how to obtain a plane projection of a `StreamPlot3D`

Question

StreamPlot3D are rather difficult to decipher. Is there a simple way to obtain the projection of such a plot onto a given plane? Ideally, I am looking for a method which does not even create the StreamPlot3D, since I am still in the process of getting the version 12.3 where this command appeared.

Answer 1

Based on @MichaelE2's comment:

As far as I know there exists only VectorPlot3D in Mathematica v12.1

First one has to define the normal n of the plane

n={1,1,1}

The projection of an arbitrary vector v into this plane follows to v-v.n/n.n n

Now we still have to define a coordinate system e1,e2 inside the plane, examplary

e1 = (# - # . n/n . n n) &[{1, 0, 0}];
e2 = (# - # . n/n . n n) &[{0, 1, 0}];

pic=VectorPlot3D[{y^2, 1, x}, {x, -1, 1}, {y, -1,1}, {z, -1, 1}] /.   Arrow[Tube[{a_, b_}, c_]] :> Arrow[{a, b}] (*eliminate Tube*) 
pic /. {Graphics3D ->Graphics, 
{x_Real, y_Real,z_Real} :> {# . e1, # . e2} &[{x, y,z} - {x, y, z} . n/n . n n]}