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I am trying to draw the following figure with 3d Tikz but with little success.

My codes will be posted here after some corrections.

I am a newbie and would appreciate any hint from the forum.

Many thanks in advance! Best regards, Shelmy

enter image description here

  • 2
    Welcome to TeX.SX! If you already have some code as you say you do, please post it here so that people don't need to do all the work for you. – Phelype Oleinik May 27 at 22:05
2

This is to give you a start. You can use tikz-3dplot to obtain orthographic projections of some 3d setting.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{intersections,backgrounds}
\begin{document}
\tdplotsetmaincoords{110}{30}
\begin{tikzpicture}[tdplot_main_coords,>=stealth,declare function={xmax=30;},
line join=bevel]
 \draw[->] (-2,0,0) coordinate (A) -- (1,0,0) node[pos=1.1]{$x$};
 \draw (-2,-5,0) coordinate (P) plot[smooth,variable=\x,domain=0:xmax] 
 (-2,{-5*cos(\x)},{5*sin(\x)})
 coordinate(p1);
 \fill[name path=plane,gray,opacity=0.3] (-2.2,{-5*cos(xmax)},-0.5) --
  (2.2,{-5*cos(xmax)},-0.5) -- (2.2,{-5*cos(xmax)},3) -- (-2.2,{-5*cos(xmax)},3) -- cycle;
 \path[name path=left] (P) -- (A);
 \draw[name intersections={of=left and plane,by={B,B'}}]
  (P) -- (B') (-2,{-5*cos(xmax)},0) coordinate(B) -- (A) -- (p1) --(B);
 \path[name path=right] (P) -- (0,0,0) coordinate (O);
 \draw[name intersections={of=right and plane,by={C,C'}}]
  (P) -- (C') 
  ({-2+2*(1-cos(xmax))},{-5*cos(xmax)},0) coordinate (C) -- (O) -- (p1) -- (C)
  (O) -- (0,{{-5*cos(xmax)}},0) coordinate (O');
 \begin{scope}[on background layer]
  \draw[gray!30] (B') -- (B) (C') -- (C); 
 \end{scope}
 \draw[->] (O) -- (0,1,0) node[pos=1.3]{$y$};
 \draw[->] (O) -- (0,0,1) node[pos=1.3]{$z$};
 \foreach \X in {O,P,A,B,C}
 {\node[inner sep=1pt,circle,fill,label=below:$\X$] at (\X){};}
\end{tikzpicture}
\end{document}

enter image description here

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