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[几何] 过空间四交点(由给定平面所产生)的球面

尝试将下面二维情形推广到三维:
\begin{align*}
&({A_1}^2+{B_1}^2)(A_2B_3-A_3B_2)(A_2 x + B_2 y + C_2)(A_3 x + B_3 y + C_3)\\
\,\,+\,\,&({A_2}^2+{B_2}^2)(A_3B_1-A_1B_3)(A_3 x + B_3 y + C_3)(A_1 x + B_1 y + C_1)\\
\,\,+\,\,&({A_3}^2+{B_3}^2)(A_1B_2-A_2B_1)(A_1 x + B_1 y + C_1)(A_2 x + B_2 y + C_2)=0.
\end{align*}
问题源自:
[几何] 如何求三点的外接圆?
http://kuing.orzweb.net/redirect ... =5244&pid=25769
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`{A_1}^2` 真比 `A_1^2` 好看吗?
还有那俩加号两边我原代码 {}+{} 已经能产生正确的间距,为啥还要去改成 \,\,+\,\, ?

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本帖最后由 青青子衿 于 2019-6-26 09:05 编辑

回复 1# 青青子衿

\begin{align*}
l_1 &= A_1x+B_1y+C_1z+D_1\\
l_2 &= A_2x+B_2y+C_2z+D_2\\
l_3 &= A_3x+B_3y+C_3z+D_3\\
l_4 &= A_4x+B_4y+C_4z+D_4\\
f&=\,\begin{split}
\lambda_{12}l_1l_2+\lambda_{13}l_1l_3+\lambda_{14}l_1l_4\\
+\lambda_{23}l_2l_3+\lambda_{24}l_2l_4\\
+\lambda_{34}l_3l_4
\end{split}\\
\end{align*}
显然任意三个\(\,l_i\,\)为零则\(\,f\,\)为零,可见曲面\(\,f=0\,\)必过三张平面\(\,l_1=0\,\),\(\,l_2=0\,\),\(\,l_3=0\,\)的三个交点,
欲求出过四个交点的球面,只需求出使 $f=0$ 为球面的方程时的一组 \(\,(\lambda_{12}, \lambda_{13}, \lambda_{14},\lambda_{23}, \lambda_{24}, \lambda_{34})\,\) 即可。

