^2+c^2}\begin{bmatrix}
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R = \frac{1}{1+a^2+b^2+c^2}\begin{bmatrix}
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1 + a^2 - b^2 - c^2, -2c - 2ab, -2b + 2ac \\
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1 + a^2 - b^2 - c^2, -2c - 2ab, -2b + 2ac \\\
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2c - 2ab, 1 - a^2 + b^2 - c^2, -2a - 2bc \\
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2c - 2ab, 1 - a^2 + b^2 - c^2, -2a - 2bc \\\
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2b + 2ac, 2a - 2bc, 1 - a^2 - b^2 + c^2
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2b + 2ac, 2a - 2bc, 1 - a^2 - b^2 + c^2
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\end{bmatrix}\cdots(2)
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\end{bmatrix}\cdots(2)
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$$
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$$
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@ -252,23 +252,23 @@ $$
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将两个公共点代入$(1)$,相减则可消去平移项,可得到:
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将两个公共点代入$(1)$,相减则可消去平移项,可得到:
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$$
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$$
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\begin{bmatrix}
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\begin{bmatrix}
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x_2-x_1\\ y_2-y_1\\ z_2-z_1
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x_2-x_1\\\ y_2-y_1\\\ z_2-z_1
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\end{bmatrix} = R \begin{bmatrix}
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\end{bmatrix} = R \begin{bmatrix}
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u_2-u_1\\ v_2-v_1\\ w_2-w_1
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u_2-u_1\\\ v_2-v_1\\\ w_2-w_1
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\end{bmatrix}
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\end{bmatrix}
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$$
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$$
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代入$(2)$则可得到($u_{21}=u_2-u_1$):
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代入$(2)$则可得到($u_{21}=u_2-u_1$):
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$$
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$$
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\begin{bmatrix}
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\begin{bmatrix}
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0,& w_{21}+z_{21},& v_{21}+y_{21} \\
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0,& w_{21}+z_{21},& v_{21}+y_{21} \\\
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w_{21}+z_{21},& 0,& -u_{21}-x_{21} \\
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w_{21}+z_{21},& 0,& -u_{21}-x_{21} \\\
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-v_{21}-y_{21},& -u_{21}-x_{21},& 0
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-v_{21}-y_{21},& -u_{21}-x_{21},& 0
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\end{bmatrix}\begin{bmatrix}
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\end{bmatrix}\begin{bmatrix}
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a\\ b\\ c
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a\\\ b\\\ c
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\end{bmatrix} = \begin{bmatrix}
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\end{bmatrix} = \begin{bmatrix}
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u_{21} - x_{21} \\
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u_{21} - x_{21} \\\
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v_{21} - y_{21} \\
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v_{21} - y_{21} \\\
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w_{21} - z_{21}
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w_{21} - z_{21}
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\end{bmatrix}
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\end{bmatrix}
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$$
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$$
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@ -276,14 +276,14 @@ $$
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代入本题,则有:
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代入本题,则有:
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$$
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$$
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\begin{bmatrix}
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\begin{bmatrix}
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0,& -\rho+z_0,& y_0 \\
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0,& -\rho+z_0,& y_0 \\\
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-\rho+z_0,& 0,& -x_0 \\
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-\rho+z_0,& 0,& -x_0 \\\
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-y_0,& -x_0,& 0
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-y_0,& -x_0,& 0
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\end{bmatrix}\begin{bmatrix}
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\end{bmatrix}\begin{bmatrix}
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a\\ b\\ c
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a\\\ b\\\ c
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\end{bmatrix} = \begin{bmatrix}
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\end{bmatrix} = \begin{bmatrix}
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-x_0 \\
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-x_0 \\\
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-y_0 \\
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-y_0 \\\
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-\rho-z_0
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-\rho-z_0
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\end{bmatrix}
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\end{bmatrix}
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$$
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$$
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@ -291,7 +291,7 @@ $$
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令$a=1$,则可求得:
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令$a=1$,则可求得:
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$$
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$$
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\begin{cases}
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\begin{cases}
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b = \frac{z_0+\rho-y_0}{x_0} \\
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b = \frac{z_0+\rho-y_0}{x_0} \\\
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c = \frac{z_0-\rho+y_0}{x_0}
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c = \frac{z_0-\rho+y_0}{x_0}
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\end{cases}
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\end{cases}
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$$
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$$
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