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Plot 2 co-dependent vectors

Posted 9 years ago

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

POSTED BY: Alan Smith
2 Replies
Posted 9 years ago

That.is.awesome!

POSTED BY: Alan Smith

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}]
POSTED BY: Gianluca Gorni
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