3D 交叉口

Question

作为练习，我尝试绘制棱柱 [0,2] x [0,4] x [0,6] 与平面 x + y + z = 5 的交点。

我的结果是：

\documentclass{article}

\usepackage{pgfplots}
\pgfplotsset{compat=newest}

\begin{document}
    
\begin{tikzpicture}[x={(-0.45cm,-0.385cm)},y={(1cm,-0.1cm)},z={(0,1cm)}]
    \draw [->] (0,0,0) -- (6,0,0) node [below left] {$x$};
    \draw [->] (0,0,0) -- (0,6,0) node [right] {$y$};
    \draw [->] (0,0,0) -- (0,0,6) node [right] {$z$};
    
    \filldraw [thick, orange, fill opacity=0.3] (0,0,5) -- (0,4,1) -- (1,4,0) -- (2,3,0) -- (2,0,3) -- cycle;
    \filldraw [thick, blue, fill opacity=0.2] (2,3,0) -- (2,0,3) -- (5,0,0) -- cycle;
    \filldraw [thick, blue, fill opacity=0.2] (1,4,0) -- (0,5,0) -- (0,4,1) -- cycle;
    
    \filldraw [thick, orange, fill opacity=0.3] (2,3,0) -- (2,0,0) -- (2,0,3) -- cycle;
    \filldraw [thick, orange, fill opacity=0.3] (1,4,0) -- (0,4,0) -- (0,4,1) --cycle;
    
\end{tikzpicture}

\end{document}

我现在有几个问题：

• 我认为有很多代码只是为了将一个简单的数学体积表示为 [0,2] x [0,4] x [0,6]。有没有更有效的方法来绘制它？
• 我是否需要手动计算交点然后将其表示出来？或者有没有直接的方法？
• 如何使用axis环境和\addplot命令而不是 来获得相同的结果\draw？我试过了，但我是新手\addplot3，而且我在轴位置 ( view={}{}) 方面遇到了麻烦，colormap颜色不均匀，表面有一个网格，难以理解图片，我对交叉点有同样的疑问，我需要手动计算它们吗？

全棱镜是：

\draw [fill=orange, fill opacity=0.3] (0,0,6) -- (2,0,6) -- (2,4,6) -- (0,4,6) -- cycle ;
\draw [fill=orange, fill opacity=0.3] (2,0,0) -- (2,0,6) -- (2,4,6) -- (2,4,0) -- cycle ;
\draw [fill=orange, fill opacity=0.3] (2,4,0) -- (0,4,0) -- (0,4,6) -- (2,4,6) -- cycle ;

Answer 1

不管你做什么，请考虑以更系统的方式安装 3D 视图。实现此目的的最佳方式可能是使用asymptote，它确实有工具来计算 3D 中的交点。如果你想使用pgfplots，请使用patch plots。但是，为此你仍然需要自己计算交点。这篇文章是提到一些实验钛钾Z 库这也允许我们计算3d中的交点。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{3dtools}%https://github.com/marmotghost/tikz-3dtools
\begin{document}
\pgfdeclarelayer{background} 
\pgfdeclarelayer{foreground}
\pgfdeclarelayer{behind}  
\pgfsetlayers{behind,background,main,foreground}   
\begin{tikzpicture}[>=stealth,
    3d/install view={theta=70,phi=110},
    line cap=round,line join=round,
    visible/.style={draw,thick,solid},
    hidden/.style={draw,very thin,cheating dash},
    3d/polyhedron/.cd,fore/.style={visible,fill opacity=0.6},
    back/.style={fill opacity=0.6,hidden,3d/polyhedron/complete dashes},
    fore layer=foreground,
    back layer=background
    ]
  \draw [->] (0,0,0) coordinate (O) -- (6,0,0) coordinate (ex) node [below left] {$x$};
  \draw [->] (0,0,0) -- (0,6,0) coordinate (ey) node [right] {$y$};
  \draw [->] (0,0,0) -- (0,0,6) coordinate (ez) node [right] {$z$};
  \path (5,0,0) coordinate (A) (0,5,0) coordinate (B) (0,0,5) coordinate (C)
   (2.5,0,0) coordinate (a) (0,3.5,0) coordinate (b) (0,0,2) coordinate (c) ;
  \path[3d/.cd,plane with normal={(ex) through (a) named px},
      plane with normal={(ey) through (b) named py},
      line through={(A) and (B) named lAB},
      line through={(A) and (C) named lAC},
      line through={(B) and (C) named lBC}];
  \path[3d/intersection of={lAB with px}] coordinate (pABx)
   [3d/intersection of={lAB with py}] coordinate (pABy)
   [3d/intersection of={lAC with px}] coordinate (pACx)
   [3d/intersection of={lBC with py}] coordinate (pBCy);
  \pgfmathsetmacro{\mybarycenterA}{barycenter("(A),(pABx),(pACx),(a)")} 
  \pgfmathsetmacro{\mybarycenterB}{barycenter("(B),(pABy),(pBCy),(b)")} 
  \tikzset{3d/polyhedron/.cd,O={(\mybarycenterA)},color=blue,
      draw face with corners={{(A)},{(pABx)},{(pACx)}},
      draw face with corners={{(A)},{(pABx)},{(a)}},
      draw face with corners={{(A)},{(a)},{(pACx)}},
      O={(\mybarycenterB)},
      draw face with corners={{(B)},{(pABy)},{(pBCy)}},
      draw face with corners={{(B)},{(pABy)},{(b)}},
      draw face with corners={{(B)},{(b)},{(pBCy)}},
      color=orange,O={(1,1,1)},
      draw face with corners={{(pABx)},{(pACx)},{(C)},{(pBCy)},{(pABy)}},
      draw face with corners={{(a)},{(pACx)},{(C)},{(O)}},
      draw face with corners={{(b)},{(pBCy)},{(C)},{(O)}},
      draw face with corners={{(b)},{(pABy)},{(pABx)},{(a)},{(O)}}
      }
\end{tikzpicture}
\end{document}

仍然需要付出很多努力。但是，有一个好处：您可以更改视图并仍然获得正确的结果。例如，3d/install view={theta=70,phi=60},您将获得

当然，对于asymptote和patch plot解决方案来说这也是正确的（也许除了自动将隐藏线变为虚线的可能性之外）。