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The following code plots the conic combination of 3 vectors in R^3 but it is undesirable since I need to tweak the inequalities by hand:

v1 = {1, 0.2, 0.2}; v2 = {0.2, 2, 0.2}; v3 = {0.2, 0.2, 3};
a = Graphics3D[Arrow[{{0, 0, 0}, #}] & /@ {v1, v2, v3}];
r = RegionPlot3D[
   Cross[v1, v3].{x, y, z} <= 0 && Cross[v1, v2].{x, y, z} >= 0 && 
    Cross[v2, v3].{x, y, z} >= 0, {x, 0, 5}, {y, 0, 5}, {z, 0, 5}, 
   AxesLabel -> Automatic];
Show[r, a, BoxRatios -> Automatic]

Is there any way to plot this more efficiently/neatly?

Thanks.

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4 Answers 4

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Looks like a case for ConicHullRegion: it produces a clean image almost instantly:

v =  {{1, 0.2, 0.2}, {0.2, 2, 0.2}, {0.2, 0.2, 3}};

chr = ConicHullRegion[{{0, 0, 0}}, v]; 

Graphics3D[{Arrow[{{0, 0, 0}, #}] & /@ v,
  FaceForm @@ ({#, #}& @ {Opacity[.3], Red}), EdgeForm @ Blue, chr}, 
 PlotRangePadding -> 0, Axes -> True, PlotRange -> Table[{0, 5}, 3]]

enter image description here

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  • 1
    $\begingroup$ Excellent that command does a neat and clean job. I am wondering how I can obtain the same smooth results by using ImplicitRegion and LinearSolve (see answer below). Thanks. $\endgroup$
    – root
    Jan 6, 2022 at 0:21
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v1 = {1, 0.2, 0.2}; v2 = {0.2, 2, 0.2}; v3 = {0.2, 0.2, 3};

Define an implicit region that satisfies your constraints:

V = Transpose[{v1, v2, v3}];
R = ImplicitRegion[Thread[LinearSolve[V, {x, y, z}] >= 0], {x, y, z}]

(*    ImplicitRegion[
        1.03186 (1. x - 0.0939597 y - 0.0604027 z) >= 0 &&
        -0.0969529 (1. x - 5.28571 y + 0.285714 z) >= 0 &&
        -0.0623269 (1. x + 0.444444 y - 5.44444 z) >= 0,
        {x, y, z}]                                            *)

RegionPlot3D[R,
             PlotPoints -> 100,
             Axes -> True, AxesLabel -> {x, y, z}]

enter image description here

Alternatively, define a parametric region:

S = ParametricRegion[V . {a1, a2, a3},
      {{a1, 0, ∞}, {a2, 0, ∞}, {a3, 0, ∞}}]

(*    ParametricRegion[{{a1 + 0.2 a2 + 0.2 a3, 
                         0.2 a1 + 2 a2 + 0.2 a3,
                         0.2 a1 + 0.2 a2 + 3 a3}, 
                        a1 >= 0 && a2 >= 0 && a3 >= 0},
                       {a1, a2, a3}]

and plot it in the same way.

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  • $\begingroup$ Very interesting. How can I remove the sawtooth effect? Looks awful even using >100 points which takes quite some time to render. Thanks. $\endgroup$
    – root
    Jan 5, 2022 at 23:54
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    $\begingroup$ I'd recommend using @kglr's solution with ConicHullRegion. ConicHullRegion allows plotting with Graphics3D, which ImplicitRegion and ParametricRegion don't. $\endgroup$
    – Roman
    Jan 6, 2022 at 7:49
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If this is good enough, you could create the convex hull like:

v1 = {1, 0.2, 0.2}; v2 = {0.2, 2, 0.2}; v3 = {0.2, 0.2, 3};
Region[ConvexHullRegion[{{0, 0, 0}, v1, v2, v3}], Axes -> True, 
 Boxed -> True]

enter image description here

If you want to show more of the infinite region, you may include sums of the given vectors like:

Region[ConvexHullRegion[{{0, 0, 0}, v1, v2, v3, v1 + v2, v1 + v3, 
   v2 + v3, v1 + v2 + v3}], Axes -> True, Boxed -> True]

enter image description here

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  • $\begingroup$ Thanks for your answer. I have Mathematica v11.0 and it seems the command "ConvexHullRegion" is not defined. In fact, I was interested in programming a command like that one by using standard vector operations, which is done in the next answer. $\endgroup$
    – root
    Jan 5, 2022 at 22:48
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RegionPlot3D can be slow, I think because each mesh point gets checked against the constraints. Setting terms for the cross products so that they only get calculated once speeds up the plotting by a factor of 15 on my system.

v1 = {1, 0.2, 0.2}; v2 = {0.2, 2, 0.2}; v3 = {0.2, 0.2, 3};
a = Graphics3D[Arrow[{{0, 0, 0}, #}] & /@ {v1, v2, v3}];
v1xv3 = Cross[v1, v3];
v1xv2 = Cross[v1, v2];
v2xv3 = Cross[v2, v3];
r = RegionPlot3D[
  v1xv3 . {x, y, z} <= 0 && v1xv2 . {x, y, z} >= 0 && 
   v2xv3 . {x, y, z} >= 0, {x, 0, 5}, {y, 0, 5}, {z, 0, 5}, 
  PlotPoints -> 100, AxesLabel -> Automatic];
Show[r, a, BoxRatios -> Automatic]
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