Plot 2 co-dependent vectors

Questions
1

The equation ( \vec{r} - \vec{a}).\vec{a} = 0 gives a plane perpendicular to \vec{a} and passing through a (by a I mean \vec{a} viewed as a point in three-space). \vec{a} is a constant vector.

I want to plot this in mathematica, need a little help please

  1. I assume (tell me if wrong) I can only do this for known coefficients, not in the general form?

  2. So choosing an arbitrary \vec{a} = {1,2,3} I set a_1, a_2, a_3 and d , rewrote the plane eqtn in terms of z and plotted the plane using
    Plot3D[{Subscript[a, 1]/Subscript[a, 3] x + Subscript[a, 2]/Subscript[a, 3] y - d}, {x, -10, 10}, {y, -10, 10}] … which works.

  3. Now I want to add the vector \vec{a} into the above - I tried [Line[{0, 0, 0, Null}, {Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}] - which works on it’s own, its just a line segment with no variables. I can’t figure out how to now combine it into Plot3D with the plane - I’m just struggling with the syntax I think…

I tried a few variations, not really sure what I’m doing, - for example

Plot3D[{Subscript[a, 1]/Subscript[a, 3] x + Subscript[a, 2]/Subscript[a, 3] y - d}, 
Line[{0, 0, 0, Null}, {Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}, {x, -10, 10}, {y, -10, 10}]

If someone could mentor me through this, I’d really appreciate it. Thanks

2

This is one way:

Manipulate[With[{a = {a1, a2, a3}, r = {x, y, z}},
  Show[ContourPlot3D[(r - a).a == 0,
    {x, -2, 2}, {y, -2, 2}, {z, -2, 2}], 
   Graphics3D[{PointSize[Large], Arrow[{{0, 0, 0}, a}], Point[a]}]]],
 {a1, -1, 1}, {a2, -1, 1}, {a3, -1, 1}]
3

That.is.awesome!