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I am plotting a curl, and I only want VectorPlot3D to show one arrow, I have tried adjusting VectorPoints-> 1, but the plot show's no arrows at all. Is 2 the minimum VectorPoints I can have?

If I cannot make use of VectorPlot3d in this way, how can I use the Arrow function to point in the direction of a vector field at a certain point?


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I don't understand what you mean by "show one arrow". Where should this "one arrow" be in your plot? –  J. M. Aug 1 '12 at 14:36
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3 Answers 3

up vote 1 down vote accepted

I don't think it's possible. I assume the vector scaling routine needs at least two vectors. You can fake it though:

VectorPlot3D[{x, y, z}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, 
 VectorPoints -> {0.9999999 {0.5, 0.5, 0.5}, {0.5, 0.5, 0.5}}]

Mathematica graphics

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thanks that works quite nicely –  Mel Aug 2 '12 at 16:30

The error message that

VectorPlot3D[{x, y, z}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, VectorPoints -> {1, 1, 1}]


enter image description here

suggests that two is indeed the minimum number of VectorPoints.

EDIT: Using

 vp1=VectorPlot3D[{x, y, z}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, VectorPoints -> {2, 2, 2}]

enter image description here

a workaround is possible by manipulating the components of the Graphics3D object vp1:

(vp2 = vp1; vp2[[1, 2, 1, 2]] = vp2[[1, 2, 1, 2, #]];vp2) & /@ Range[8],

enter image description here

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You can specify two vectors, one of which is the vector you want, then use VectorColorFunction to hide the undesired one.

VectorPlot3D[{x, y, z}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1},
 VectorPoints -> {{.5, 0, .3}, {.4, .1, .4}},
 VectorColorFunctionScaling -> False,
 VectorColorFunction -> Function[{x, y, z, vx, vy, vz, n},
   If[{x, y, z} == {.4, .1, .4}, Black, Directive[Opacity[0]]]

Mathematica graphics

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thanks for the nice example with colorfunction, I haven't come across something like this in the wolfram reference page –  Mel Aug 2 '12 at 16:32
@Mel you are welcome :) –  Silvia Aug 2 '12 at 19:20

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