I think that the important point is that the vector field data needs to be given on a regular 3-D array (i.e., a completely filled in rectangular array of positions) of vector values. The documentation says:

"ListVectorPlot3D by default interpolates the data given and plots vectors for the vector field at a regular 3D grid of positions. "

So, in the case of the first example,the values of the positions are assumed and in the second example the values of the positions are specified.

Note also that in your example of

{{{1, 2, 3}, {4, 5, 6}}, {{1, 2, 3}, {3, 2, 4}}, {{1, 2, 3}, {7, 2, 5}}}

you have repeated the point {1,2,3} in each case leading to ambiguity of what the vector field value is at that point.

But the main point in all of this is that the data needs to be specified on a regular cartesian grid with no points left out.

One more thing:

Why doesn't this work?:

ListVectorPlot3D[{{{1, 1, 1}, {2, 3, 5}}, {{2, 3, 1}, {6, 3, 1}}}]

Visualization`Core`ListVectorPlot3D::vfldata: {{{1.,1.,1.},{2.,3.,5.}},{{2.,3.,1.},{6.,3.,1.}}} is not a valid vector field dataset or a valid list of datasets.

Where can you see examples?

Because my idea was to plot the vector field using VectorListPlot3D to check if the illustration can be any better. :/

Yes, I know, I was looking at this example too, but I still don't understand why this doesn't want to work with my data?

Can you please correct this example? Maybe than the things will be obvious.

ListVectorPlot3D[{{{1, 1, 1}, {2, 3, 5}}, {{2, 3, 1}, {6, 3, 1}}}]

You see, that's so confusing. Why does it have to be so complicated? I highly regret never learning Mathematica. Now that this is clear, I have no idea how to correct my data in Mathematica :/

Table[vec, {x, 0, 1, 1}, {y, 0, 1, 1}, {z, 0, 1, 1}]

What are those x,y,z? Is that {x, xmin, xmax, step} ? If yes, ok, But how do I create the same table with the data I already have? :(

Apply Dimensions to the array to see its structure.

First example at ref/ListVectorPlot3D

In[73]:= Dimensions[
 Table[{x, y, z}, {x, -1, 1, .1}, {y, -1, 1, .1}, {z, -1, 1, .1}]]

Out[73]= {21, 21, 21, 3}

Second example there:

In[74]:= Dimensions[
 Table[{{x, y, z}, {y, x - x^3, z}}, {x, -1.5, 1.5, 0.2}, {y, -2, 2, 
   0.2}, {z, -1, 1, 0.1}]]

Out[74]= {16, 21, 21, 2, 3}

http://reference.wolfram.com/language/ref/Dimensions.html

I copied the second data structure shown in the attachment and I added one additional tuplet of vectors to replace the ellipsis. I replaced the symbols with numbers, getting

{{{1, 2, 3}, {4, 5, 6}}, {{1, 2, 3}, {3, 2, 4}}, {{1, 2, 3}, {7, 2, 5}}}

When I applied ListVectorPlot3D to this I got the error message

“… {{{1.,2.,3.},{4.,5.,6.}},{{1.,2.,3.},{3.,2.,4.}},{{1.,2.,3.},{7.,2.,5.}}} is not a valid vector field dataset or a valid list of datasets. “

I applied Dimensions to the data structure, getting:

 {3, 2, 3}

which appears to show three instances of two triplets each. Thus, it should plot 3 vectors. Doesn’t this show there is something either wrong or incomplete in the template from the documentation? Or are we still having a conceptual difficulty?

I am using Mathematica 10.

Attachments:

I just wonder what several million arrows on a typical window size plot would look like? If they are more or less evenly distributed on the domain you are just going to get a black plot. Otherwise you are going to get clumpy smears. It will probably need a better treatment than just plotting all the arrows - unless you have some super large display with extra fine detail. What do you expect the field to look like?

One technique might be to select a few hundred arrows at random and try multiple samples.

If there are critical regions you may have to zoom in on them.

One surprisingly effective technique is to generate an Interpolation function (if you can) and plot only a single arrow! But attach it to a movable Locator that you can drag around to explore the field.

Make sure that seznamcek[[i]] and the following point both have the structure of a vector, that is, the argument to Arrow has the structure

 {{a, b,}, {c, d}} or {{a, b, c}, {d, e, f}}.

Even so, For seems to suppress the Graphics output. I don't know why. Try something like this:

t = Table[Arrow[{{1, 4}, {6, x}}], {x, 1, 10}];

Graphics[t]

If you want the arrows to appear sequentially, use Manipulate or Animate.

Even than, it doesn't work.

Show is misspelled as Showt.