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球面与平面的交点已知方程

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球面与平面的交点已知方程

我正在尝试绘制一个圆与平面的交点有方程2 x − 2 y + z − 15 = 0，而球体的方程是( x − 1)^2 + ( y + 1)^ 2 + ( z − 2)^ 2 − 25 = 0。 在此处输入图片描述

平面切球体为一个圆，其圆心为(3,-3,3，半径为r = 4。

我不能画圆。我试过了

\documentclass[12pt,border = 2 mm]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords]
\path
      coordinate (T) at (3,-3,3)
      coordinate (I) at (1,-1,2);

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}

\draw[dashed] (T) circle[radius={4}];

\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R); 
\end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

我怎样才能画圆圈？

答案1

我们知道n平面的法线和圆的半径。设两个n相互正交的向量分别为u和v。则圆的半径为

 gamma(t) = I + n + r * cos(t) * u + r* sin(t) * v,

其中n满足长度约束n^2+r^2=R^2，其中R是球体的半径和r圆的半径。您的向量n已经满足此约束，因此我们不必更改其规范化。

现在我们需要弄清楚可见和不可见的拉伸是什么。球体上的任何一点在屏幕上的法向量上都有正投影或负投影

n_screen =({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},{-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});

因此我们需要找到投影的零点gamma(t).n_screen。这可以通过让 Ti 来实现钾Z 找到交点。当然，我们在这里并不真正绘制路径，而是使用overlay它们来避免弄乱边界框。请注意，当前版本假设有两个零，因此如果您大幅改变视角，此版本将不再起作用。

\documentclass[12pt,border=2mm,tikz]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds,intersections}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords,
declare function={dicri(\t,\th,\ph,\R)=
sin(\th)*sin(\ph)*(2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18))-
sin(\th)*cos(\ph)*(-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18))+
cos(\th)*(1-4*\R*sin(\t)*1/sqrt(18));}]
\path
      coordinate (T) at (3,-3,3)
      coordinate (I) at (1,-1,2)
      coordinate (n) at (2,-2,1)
      coordinate (u) at ({1/sqrt(2)},{1/sqrt(2)},0)
      coordinate (v) at ({1/sqrt(18)},{-1/sqrt(18)},{-4/sqrt(18)});
      % the coordinatesn, u and v are not really used here

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}
% \draw[red,thick,-latex] (0,0,0) --
% ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},
% {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});
% normal to screen
\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R);
  % determine the zeros of dicri
  \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73]
  ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)});
  \path[overlay,name path=zero] (0,0) -- (360pt,0);
  \path[name intersections={of=dicri and zero,total=\t}]
  let \p1=(intersection-1),\p2=(intersection-2) in
  \pgfextra{\xdef\xmin{\x1}\xdef\xmax{\x2}};
\end{scope}
\pgfmathsetmacro{\R}{4}
\draw[dashed] plot[variable=\t,domain=\xmin:\xmax,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\draw[thick] plot[variable=\t,domain=\xmax:\xmin+360,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\end{tikzpicture}
\end{document}

在此处输入图片描述

这是一架飞机，使用你更好的矢量u和v从聊天。

\documentclass[12pt,border=2mm,tikz]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{arrows,calc,backgrounds,intersections} 
\makeatletter % https://tex.stackexchange.com/a/38995/121799
\tikzset{
  use path/.code={\pgfsyssoftpath@setcurrentpath{#1}}
}
\makeatother
\begin{document} 
\tdplotsetmaincoords{60}{110} 
\begin{tikzpicture}[tdplot_main_coords, 
  declare function={dicri(\t,\th,\ph,\R)=% 
  sin(\th)*sin(\ph)*(2+\R*cos(\t)/3+2*\R*sin(\t)/3)-%
  sin(\th)*cos(\ph)*(-2 +2*\R*cos(\t)/3 + \R*sin(\t)/3)+%
  cos(\th)*(1+2*\R*cos(\t)/3-2*\R*sin(\t)/3);}] 
  \pgfmathsetmacro{\R}{5}% 
  \path  coordinate (T) at (3,-3,3) 
   coordinate (I) at (1,-1,2) 
   coordinate (n) at (2,-2,1) 
   coordinate (u) at (1, 2, 2) 
   coordinate (v) at (2, 1, -2); 
  % the coordinatesn, u and v are not really used here 
   \path[tdplot_screen_coords,shift={(I)},use as bounding box] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   

