Consider a simple plane equation


I just want some position vector on that plane. For example, {7/2,0,0} would be great.

How can I ask Mathematica for 'some point' on this plane? The best I've got right now is:

Solve[plane && y == 0 && z == 0, x]

But I don't actually care what the values of y and z are, I just want some random point that satisfies the plane equation. Also that just returns the value of x - I want all three values in a vector (list?) form {a, b, c}.

  • $\begingroup$ There are quite literally an infinite number of points on that plane. You have to give some criteria for which point you want. $\endgroup$ – Sparr Feb 1 '14 at 4:45
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    $\begingroup$ And if I asked you to pick a random integer, could you do that? Could a computer? I think my question is pretty reasonable, and if the answer insists I put some bounds on the values, that's fine too. $\endgroup$ – Henry Feb 1 '14 at 4:51
  • $\begingroup$ Related: (9734) $\endgroup$ – Mr.Wizard Feb 2 '14 at 8:44

You can use FindInstance:

{x,y,z} /. FindInstance[2 x - 3 y + 4 z == 7, {x, y, z}, Reals, 5]
{{-22, 8/5, 279/20},
 {-(109/5), 11/5, 143/10},
 {8/5, 2, 49/20},
 {42/5, 82/5, 197/20},
 {149/10, -(11/2), -(393/40)}}
  • $\begingroup$ Thanks, that's exactly what I needed! $\endgroup$ – Henry Feb 1 '14 at 5:02
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    $\begingroup$ Since Henry said "I don't actually care what the values of y and z are" I am editing this to leave out those conditions, and show the output. (+1 of course.) $\endgroup$ – Mr.Wizard Feb 1 '14 at 8:03


In Euclidean space, the point on a plane ax + by + cz = d that is closest to the origin has the Cartesian coordinates (x,y,z), where:

x = ad/(a^2+b^2+c^2)

y = bd/(a^2+b^2+c^2)

z = cd/(a^2+b^2+c^2)

Since you know a=2, b=-3, c=4, and d=-7, you can get {x,y,z} on your plane using these three equations.

  • $\begingroup$ I guess that does give a point, and I could translate it into Mathematica. But this is not really what the question was asking. This works great though: FindInstance[2 x - 3 y + 4 z == 7, {x, y, z}] $\endgroup$ – Henry Feb 1 '14 at 5:01

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