易知\(\,f\,\)中\(\,x^2\,\),\(\,y^2\,\),\(\,z^2\,\),\(\,xy\,\),\(\,yz\,\),\(\,xz\,\)的系数分别为
\begin{gather*}
\operatorname{coef}(x^2)
=\,\begin{split}
\lambda_{12}A_1A_2+\lambda_{13}A_1A_3+\lambda_{14}A_1A_4\\
+\lambda_{23}A_2A_3+\lambda_{24}A_2A_4\\
+\lambda_{34}A_3A_4
\end{split}\\
\operatorname{coef}(y^2)
=\,\begin{split}
\lambda_{12}B_1B_2+\lambda_{13}B_1B_3+\lambda_{14}B_1B_4\\
+\lambda_{23}B_2B_3+\lambda_{24}B_2B_4\\
+\lambda_{34}B_3B_4
\end{split}\\
\operatorname{coef}(z^2)
=\,\begin{split}
\lambda_{12}C_1C_2+\lambda_{13}C_1C_3+\lambda_{14}C_1C_4\\
+\lambda_{23}C_2C_3+\lambda_{24}C_2C_4\\
+\lambda_{34}C_3C_4
\end{split}\\
\operatorname{coef}(xy)
=\,\begin{split}
\lambda_{12}(A_1B_2+A_2B_1)+\lambda_{13}(A_1B_3+A_3B_1)+\lambda_{14}(A_1B_4+A_4B_1)\\
+\lambda_{23}(A_2B_3+A_3B_2)+\lambda_{24}(A_2B_4+A_4B_2)\\
+\lambda_{34}(A_2B_4+A_4B_2)
\end{split}\\
\operatorname{coef}(yz)
=\,\begin{split}
\lambda_{12}(B_1C_2+B_2C_1)+\lambda_{13}(B_1C_3+B_3C_1)+\lambda_{14}(B_1C_4+B_4C_1)\\
+\lambda_{23}(B_2C_3+B_3C_2)+\lambda_{24}(B_2C_4+B_4C_2)\\
+\lambda_{34}(B_2C_4+B_4C_2)
\end{split}\\
\operatorname{coef}(xz)
=\,\begin{split}
\lambda_{12}(A_1C_2+A_2C_1)+\lambda_{13}(A_1C_3+A_3C_1)+\lambda_{14}(A_1C_4+A_4C_1)\\
+\lambda_{23}(A_2C_3+A_3C_2)+\lambda_{24}(A_2C_4+A_4C_2)\\
+\lambda_{34}(A_2C_4+A_4C_2)
\end{split}\\
\end{gather*}
那么只需令前三个相等且后三个为零即可,最终解出
\[
\begin{pmatrix}
\lambda_{12}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}\\
\lambda_{13}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}\\
\lambda_{14}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}\\
\lambda_{23}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}\\
\lambda_{24}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}\\
\lambda_{34}\vphantom{\Big|P(A,B,C)_{(0,\,0,\,0)}\Big|}
\end{pmatrix}
=
{\Large{t}}\begin{pmatrix}
\Big|P(A,B,C)_{(1,\,2,\,3)}\Big|\cdot\Big|P(A,B,C)_{(1,\,2,\,4)}\Big|\cdot\,\!Q(A,B,C)_{(3,\,4)}\\
\Big|P(A,B,C)_{(1,\,3,\,2)}\Big|\cdot\Big|P(A,B,C)_{(1,\,3,\,4)}\Big|\cdot\,\!Q(A,B,C)_{(2,\,4)}\\
\Big|P(A,B,C)_{(1,\,4,\,2)}\Big|\cdot\Big|P(A,B,C)_{(1,\,4,\,3)}\Big|\cdot\,\!Q(A,B,C)_{(2,\,3)}\\
\Big|P(A,B,C)_{(2,\,3,\,1)}\Big|\cdot\Big|P(A,B,C)_{(2,\,3,\,4)}\Big|\cdot\,\!Q(A,B,C)_{(1,\,4)}\\
\Big|P(A,B,C)_{(2,\,4,\,1)}\Big|\cdot\Big|P(A,B,C)_{(2,\,4,\,3)}\Big|\cdot\,\!Q(A,B,C)_{(1,\,3)}\\
\Big|P(A,B,C)_{(3,\,4,\,1)}\Big|\cdot\Big|P(A,B,C)_{(3,\,4,\,2)}\Big|\cdot\,\!Q(A,B,C)_{(1,\,2)}
\end{pmatrix}\]
其中\(\,\Big|P(A,B,C)_{(i,\,j,\,k)}\Big|=\begin{vmatrix}
A_i&B_i&C_i\\
A_j&B_j&C_j\\
A_k&B_k&C_k\\
\end{vmatrix}\)以及\(\,Q(A,B,C)_{(i,\,j)}=\begin{vmatrix}
A_i&B_i\\
A_j&B_j\\
\end{vmatrix}^2+\begin{vmatrix}
B_i&C_i\\
B_j&C_j\\
\end{vmatrix}^2+\begin{vmatrix}
A_i&C_i\\
A_j&C_j\\
\end{vmatrix}^2>0\)
所以过四个交点的球面方程为
\begin{align*}
&\lambda_{12}(A_1x+B_1y+C_1z+D_1)(A_2x+B_2y+C_2z+D_2)\\
\,\,+\,\,&\lambda_{13}(A_1x+B_1y+C_1z+D_1)(A_3x+B_3y+C_3z+D_3)\\
\,\,+\,\,&\lambda_{14}(A_1x+B_1y+C_1z+D_1)(A_4x+B_4y+C_4z+D_4)\\
\,\,+\,\,&\lambda_{14}(A_2x+B_2y+C_2z+D_2)(A_3x+B_3y+C_3z+D_3)\\
\,\,+\,\,&\lambda_{14}(A_2x+B_2y+C_2z+D_2)(A_4x+B_4y+C_4z+D_4)\\
\,\,+\,\,&\lambda_{14}(A_3x+B_3y+C_3z+D_3)(A_4x+B_4y+C_4z+D_4)=0\\
\end{align*}

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