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3

You did not specify which value x, y, or z you wish to consider, or some combination thereof. Nevertheless this may help: data = Import["DATA - Sheet1.csv"]; ve = Partition[data, 5]; vecdata = Partition[ve, 7]; ListVectorPlot3D[vecdata, DataRange -> {{-15, 15}, {-15, 15}, {20, 30}}, VectorPoints -> 5, VectorScale -> {0.1, Scaled[0.5]}, ...


4

RandomPoint makes this a one liner (as of version 10.2): RandomPoint[Sphere[{0, 0, 0}, 4]] {1.80874, 3.43311, -0.970669} Or get multiple points at once: RandomPoint[Sphere[{0, 0, 0}, 4], 10] {{-2.15717, -0.871558, -3.25377}, {1.65153, 2.06714, -2.99989}, {-0.798, 3.91915, -0.0591287}, {-2.19946, 3.2677, 0.696074}, {3.21113, 2.00556, 1.29087}, {...


3

Here is a way to do the manipulation. First I define a convenient way to set up vector fields, then I do the dot product and gradient. The main thing is that the application to the vector argument on the right is done element-wise, so that the natural Mathematica operation is Map (/@). To make the order of operations clearer, I also use two different ...


4

The confusion simply arises from mistaking a typesetting information for \[CenterDot] with the meaning of a symbol. Replacing · with . will solve the issue: ClearAll["Global`*"]; r = (6.3674447) (10^6); θm = (90 - 21.43) Degree; ϕm = 39.82 Degree; θe = (90 - 56.85) Degree; ϕe = 60.6 Degree; RM = {r Sin[θm] Cos[ϕm] , r Sin[θm] Sin[ϕm], r Cos[θm]}; RE = {r ...


0

Here are two possible approaches: You could inactivate Plus as follows: expr = Inactive[Plus][x, IdentityMatrix[2]] x = {{1, 1}, {0, 0}}; expr Activate@expr As you can see, the inactive form will format very similarly to a "normal" plus sign, with a slight color change to indicate its inactive status. Alternatively, you could use an operator ...



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