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Equation of the plane that contains a point

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Let p1 = (4,0,4), p2 = (2,-1,8) and p3 = (1, 2, 3).

    Specify the condition that p = (x,y,z) lies in the plane of p1, p2, p3 (as an equation in x, y, z). Equation of a plane has the form of Ax + By + Cz + D = 0.

    (There were some other questions before this about finding the normal to the plane and area of the triangle formed by the 3 points which I have completed).

    2. The attempt at a solution
    I'm not sure where to start here. The normal vector I have is <s>(-7, -9, -7)</s> (-7, -14, -7) which I think is correct. I think I somehow need to incorporate it into the general equation Ax+By+Cz+D=0, but don't know how.

    Any help is very much appreciated.
     
    Last edited: Feb 17, 2009
  2. jcsd
  3. Feb 17, 2009 #2

    Dick

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    I don't think you have the normal right. How did you get it?
     
  4. Feb 17, 2009 #3
    I took the cross product of (p2-p1) and (p3-p1). I found it using the determinant.
     
  5. Feb 17, 2009 #4

    Dick

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    I did too. But I didn't get the same thing. Either there's a typo in the statement. Possible. Or I made a mistake. Also possible. Or you made a mistake. Also possible. It's hard to tell since you didn't show how you worked the cross product in any detail. If you do that people can suggest where you might have gone wrong. Or I can figure out where I did. I got the -7's ok. I wonder about the -9 in the middle.
     
  6. Feb 17, 2009 #5
    Thanks for the quick response. I just double checked and it seems I was wrong. I'm now getting (-7, -14, -7) for the normal. Is that what you got?

    v = (p2-p1) = -2, -1, 4
    w = (p3-p1) = -3, 2, -1

    v X w = (-7, -14, -7)
    I then did the determinant of the matrix.
    http://en.wikipedia.org/wiki/Cross_product#Matrix_notation

    I found a cross product calculator which also seems to give the answer I got.
    http://www.analyzemath.com/vector_calculators/vector_cross_product.html

    If you need more detail, I'll be happy to explain myself.
     
  7. Feb 17, 2009 #6

    Dick

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    No, s'ok. That's what I got. Now that you have the normal n right, the equation of the plane is n.(x,y,z)+D=0 (dot product). Put any one of p1, p2 or p3 in for (x,y,z) to get the value of D. If everything is kosher, you get the same value for any of them.
     
  8. Feb 17, 2009 #7
    Ok, let's see. Using p2 (2,-1,8)
    -7*(2) + -14*(-1) + -7*(8) = -56
    I get -56 when using p1 and p2, as well.
    That's D.

    For A, B, and C in the equation, I just use the normal components? So the final equation becomes...
    -7x + -14y + -7z - 56 = 0.

    If that's it, then thanks a bunch!
     
  9. Feb 17, 2009 #8

    Dick

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    Almost. Your first stuff seems to say -7x-14y-7z=-56. Then you say -7x-14y-7z-56 = 0. They are just a little different. Right? Which one is correct?
     
  10. Feb 17, 2009 #9
    Hmm, so the -56 is -D. So it should be -7x-14y-7z + 56 = 0. Thanks for catching that! :)
     
  11. Feb 17, 2009 #10

    Dick

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    No problem.
     
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