1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation of a plane through three points

  1. Oct 1, 2009 #1

    srn

    User Avatar

    edit: should probably be in precalc sorry about that

    1. The problem statement, all variables and given/known data
    I have three points, two known and one partially known, only the x coordinate (3D)

    Point one (A): (-0,2; -0,2; -0,2)
    Point two (B): (-0.13; -0.15; 0)
    Point three (C): (-0,2; unknown; unknown)

    I also know that the distance from A to C is 0,2 and that the distance from B to C is 0,15.

    Question: what are the other two components of point C.

    3. The attempt at a solution
    As I see it, the lines going from A and from B with the above magnitudes and crossing each other, describe a circle that is perpendicular to AB. So far, there are infinitely many solutions. However, with the x coordinate known, I think that it brings the amount of solutions down to exactly two.

    I've composed the vector AC and AB and their cross product. This should then be perpendicular to the plane and I've used that with the equation

    a(x-x0) + b(y-y0) + c(z-z0) = 0

    Where a, b and c are the components of AC X AB and x0, y0, z0 a random point which I took as A.

    I then replaced the x with x = -0,2.

    Then I took |AC| = 0,2 and |BC| = 0,15. I worked this out and set it equal to each other, then I got something in y and z, set one of them equal to the other and then filled it in in the above equation of the plane.

    This didn't yield anything at all. Can someone please explain where I have gone wrong?
     
  2. jcsd
  3. Oct 1, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    I think you should just have C= (-0.2,y1,z1) and then find AC and BC in terms of y1 and z1.

    Then use the fact that |AC|=0.2 and |BC|=0.15. You will get two equations in y1 and z1 which you can solve.
     
  4. Oct 1, 2009 #3

    srn

    User Avatar

    Ah, not so hard after all :(

    j := (y+.2)^2+(z+.2)^2 = 0.04
    k := 0.07^2+(y+.15)^2+z^2 = .02255;

    > solve({j, k}, {z, y});

    {y = -.2815230378, z = -0.01736924054}, {y = -0.4212402098e-1, z = -0.07721899475}

    That should be it then? Does anyone know how to draw a plane through 3 points in maple? Can't find anything on google.
     
    Last edited: Oct 1, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equation of a plane through three points
Loading...