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# Simplifying visualization of VectorPlot3D

Posted 11 years ago
 I'm trying to visualize a vector field, but it's very crowded.VectorPlot3D[{0, 0,   Cos[x] Cos[y]}, {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}, {z,   0, 1}]In order to make it look less crowded, I'd like to see only one plane of vectors, say for the plane z=0. I've tried using the VectorPoints option, but I can't specify an integer lower than 2 (I'd prefer 1) for my plot because of the error below.VectorPlot3D[{0, 0,   Cos[x] Cos[y]}, {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}, {z,   0, 1}, VectorPoints -> {10, 10, 1}]VectorPlot3D::ppts: Value of option PlotPoints -> {10,10,1} is not an integer >= 2. >>Instead, I'm forced into the following, which gives me 2 planes of vectors when I only want 1 (although it is slightly clearer). Any thoughts? I thought of reverting to 2D, but that is tricky because there is no x,y component for either of the vectors, for which the magnitude depends only on x and y.VectorPlot3D[{0, 0,   Cos[x] Cos[y]}, {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}, {z,   0, 1}, VectorPoints -> {10, 10, 2}]How can I get rid of all but one of the planes of vectors?
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Posted 11 years ago
 Hi everyone,It was a great program, really enjoyed it.
Posted 11 years ago
 Give the location of centers of the vectors to be drawn to VectorPlot3D using VectorPoints option as a list of points.points = Join @@ Table[{i, j, .5}, {i, -1, 1, .25}, {j, -1, 1, .25}];VectorPlot3D[{0, 0, Cos[x] Cos[y]}, {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}, {z, 0, 1}, VectorPoints -> points]