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Equation of plane through points on x,y,z axes

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the plane through the points on the x, y, and z-axes where x=a, y=b, z=c, respectively.


    2. Relevant equations
    If <A,B,C> is a vector perpendicular to the plane, then A(x-a)+B(y-b)+C(z-c)=0 is the equation of the plane.


    3. The attempt at a solution
    I know this is extremely simple, but I can't get the right answer even using different methods.

    Method 1:
    The vectors <0,b,-c> and <a,-b,0> are in the plane. The cross product <0,b,-c> X <a,-b,0> gives <-bc,-ac,-ab>, which is normal to the plane. Then the equation of the plane is (-bc)(x-a)+(-ac)(y-b)+(-ab)(z-c)=0, and simplifying I get (x/a)+(y/b)+(z/c)=3. Wrong.

    Method 2:
    Substituting the 3 points (a,0,0), (0,b,0), (0,0,c) into A(x-a)+B(y-b)+C(z-c)=0 gives

    -Bb-Cc=0
    -Aa-Cc=0
    -Aa-Bb=0.

    Thus C=-Aa/c, B=-Aa/b and substituting this back into A(x-a)+B(y-b)+C(z-c)=0 and simplifying I get (-x/a)+(y/b)+(z/c)=1, which is different than the answer from Method 1 and still wrong.
     
  2. jcsd
  3. Jul 21, 2009 #2

    tiny-tim

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    Homework Helper

    Hi iomtt6076! :wink:
    eugh … you've made it so complicated. :yuck:

    Just call the plane px + qy + rz = 1, and solve! :smile:
     
  4. Jul 21, 2009 #3
    tiny-tim, thank you for your reply. I see how to get from px + qy + rz = 1 to the correct answer and at last realize my (embarrassing) mistake.

    In Method 1, it was incorrect to say that "the equation of the plane is (-bc)(x-a)+(-ac)(y-b)+(-ab)(z-c)=0." This equation comes from setting the dot product <-bc,-ac,-ab>*<x-a,y-b,z-c> equal to 0. But (a,b,c) is NOT on the plane. It should be <-bc,-ac,-ab>*<x-a,y,z>.

    In Method 2, my mistake can be traced to my statement under "2. Relevant equations": A(x-a)+B(y-b)+C(z-c)=0 is NOT the equation of the plane because again, (a,b,c) is not on the plane.
     
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