TiKz 中两个向量和一个平面之间的角度

Question

我只能在 TiKz 中绘制下图。如何跨越平面并绘制所示角度？在此先谢谢了！

Answer 1

这对于tikz-3dplot软件包来说很简单，并且我在这里使用的所有内容都在手册中通过示例进行了说明。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\begin{document}
\begin{tikzpicture}[font=\sffamily]
\tdplotsetmaincoords{70}{210}
\begin{scope}[tdplot_main_coords]
  \draw[-latex] (0,0,0) -- (4,0,0) node[pos=1.1] {$x$};
  \draw[-latex] (0,0,0) -- (0,4,0) node[pos=1.1] {$y$};
  \draw[-latex] (0,0,0) -- (0,0,4) node[pos=1.1] {$z$};
  \draw (2,0,0) -- (2,-5,0) -- node[pos=0.05,above right,sloped]{Leading edge} 
  (-2,-5,0) -- (-2,0,0) -- 
  node[pos=0.5,below right,sloped]{Trailing edge} 
  cycle;
  \draw[dashed] (0,0,0) -- (1,0,4) -- (1,3,4) coordinate (O)
  -- (0,3,0) coordinate (A) -- (2,3,0) coordinate (B);
  \draw[-latex] (0,0,0) -- (O) node[pos=1.03,circle,inner sep=1pt,fill,label=above left:$O$]{};
  %
  \tdplotdefinepoints(0,0,0)(0,2,0)(1,3,4)
  \tdplotdrawpolytopearc[thick,-latex]{1.5}{anchor=west}{$\theta_e$}
  %
  \tdplotdefinepoints(0,3,0)(2,3,0)(1,3,4)
  \tdplotdrawpolytopearc[dashed,-latex]{1.5}{anchor=west}{$\psi_e$}
  %
  \draw[latex-] (-1.5,-2.5,0) -- ++ (-1,0,0) node[right]{Blade};
  \draw[latex-] (-1.5,-5,0) -- ++ (0,-1,0);
  \path (-1.5,-6,0) -- ++ (0,-2,0) node[pos=0.4,sloped]{Flow};
\end{scope}
\end{tikzpicture}
\end{document}