  \foreach \v/\position in {T/above,I/below} { 
   \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$}; 
  } 
  % \draw[red,thick,-latex] (0,0,0) -- 
  % ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)}, 
  % {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)}); 
  % normal to screen 
  \begin{scope}[tdplot_screen_coords, on background layer] 
   \fill[ball color=green, opacity=0.8] (I) circle (\R); 
   % determine the zeros of dicri 
   \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73] 
   ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)}); 
   \path[overlay,name path=zero] (0,0) -- (360pt,0); 
   \path[name intersections={of=dicri and zero,total=\t}] 
   let \p1=(intersection-1),\p2=(intersection-2) in 
   \pgfextra{\xdef\tmin{\x1}\xdef\tmax{\x2}}; 
  \end{scope} 
  \pgfmathsetmacro{\SmallR}{4} 
  \draw[dashed] plot[variable=\t,domain=\tmin:\tmax,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \draw[thick,save path=\pathA] plot[variable=\t,domain=\tmax:\tmin+360,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \path ({1+2+\SmallR*cos(\tmin)/3+2*\SmallR*sin(\tmin)/3}, 
   {-1-2 +2*\SmallR*cos(\tmin)/3+ \SmallR*sin(\tmin)/3}, 
   {2+1+2*\SmallR*cos(\tmin)/3 - 2*\SmallR*sin(\tmin)/3 }) coordinate (pmin)
   ({1+2+\SmallR*cos(\tmax)/3+2*\SmallR*sin(\tmax)/3}, 
   {-1-2 +2*\SmallR*cos(\tmax)/3+ \SmallR*sin(\tmax)/3}, 
   {2+1+2*\SmallR*cos(\tmax)/3 - 2*\SmallR*sin(\tmax)/3 }) coordinate (pmax);
  \begin{scope}[tdplot_screen_coords]
   \clip[shift={(I)}] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   
   \path[fill=gray,fill opacity=0.4,even odd rule] let \p1=($(pmin)-(I)$),\p2=($(pmax)-(I)$),
   \p3=($(pmax)-(pmin)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)},
   \n3={atan2(\y3,\x3)}
    in [use path=\pathA]  (pmin) arc(\n1:\n2-360:\R) 
    (0,-6) -- ++(\n3:{12cm/sin(\n3)}) -- ++(\n3+90:{12cm/sin(\n3)})
    -- ++(\n3+180:{12cm/sin(\n3)}) -- cycle;
  \end{scope}
\end{tikzpicture} 
\end{document}

在此处输入图片描述

答案2

非常感谢薛定谔的猫3dtools。在这个答案中，我使用数学找到三个点的坐标A, B, C并画圆（ABC）。我tikz-3dplot-circleofsphere使用 这里绘制圆的线条样式（ABC）。

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usetikzlibrary{3dtools}
\begin{document}
    \tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=5;
}]
    \path   (3,-3,3) coordinate (T)
    (1,-1,2) coordinate (I) 
    (1, {1/5 *(-23 - 2* sqrt(11))}, {1/5 *(19 - 4 *sqrt(11))}) coordinate (A)
    (1, {1/5 *(-23 + 2* sqrt(11))}, {1/5 *(19 + 4 *sqrt(11))}) coordinate (B)
    ({1/4* (13 + sqrt(119))}, {1/4 *(-13 + sqrt(119))}, 2) coordinate (C);
\begin{scope}[tdplot_screen_coords] 
\fill[ball color=green, opacity=0.8] (I) circle (R);
    \end{scope}
\begin{scope}[shift={(I)}]
\path[overlay] [3d coordinate={(A-B)=(A)-(B)},
3d coordinate={(A-C)=(A)-(C)},
3d coordinate={(myn)=(A-B)x(A-C)},
3d coordinate={(A-T)=(A)-(T)}];
\pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")}
\pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]}
\pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]}
\pgfmathsetmacro{\mygamma}{-acos(sqrt(TD("(A-T)o(A-T)"))/R)}
\tdplotCsDrawCircle[tdplotCsFront/.style={thick}]{R}{\myalpha}{\mybeta}{\mygamma} 
\end{scope} 

\foreach \p in {I,T,B}
\draw[fill=black] (\p) circle (1.5 pt);
\foreach \p/\g in {I/0,T/-90,B/30}
\path (\p)+(\g:3mm) node{$\p$};
\draw[dashed] (I) -- (T) -- (B) -- cycle;
    \end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

我之所以添加这个，是因为你让我在这里使用3d circle through 3 points图片。我们有三个点后就可以使用它了。这些点可以通过添加长度为 4 的向量（半径）来构造，这些T向量位于圆的平面上。这些向量取自另一个答案，并重新缩放。

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3dtools,backgrounds}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords]
\pgfmathsetmacro{\mysq}{4/3}
\path   (3,-3,3) coordinate (T)
      (1,-1,2) coordinate (I) 
      (1, 2, 2) coordinate (u)  
      (2, 1, -2) coordinate (v)
      [3d coordinate={(A)=(T)+\mysq*(u)}]
      [3d coordinate={(B)=(T)+\mysq*(v)}]
      [3d coordinate={(C)=(T)-\mysq*(u)}];

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}

\path[dashed] pic{3d circle through 3 points={A={(A)},B={(B)},C={(C)}}};

\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R); 
\end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案4

我是 3dtools这样使用的。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
    \begin{tikzpicture}[3d/install view={phi=110,theta=70},line cap=butt,line join=round,declare function={R=5;},c/.style={circle,fill,inner sep=1pt}] 
    \path[overlay]
        (0,0,15) coordinate (A)
        (1,-1,2)  coordinate (I);
        \draw[3d/screen coords] (I) circle[radius=R]; 
        \path[overlay,3d/plane with normal={(2,-2,1) through (A) named p}];
        \path[3d/project={(I) on p}] coordinate (T);
\pic{3d/circle on sphere={R=R,C={(I)}, P={(T)}}};
 \path foreach \p/\g in {I/-90,T/90}
 {(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
\end{tikzpicture}
\end{document}

在此处输入图片描述

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