Right angle in 3D-pstricks

Question

I'd like to draw a right angle in a 3D-pstricks-graphic. I can't find a way to do that. This is what I want to do (but in 3D):

Here is my code:

\documentclass{standalone}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[ngerman]{babel}
\usepackage{pst-all}
\usepackage{pst-3dplot}
\begin{document}
\psset{coorType=2}
\begin{pspicture}(-3,-3.1)(3.5,3.5)
\pstThreeDCoor[xMin = -3, xMax = 3, yMin = -3, yMax = 3, zMin = -3, zMax = 3, arrows = ->, linecolor=black]
\pstThreeDTrianglefillcolor=yellow,fillstyle=solid,linecolor=black,opacity=0.5(0,2.5,0)(0,0,1)
\pstThreeDLinelinestyle=dashed(1.5,1.5,2.5)
\pstThreeDDot(1.5,1.5,2.5)
\pstThreeDPut(1.5,1.8,2.5){$P$}
\pstThreeDDot(0.489361,0.691489,0.478723)
\pstThreeDPut(0.489361,0.991489,0.478723){$P'$}
\pstThreeDDot(2,0,0)
\pstThreeDPut(2,-0.3,0){2}
\pstThreeDDot(0,2.5,0)
\pstThreeDPut(0,2.5,0.3){2.5}
\pstThreeDDot(0,0,1)
\pstThreeDPut(0,-0.3,1){1}
\end{pspicture}
\end{document}

In the picture you can see what I have got. I need to show two right angles so the projection is better visible.

Is there an equivalent of \psarc in 3D?

Welcome to TeX.SE. We usually use the comments for thanking users. Have fun here. — Jan, Jan 28 '17 at 18:56

score 4 · Answer 1 · 2017-01-29T19:52:22.733

I'd suggest to use \ThreeDput :

 \documentclass{standalone}
 \usepackage[T1]{fontenc}
 \usepackage[utf8]{inputenc}
 \usepackage[ngerman]{babel}
 \usepackage{pst-all}
 \usepackage{pst-3dplot}

 \begin{document}

 \psset{coorType=2}
 \begin{pspicture}(-3,-3.1)(3.5,3.5)
 \pstThreeDCoor[xMin = -3, xMax = 3, yMin = -3, yMax = 3, zMin = -3, zMax = 3, arrows = ->, linecolor=black]
    \ThreeDput[normal=1.3 1 3](0,0,1.66){%
    \pspolygon[fillcolor=yellow,fillstyle=solid,linecolor=black,opacity=0.8]%
     (0,0)(-3,0)(0,-3)
     \psarc{-}(0,0){0.6}{-180}{-90}
     \psdot[dotsize=1mm](-0.2,-0.2)
     \psdot[dotsize=2mm](0,0)
     \psdot[dotsize=2mm](-3,0)
     \psdot[dotsize=2mm](0,-3)}
 \pstThreeDLine[linestyle=dashed](0.33,0.5,0.5)(1.5,1.5,2.5)
 \pstThreeDDot(1.5,1.5,2.5)
 \pstThreeDPut(1.5,1.8,2.5){$P$}
 \pstThreeDDot(0.33,0.5,0.5)
 \pstThreeDPut(0.33,0.8,0.5){$P'$}
 \rput[r](-0.2,1.4){1}
 \rput[r](-0.5,-0.4){2}
 \rput[b](2.6,0.1){2.5}
\end{pspicture}

\end{document}

Thank you very much. So I have to draw the triangle in 2D an tilt it so it gets three dimensional. Can you explain what you have to write in the brackets after \ThreeDput. — iom10, Jan 29 '17 at 19:12
@P.W. Made an update but it might require further fine-tuning. — , Jan 29 '17 at 19:53

score 3 · Accepted Answer · answered Jan 29 '17 at 20:31

3

\pstThreeDEllipse[beginAngle=0,endAngle=90](0,0,1)(0.5,0,-0.25)(0,0.625,-0.25)
\pstThreeDDot(0.3,0.3,0.85)

answered Jan 29 '17 at 20:31

Thank you very much. This was the solutions I was waiting for. – iom10 Jan 29 '17 at 20:49

Right angle in 3D-pstricks

2 Answers2

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