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2

To implement your first identity you can use: vec /: Cross[vec[x_], Cross[vec[y_], vec[z_]]] := Cross[Cross[vec@x, vec@y], vec@z] -Cross[Cross[vec@x, vec@z], vec@y] Resulting in: In[1] := vec[x]\[Cross](vec[y]\[Cross]vec[z]) Out[1] = (vec[x]\[Cross]vec[y])\[Cross]vec[z] - (vec[x]\[Cross]vec[z])\[Cross]vec[y] The other identity is implemented ...


3

Manipulate[ Graphics3D[{{Opacity[0.5], Sphere[]}, {EdgeForm[{Dashed, Red}], FaceForm[None], Cylinder[{{0, 0, -.001}, {0, 0, .001}}]}, {Dashed, Line[{{0, 0, 0}, {1, 0, 0}}]}, {Dashed, Line[{{0, 0, 0}, {0, 1, 0}}]}, {Dashed, Line[{{0, 0, 0}, {0, 0, 1}}]}, Arrow[{{1, 0, 0}, {1.3, 0, 0}}], Arrow[{{0, 1, 0}, {0, 1.3, 0}}], Arrow[{{0, 0, ...


1

In my opinion the easy way to plot vectors over 1D curve is to used VectorPoints option: points = Table[{i, 1}, {i, 0, 1, .1}]; VectorPlot[{Sin[x], Cos[y]}, {x, 0, 1}, {y, -1, 2}, VectorPoints -> points, VectorScale -> {0.1, .2}, Epilog -> Point[points]]


2

Well, here's a way that works when the number of seconds since Jan. 1, 1970 is odd (that is, it crashes the kernel every other time I execute it): reg = MeshRegion[{{0, 1}, {1, 1.001}}, Line[{1, 2}]]; points = MeshCoordinates@ DiscretizeRegion[reg, MaxCellMeasure -> {"Length" -> 0.1}]; vf = Table[{p, {Sin[x], Cos[y]} /. Thread[{x, y} -> p]}, {p, ...


4

I don't know how to make this with RegionFunctions but you could show the vectors along the Line[{{0, 1}, {1, 1}}] like this: VectorPlot[{Sin[x], Cos[y]}, {x, 0, 1}, {y, 0.5, 1.5}, AspectRatio -> 1/5, FrameTicks -> {True, {0.95, 1, 1.05}, False, False}, GridLines -> {None, {1}}, GridLinesStyle -> {Blue, Dashed}, PlotRange -> {Automatic, ...



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