Start with a quarter-circle of radius 1 centered at the origin and lying in the $xz$-plane:

 arc = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]

I want to dilate this by a factor of 2 and shift the center to {3, 0, 0}, then show the result graphically.

The following does not work:

shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}]

Graphics3D[GeometricTransformation[arc3D, shiftAndDilate3D]]

The error I get is that Graphics3DBox is not a Graphics3D primitive or directive.

What am I doing wrong?

  • $\begingroup$ At least for this case, it is much better to just apply the affine transformation to the parametric equations directly: ParametricPlot3D[AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}] @ {Cos[t], 0, Sin[t]} // Evaluate, {t, 0, π/2}]. $\endgroup$ Mar 8, 2019 at 1:46
  • $\begingroup$ @J.M.: That's a refreshingly different approach! I have to look at in the entire context of the more complicated thing I'm actually trying to do. It's worth making an answer! $\endgroup$
    – murray
    Mar 8, 2019 at 1:55
  • $\begingroup$ I think I kind of understand your confusion now, in light of this and your other question. One problem is that the docs do not give a complete and unambiguous list of primitives that one can point to and say: "these are primitives, and they are the only ones supported by GeometricTransformation[]"; this list mixes up directives and primitives, so that doesn't count. $\endgroup$ Mar 8, 2019 at 23:56
  • $\begingroup$ Possibly relevant: mathematica.stackexchange.com/questions/10957/… $\endgroup$ Mar 9, 2019 at 14:07

3 Answers 3


You could work with regions instead. Your arc:

arc = ParametricRegion[{Cos[t], 0, Sin[t]}, {{t, 0, \[Pi]/2}}];

The transformed arc:

shiftAndDilate3D = AffineTransform[{2 IdentityMatrix[3], {3, 0, 0}}];
new = TransformedRegion[arc, shiftAndDilate3D];


    Region[arc, BaseStyle->Red],
    Region[new, BaseStyle->Blue],

enter image description here

  • $\begingroup$ How should one know from the Mathematica documentation that ParametricRegion is a "region" suitable for the 1st argument to TransformedRegion? (The documentation involving GeometricTransformatino, Region, etc., is sadly deficient. As with Image, these things seem to have been thrown into Mathematica without sufficient exposition, or perhaps even without sufficient coherence for the various kinds of things, on the one hand, and sufficient distinctions among them, on the other hand.) $\endgroup$
    – murray
    Mar 7, 2019 at 20:22
  • $\begingroup$ On the one hand, I like this answer better than the others because it is more direct. On the other hand, it uses yet a different type of object, namely, a "region". $\endgroup$
    – murray
    Mar 7, 2019 at 20:23
  • $\begingroup$ But I'm annoyed by the different method to style the resulting graphics display, namely, through use of the BaseStyle option of Region instead of the usual PlotStyle option of ParametricPlot3D and so many other graphics and graphics 3D functions. Moreover, why should it be necessary to wrap the result of a ParametricRegion expression with a Region? Either there's something fundamental here I don't understand, or else a whole slew of graphics-like or geometric-like constructs have been thrown into recent versions of Mathematica without sufficient rationalization of the whole domain. $\endgroup$
    – murray
    Mar 7, 2019 at 20:36
  • $\begingroup$ Further rant: Why does ParametricRegion take as 2nd argument a list of lists (of parameters and their extent), whereas ParametricPlot3D uses the parameter information as a list, then another list, etc.? $\endgroup$
    – murray
    Mar 7, 2019 at 20:39
  • $\begingroup$ @murray Lots of questions. For a brief overview of regions, see reference.wolfram.com/language/guide/GeometricComputation. Basically, a region is a computable object. Region is a function that displays the object (similar to RegionPlot). Show converts a region object to a Graphics/Graphics3D object. ParametricRegion has a different syntax because it supports arguments without bounds, e.g., ParametricRegion[{Cos[t], 0, Sin[t]}, t]. $\endgroup$
    – Carl Woll
    Mar 7, 2019 at 20:48

You cannot apply those geometric transformations to the results of the plotting; instead, you should apply them to a Graphics primitive, e.g. the Line object generated by ParametricPlot, which you can extract using e.g. Cases:

arcLine = 
 First@Cases[ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}], _Line, All]

  Red, arcLine,
  Blue, GeometricTransformation[arcLine, shiftAndDilate3D]

Mathematica graphics

In red in the plot above is your original curve, in blue the transformed one.

  • $\begingroup$ Where in the documentation does it explain what is, and what is not, a "geometric object"? $\endgroup$
    – murray
    Mar 7, 2019 at 17:29
  • $\begingroup$ @murray I was being loose with words there. I really meant a "Graphics primitive", such as those discussed in this guide: Graphics objects. I fixed it in the answer. $\endgroup$
    – MarcoB
    Mar 7, 2019 at 18:20
  • $\begingroup$ Alas, the doc page guide/GraphicsObjects does not fully identify or clarify what is, and what is not, a "graphics object". For example, given the answer by Carl Woll to my question, a ParametricRegion seems to qualify as such an object, yet is not listed on that guide page. So the documentation list of graphics objects doesn't specify what they are, and certainly -- and this is perhaps a real design defect -- the Head of these objects doesn't tell you that they are "graphics objects"! $\endgroup$
    – murray
    Mar 7, 2019 at 19:53

Use arc3D[[1]] (which contains all the graphics primitives and their styles) as the first argument of GeometricTransformation:

arc3D = ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, π/2}]
 Graphics3D[GeometricTransformation[arc3D[[1]], shiftAndDilate3D] /. l_Line :> {Orange, l}],
 PlotRange -> All]

enter image description here

  • $\begingroup$ I cannot reproduce that. Rather, I get an error message: "An improperly formatted option was encountered while reading a Graphics3DBox. The left-hand side of the option was not a symbol or string." $\endgroup$
    – murray
    Mar 7, 2019 at 17:38
  • $\begingroup$ @murray, forgot a comma before PlotRange (fixed now). $\endgroup$
    – kglr
    Mar 7, 2019 at 17:41
  • $\begingroup$ Sorry, I should have caught that! $\endgroup$
    – murray
    Mar 7, 2019 at 19:47

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