How do you draw the plane on which two vectors lie?

Question

Given two arbitrary vectors $\textbf{v}_1$ and $\textbf{v}_2$, how can I draw the plane which they span?

Answer 1

SeedRandom[3];
{v1, v2} = RandomReal[{-2, 2}, {2, 3}];
n = Cross[v1, v2];
Show[{
  ContourPlot3D[n.{x, y, z} == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
    ContourStyle -> Opacity[0.5], Mesh -> False],
  Graphics3D[{Arrow[{{0, 0, 0}, v1}], Arrow[{{0, 0, 0}, v2}]}]
}]

Answer 2

Here's an example of how you could do it:

v1 = {1, -1/2, -1.9}; (* pick something *)
v2 = {0, 1, -1}; (* pick something *)
r0 = {-1/2, 1/2, 3/4}; (* point in the plane; pick something *)

nn = Normalize[Cross[v1-r0, v2-r0]];
r = {rx, ry, rz};
sol = Solve[Dot[nn, (r - r0)] == 0, {rz}] // Simplify
Plot3D[rz /. sol, {rx, -10, 10}, {ry, -10, 10}]

Answer 3

Another option is to use Graphics3D primitives

Graphics3D[Polygon[{{0,0,0},v1,v1+v2,v2}]]

Edit

J.M. already gave one way to produce a square spanned by the two vectors as a Polygon. Another way would be to use Rotate:

{v1, v2} = RandomReal[{-1, 1}, {2, 3}]

Graphics3D[{
  Rotate[Polygon[{{-1, -1, 0}, {1, -1, 0}, {1, 1, 0}, {-1, 1, 0}}],
    {{0, 0, 1}, Cross[v1, v2]}],
  Arrow[{{0, 0, 0}, v1}],
  Arrow[{{0, 0, 0}, v2}]}]

Answer 4

If your vectors are three-element lists, here's one way (though I really feel like a hack for suggesting this):

ParametricPlot3D[u v1 + v v2, {u,-1,1},{v,-1,1},
  PlotRange->{{-1,1},{-1,1},Automatic}]

You may have to adjust the limits of -1 and 1 depending on how much of the plane you want to plot.

Answer 5

I present here a modification of Heike's approach that might be attractive for some applications; it produces a square with side length c as a Polygon[] object, representing the plane spanned by v1 and v2, with origin at r0:

v1 = {1, -1/2, -19/10}; v2 = {0, 1, -1}; (* spanning vectors *)
r0 = {-1/2, 1/2, 3/4}; (* origin *)
{o1, o2, o3} = Orthogonalize[Append[#, Cross @@ #]] &[{v1 - r0, v2 - r0}];
With[{c = 6}, (* generate a c×c square *)
   Graphics3D[{{Directive[EdgeForm[], Gray], 
      Polygon[{r0 + c (o1 + o2)/2, r0 - c (o1 - o2)/2, 
               r0 - c (o1 + o2)/2, r0 + c (o1 - o2)/2}]},
              {Red, Arrow[{r0, v1}], Arrow[{r0, v2}]},
              {Blue, Arrow[{r0, r0 + Cross[v1 - r0, v2 - r0]}]}},
              Boxed -> False, Lighting -> "Neutral", PlotRange -> All]]

As can be seen from the code, the trick is in producing mutually orthogonal unit vectors via Orthogonalize[] (for older versions, use QRDecomposition[] instead and take the $\mathbf Q$ factor) that can be nicely scaled/combined afterwards.

Here's how to verify that a square is indeed produced as claimed:

FullSimplify[Map[Norm, Differences[Append[#, First[#]]]]&[{r0 + c (o1 + o2)/2, 
             r0 - c (o1 - o2)/2, r0 - c (o1 + o2)/2, r0 + c (o1 - o2)/2}], c > 0]
{c, c, c, c}

Answer 6

Using

v1 = {1, -1/2, -19/10}; v2 = {0, 1, -1}; (* spanning vectors *)
r0 = {-1/2, 1/2, 3/4}; (* origin *)

as a concrete example, I present here, for giggles, variations of Mark's and David Skulsky's answers, based on formula 18 here.

Mark:

Show[ContourPlot3D[
  Det[PadRight[{{x, y, z}, r0, v1, v2}, {4, 4}, 1]] == 0, {x, -3, 
   3}, {y, -3, 3}, {z, -3, 3}, BoxRatios -> Automatic, 
  ContourStyle -> Opacity[0.5], Mesh -> False], 
 Graphics3D[{{Red, Arrow[{r0, v1}], Arrow[{r0, v2}]}, {Blue, 
    Arrow[{r0, r0 + Cross[v1 - r0, v2 - r0]}]}}]]

David (plus Heike's suggestion):

Show[Plot3D[
  Evaluate[z /. 
    First[Solve[
      Det[PadRight[{{x, y, z}, r0, v1, v2}, {4, 4}, 1]] == 0, 
      z]]], {x, -2, 2}, {y, -2, 2}, BoxRatios -> Automatic, 
  MaxRecursion -> 1, Mesh -> 0, PlotPoints -> 2], 
 Graphics3D[{{Red, Arrow[{r0, v1}], Arrow[{r0, v2}]}, {Blue, 
    Arrow[{r0, r0 + Cross[v1 - r0, v2 - r0]}]}}]]

Answer 7

Interactive - drag orange dots around.

Manipulate[ Graphics3D[{{Blue, Opacity[.5], Polygon[{{-1, -1, 0}, {1, -1, 0}, 
{1, 1, 0}, {-1, 1, 0}}]}, {Red, Thick, Arrow[{{0, 0, 0}, Flatten@{p, 0}}]}, 
{Black, Thick, Arrow[{{0, 0, 0}, Flatten@{q, 0}}]}}, SphericalRegion -> True, 
Boxed -> False, ViewAngle -> .37],{{p, {1, 0}}, {-1, -1},{1, 1}},{{q, {0, 1}}, 
{-1, -1}, {1, 1}}, ControlPlacement -> Left]