空间中任意一点到球的截面的最短距离
假设球的球心坐标为Oball={x0,y0,z0}O_{ball}=\{x_0,y_0,z_0\}Oball={x0,y0,z0}，球的半径为rrr，球的方程为(x−x0)2+(y−y0)2+(z−z0)2=r2(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2(x−x0)2+(y−y0)2+(z−z0)2=r2球的一截面的方程为Ax+By+Cz+1=0Ax+By+Cz+1=0Ax+By+Cz+1=0该截面为一个空间中的圆，球心Oball={x0,y0,z0}O_{ball}=\{x_0,y_0,z_0\}Oball={x0,y0,z0}在截面上的垂足即为空间中圆的圆心。假设圆上的任意三点的坐标分别为J(x1,y1,z1)J(x_1,y_1,z_1)J(x1,y1,z1)，K(x2,y2,z2)K(x_2,y_2,z_2)K(x2,y2,z2)，L(x3,y3,z3)L(x_3,y_3,z_3)L(x3,y3,z3)，圆心坐标为P=(xp,yp,zp)P=(x_p,y_p,z_p)P=(xp,yp,zp)，则球心Oball={x0,y0,z0}O_{ball}=\{x_0,y_0,z_0\}Oball={x0,y0,z0}在截面上的投影为PPP，可以得到下列向量OballP→=(xp−x0,yp−y0,zp−z0)JK→=(x2−x1,y2−y1,z2−z1)JL→=(x3−x1,y3−y1,z3−z1)\overrightarrow{O_{ball}P}=(x_p-x_0,y_p-y_0,z_p-z_0)\\\overrightarrow{JK}=(x_2-x_1,y_2-y_1,z_2-z_1)\\\overrightarrow{JL}=(x_3-x_1,y_3-y_1,z_3-z_1)OballP=(xp−x0,yp−y0,zp−z0)JK=(x2−x1,y2−y1,z2−z1)JL=(x3−x1,y3−y1,z3−z1)由向量垂直关系OballP→⊥JK→\overrightarrow{O_{ball}P}\bot\overrightarrow{JK}OballP⊥JK以及OballP→⊥JL→\overrightarrow{O_{ball}P}\bot\overrightarrow{JL}OballP⊥JL可以得到(xp−x0)(x2−x1)+(yp−y0)(y2−y1)+(zp−z0)(z2−z1)=0(xp−x0)(x3−x1)+(yp−y0)(y3−y1)+(zp−z0)(z3−z1)=0(x_p-x_0)(x_2-x_1)+(y_p-y_0)(y_2-y_1)+(z_p-z_0)(z_2-z_1)=0\\(x_p-x_0)(x_3-x_1)+(y_p-y_0)(y_3-y_1)+(z_p-z_0)(z_3-z_1)=0(xp−x0)(x2−x1)+(yp−y0)(y2−y1)+(zp−z0)(z2−z1)=0(xp−x0)(x3−x1)+(yp−y0)(y3−y1)+(zp−z0)(z3−z1)=0由J,K,LJ,K,LJ,K,L三点均在截面上，则有(x1y1z1x2y2z2x3y3z3)(ABC)+(111)=0\begin{pmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{pmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}+\begin{pmatrix}1\\1\\1\end{pmatrix}=0x1x2x3y1y2y3z1z2z3ABC+111=0解得(ABC)=−(x1y1z1x2y2z2x3y3z3)−1(111)\begin{pmatrix}A\\B\\C\end{pmatrix}=-\begin{pmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{pmatrix}^{-1}\begin{pmatrix}1\\1\\1\end{pmatrix}ABC=−x1x2x3y1y2y3z1z2z3−1111因为点PPP在截面上，所以Axp+Byp+Czp+1=0Ax_p+By_p+Cz_p+1=0Axp+Byp+Czp+1=0，联立方程组得{(xp−x0)(x2−x1)+(yp−y0)(y2−y1)+(zp−z0)(z2−z1)=0(xp−x0)(x3−x1)+(yp−y0)(y3−y1)+(zp−z0)(z3−z1)=0Axp+Byp+Czp+1=0\begin{equation*}\begin{cases} (x_p-x_0)(x_2-x_1)+(y_p-y_0)(y_2-y_1)+(z_p-z_0)(z_2-z_1)=0\\ (x_p-x_0)(x_3-x_1)+(y_p-y_0)(y_3-y_1)+(z_p-z_0)(z_3-z_1)=0 \\ Ax_p+By_p+Cz_p+1=0 \end{cases}\end{equation*}⎩⎨⎧(xp−x0)(x2−x1)+(yp−y0)(y2−y1)+(zp−z0)(z2−z1)=0(xp−x0)(x3−x1)+(yp−y0)(y3−y1)+(zp−z0)(z3−z1)=0Axp+Byp+Czp+1=0解得(xpypzp)=(x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1ABC)−1(x0(x2−x1)+y0(y2−y1)+z0(z2−z1)x0(x3−x1)+y0(y3−y1)+z0(z3−z1)−1)\begin{pmatrix}x_p\\y_p\\z_p\end{pmatrix}=\begin{pmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\\A&B&C\end{pmatrix}^{-1}\begin{pmatrix}x_0(x_2-x_1)+y_0(y_2-y_1)+z_0(z_2-z_1)\\x_0(x_3-x_1)+y_0(y_3-y_1)+z_0(z_3-z_1)\\-1\end{pmatrix}xpypzp=x2−x1x3−x1Ay2−y1y3−y1Bz2−z1z3−z1C−1x0(x2−x1)+y0(y2−y1)+z0(z2−z1)x0(x3−x1)+y0(y3−y1)+z0(z3−z1)−1球心到截面的距离为d1=∣Ax0+By0+Cz0+1∣A2+B2+C2d_1=\frac{|Ax_0+By_0+Cz_0+1|}{\sqrt{A^2+B^2+C^2}}d1=A2+B2+C2∣Ax0+By0+Cz0+1∣空间中圆的半径为rcircle=r2−d12r_{circle}=\sqrt{r^2-d_1^2}rcircle=r2−d12。
假设空间中任意一点m(xm,ym,zm)m(x_m,y_m,z_m)m(xm,ym,zm)，该点到截面的距离为d2=∣Axm+Bym+Czm+1∣A2+B2+C2d_2=\frac{|Ax_m+By_m+Cz_m+1|}{\sqrt{A^2+B^2+C^2}}d2=A2+B2+C2∣Axm+Bym+Czm+1∣垂足为Q=(xq,yq,zq)Q=(x_q,y_q,z_q)Q=(xq,yq,zq)，则有(xqyqzq)=(x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1ABC)−1(xm(x2−x1)+ym(y2−y1)+zm(z2−z1)xm(x3−x1)+ym(y3−y1)+zm(z3−z1)−1)\begin{pmatrix}x_q\\y_q\\z_q\end{pmatrix}=\begin{pmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\\A&B&C\end{pmatrix}^{-1}\begin{pmatrix}x_m(x_2-x_1)+y_m(y_2-y_1)+z_m(z_2-z_1)\\x_m(x_3-x_1)+y_m(y_3-y_1)+z_m(z_3-z_1)\\-1\end{pmatrix}xqyqzq=x2−x1x3−x1Ay2−y1y3−y1Bz2−z1z3−z1C−1xm(x2−x1)+ym(y2−y1)+zm(z2−z1)xm(x3−x1)+ym(y3−y1)+zm(z3−z1)−1垂足QQQ到圆心PPP的距离为d3=(xp−xq)2+(yp−yq)2+(zp−zq)2d_3=\sqrt{(x_p-x_q)^2+(y_p-y_q)^2+(z_p-z_q)^2}d3=(xp−xq)2+(yp−yq)2+(zp−zq)2则垂足QQQ到空间圆上的最短距离为d4=rcircle−d3d_4=r_{circle}-d_3d4=rcircle−d3，对应的圆上的点的坐标为T=(xt,yt,zt)T=(x_t,y_t,z_t)T=(xt,yt,zt)，则该点的坐标满足以下的方程组{(xt−x0)2+(yt−y0)2+(zt−z0)2=r2Axt+Byt+Czt+1=0(xq−xt)2+(yq−yt)2+(zq−zt)2=d42\begin{equation*}\begin{cases} (x_t-x_0)^2+(y_t-y_0)^2+(z_t-z_0)^2=r^2 \\ Ax_t+By_t+Cz_t+1= 0 \\ (x_q-x_t)^2+(y_q-y_t)^2+(z_q-z_t)^2=d_4^2 \end{cases}\end{equation*}⎩⎨⎧(xt−x0)2+(yt−y0)2+(zt−z0)2=r2Axt+Byt+Czt+1=0(xq−xt)2+(yq−yt)2+(zq−zt)2=d42方程1满足点在球面上，方程2满足点在截面上，方程3满足点到垂足QQQ的距离为d4d_4d4。