# bug in ListPlot3D ex) ListPlot3D[{{1, 1, 1}, {0, 1, 0}, {0, 1, 1}}]

Posted 9 years ago
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 This function does not work under this kind of points below. I think it should be plotted triangular mesh, but It doesn't work... Is it bug? Is there any solution? (Example:)  ListPlot3D[{{1, 1, 1}, {0, 1, 0}, {0, 1, 1}}] (version of mathematica is 9.0.1.0)
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Posted 9 years ago
 If you are looking to plot the points as independent locations, or the triangle defined by them, you could try these functions: In[1]:= points = {{1, 1, 1}, {0, 1, 0}, {0, 1, 1}} Out[1]= {{1, 1, 1}, {0, 1, 0}, {0, 1, 1}} In[2]:= ListPointPlot3D[points] In[3]:= Polygon[points] // Graphics3D 
Posted 9 years ago
 Hi,I don't think that it would do that. The documentation saysThe last two points in {{1, 1, 1}, {0, 1, 0}, {0, 1, 1}} have the same x/y coordinates but different values for z. If you modify your input so that it reads ListPlot3D[{{1, 1, 1}, {0, 1/5, 0}, {0, 1, 1}}, PlotStyle -> Thick] it shows a result. If you want the two last points to have at nearly the same y coordinate ListPlot3D[{{1, 1, 1}, {0, 0.99, 0}, {0, 1, 1}}, PlotRange -> {{0, 1}, {0, 1}, {0, 1}}] Cheers,Marco
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