绘制立方体中通过两个给定点的所有平面

Question 1

我可能不会做动画，而只是使用一个具有非平凡不透明度的平面，这样人们就可以看到虚线红线击中了平面。我的看法是，你想

直观证明点 R 是 ABCD 与 PQ 的公共点

因为这是问题中写的内容。如果问题不是关于这个的，请考虑编辑它以使其更清楚。（抱歉，我不久前退出了 pstricks，但我相信你能够用 pstricks 重新做到这一点。）

\documentclass[tikz,border=3.14mm]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{backgrounds} 
\newcounter{coord}
\begin{document} 
\tdplotsetmaincoords{70}{200} 
\begin{tikzpicture}[tdplot_main_coords,scale=3,bullet/.style={fill,circle,inner sep=1pt}] 
 \foreach \Z in {-1,1}
 {\foreach \Y in {-1,1}
  {\foreach \X in {-1,1} 
   {\stepcounter{coord}
   \path (\Y*\X,-1*\Y,\Z) coordinate (\Alph{coord})  -- ++ (0.2*\Y*\X,0,0) 
  node{$\Alph{coord}$} ;}}}
 \fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
 \draw[dashed] (A) -- (D) -- (C) (D) -- (H);  
 \draw (E) -- (F)  -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
 \node[bullet,label=left:$P$] (P) at (1,0,0.5){};
 \node[bullet,label=above right:$R$] (R) at (1/3,0,-1){};
 \draw[red,dashed,thick] (P) -- (R)
     node[pos=1.5,black,bullet,label={[black]below:$Q$}] (Q){}; 
 \begin{scope}[on background layer]
   \draw[red,dashed,thick] (R) -- (Q);
 \end{scope}
\end{tikzpicture}
\end{document}

如果您只是想旋转一个平面，当然可以做到。

\documentclass[tikz,border=3.14mm]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{backgrounds,calc} 
\newcounter{coord}
\begin{document} 
\tdplotsetmaincoords{70}{200} 
\foreach \Angle in {5,10,...,180}
{\setcounter{coord}{0}
\begin{tikzpicture}[tdplot_main_coords,scale=3,bullet/.style={fill,circle,inner sep=1pt}] 
 \path[tdplot_screen_coords,use as bounding box]
  (-2,-2) rectangle (1.5,1.8);
 \foreach \Z in {-1,1}
 {\foreach \Y in {-1,1}
  {\foreach \X in {-1,1} 
   {\stepcounter{coord}
   \path (\Y*\X,-1*\Y,\Z) coordinate (\Alph{coord})  -- ++ (0.2*\Y*\X,0,0) 
  node{$\Alph{coord}$} ;}}}
 \fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
 \draw[dashed] (A) -- (D) -- (C) (D) -- (H);  
 \draw (E) -- (F)  -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
 \node[bullet,label=left:$P$] (P) at (1,0,0.5){};
 \node[bullet,label=above right:$R$] (R) at (1/3,0,-1){};
 \draw[red,dashed,thick] (P) -- (R)
     node[pos=1.5,black,bullet,label={[black]below:$Q$}] (Q){}; 
 \begin{scope}[on background layer]
   \draw[red,dashed,thick] (R) -- (Q);
    \fill[opacity=0.3,red!40] 
    ($(R)+({cos(\Angle)},{sin(\Angle)},0)$) -- 
    ($(R)+({cos(\Angle+180)},{sin(\Angle+180)},0)$) --
    ($(P)+({cos(\Angle+180)},{sin(\Angle+180)},0)$) -- 
    ($(P)+({cos(\Angle)},{sin(\Angle)},0)$) -- cycle;   
 \end{scope}
\end{tikzpicture}}
\end{document}

还有裁剪的平面。我到此为止。我相信我已经回答了最初的问题。如果你不同意，我尊重这个意见，但我不愿意再重复一遍。

\documentclass[tikz,border=3.14mm]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{backgrounds,calc,intersections} 
\newcounter{coord}
\begin{document} 
\tdplotsetmaincoords{70}{200} 
\foreach \Angle in {5,10,...,180}
{\setcounter{coord}{0}
\begin{tikzpicture}[tdplot_main_coords,scale=3,bullet/.style={fill,circle,inner sep=1pt}] 
 \path[tdplot_screen_coords,use as bounding box]
  (-2,-2) rectangle (1.5,1.8);
 \foreach \Z in {-1,1}
 {\foreach \Y in {-1,1}
  {\foreach \X in {-1,1} 
   {\stepcounter{coord}
   \path (\Y*\X,-1*\Y,\Z) coordinate (\Alph{coord})  -- ++ (0.2*\Y*\X,0,0) 
  node{$\Alph{coord}$} ;}}}
 \fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
 \draw[dashed] (A) -- (D) -- (C) (D) -- (H);  
 \draw (E) -- (F)  -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
 \node[bullet,label=left:$P$] (P) at (1/3,1,0.5){};
 \node[bullet,label=above right:$R$] (R) at (1/3,0,-1){};
 \draw[red,dashed,thick] (P) -- (R)
     node[pos=1.5,black,bullet,label={[black]below:$Q$}] (Q){}; 
 \begin{scope}[on background layer]
   \draw[red,dashed,thick] (R) -- (Q);
    \begin{scope}[overlay]
     \path[name path=upper boundary]
      (1,1,0.5)--(1,-1,0.5) -- (-1,-1,0.5)--(-1,1,0.5)--cycle;
     \path[name path=lower boundary]
      (1,1,-1)--(-1,1,-1) -- (-1,-1,-1)--(1,-1,-1)--cycle;
     \path[name path=upper line]  %  ($(P)+({3*cos(\Angle)},{3*sin(\Angle)},0)$)
     (P)
     -- ($(P)+({3*cos(180+\Angle)},{3*sin(180+\Angle)},0)$);
     \path[name path=lower line]  
     ($(R)+({3*cos(\Angle)},{3*sin(\Angle)},0)$)-- ($(R)+({3*cos(180+\Angle)},{3*sin(180+\Angle)},0)$);
     \path[name intersections={of=upper boundary and upper line,name=Pint}];
     \path[name intersections={of=lower boundary and lower
     line,name=Rint}];
    \end{scope}
    \ifnum\Angle<60
     \draw[red,fill opacity=0.3,fill=red!40]  (P) -- (Pint-1) --
      (Rint-1) -- (Rint-2)   -- (1,1,0.5)-- cycle;
    \else
     \ifnum\Angle<110
     \draw[red,fill opacity=0.3,fill=red!40] (P) -- (Pint-1) -- (Rint-2) -- (Rint-1) -- cycle;
     \else
     \ifnum\Angle<150
      \draw[red,fill opacity=0.3,fill=red!40] (P) -- (Pint-1)
      --  (Rint-2)  -- (Rint-1)  -- cycle;
     \else
      \draw[red,fill opacity=0.3,fill=red!40] (P) -- (Pint-1)
      --  (Rint-2)  -- (Rint-1)  -- (-1,1,0.5) --cycle;
     \fi 
     \fi
    \fi
 \end{scope}
 \node at (0,0) {\Angle};
\end{tikzpicture}}
\end{document}

Answer

我可能不会做动画，而只是使用一个具有非平凡不透明度的平面，这样人们就可以看到虚线红线击中了平面。我的看法是，你想