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I want to plot a 2D vector function such as $F(x,y) = (a(x,y),\,b(x,y))$ in a 3D graph so that the vectors are embedded in the xy plane. I tried to do the following:

First I defined a piecewise function like this

g[z_] := Piecewise[{{1, z == 0}}, 0]

Then I converted the 2D vector function to a 3D one by setting the 3rd component to zero and multiplying 1st and 2nd components by g[z] so that the x and y components are null when z != 0:

f[x_, y_, z_] := {x g[z], y g[z], 0};

Plot the function:

VectorPlot3D[{x g[z], y g[z],0}, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}]

The issue with this solution is that VectorPlot3D won't evaluate the function in the relevant points, The above example shows an empty graph, because Mathematica jumps from z = -1 to z = 1 without evaluating z = 0.

I tried with RegionFunction (which would've rendered the definition of the above-mentioned piecewise function useless), but that only accepts inequalities, and I want to evaluate the function at any coordinate {x ,y, 0}.

I could feed it a list of vectors via VectorPoints -> { {a, b, 0}, {c, d, 0}, ...}, but that's not an elegant solution at all. Are there other ways to do this?

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2 Answers 2

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VectorPlot3D[{x, y, 0}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
 RegionFunction -> ((-.1 < #3 < .1) &),
 VectorPoints -> {8, 8, 3},
 VectorStyle -> "Arrow3D",
 VectorColorFunction -> "Rainbow",
 VectorScale -> Scaled[0.15]]

enter image description here

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  • $\begingroup$ It works fine, although I don't fully understand the syntax. What does #3 mean? Does it represent the third argument of the function? And what does the & operator do? $\endgroup$
    – DvD
    Commented Apr 2, 2013 at 23:24
  • $\begingroup$ @DvD #3 represents vertical coordinate z. If I wouldn't put limitation with RegionFunction, you would see many parallel identical planes. I limited plotting to a narrow region around a single plane with a constraint in z-coordinate (-.1 < #3 < .1) &. $\endgroup$
    – Vitaliy Kaurov
    Commented Apr 2, 2013 at 23:28
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Graphics3D[
  VectorPlot[{x, y}, {x, -1, 1}, {y, -1, 1}][[1, 2]] /. 
    Arrow[{{x1_, y1_}, {x2_, y2_}}] :> Arrow[{{x1, y1, 0}, {x2, y2, 0}}
  ]
]

Mathematica graphics

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  • $\begingroup$ Combining this result with some other graphs (such as a scalar function) is a bit troubling, but it's helpful to know nevertheless! $\endgroup$
    – DvD
    Commented Apr 2, 2013 at 23:20
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    $\begingroup$ @DvD This figure should combine with other graphics just fine. Have you tried Show? $\endgroup$
    – Sjoerd C. de Vries
    Commented Apr 3, 2013 at 5:22
  • $\begingroup$ Um, back when I formulated my comment, it just showed some huge arrows for some reason. I tried it again and it works now. I can't tell what I did differently this time. $\endgroup$
    – DvD
    Commented Apr 3, 2013 at 10:40

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