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Help!Given a common intersection point, create 3 different planes

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Given a common intersection point (3,4,5), find 3 different planes.


    2. Relevant equations
    None


    3. The attempt at a solution

    What I did is let

    a1x+a2y+a3z=a
    b1x+b2y+b3z=b
    c1x+c2y+c3z=c

    3=Dx/D 4=Dy/D 5=Dz/D

    I set D=2, therefore Dx=6 Dy=8 Dz=10,
    I was stuck then, no matter how I did I just got a bunch of unknown.
    This question is supposed to find any 3 linear equations(aka 3 planes) that satisfy the intersection point.

    Well, who doesn't know how to find the intersecting point using Cramer's rule...

    P/S: This question comes up in my calculus 3 question, and I have not taken any algebra course yet, so I only know the condition for Cramer's rule for a common intersection point to take place which is D not equal to 0.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 13, 2010 #2
    If you want 3 different planes that intersect at (3,4,5), why not just use z=5, y=4, and x=3?
     
  4. Mar 15, 2010 #3
    Thanks a lot dustin for your reply.

    If i want to justify that this is indeed the intersection point of the 3 different planes you mentioned, can I apply Cramer's rule and then do the dot and cross products with the normal vectors of the 3 planes. If the determinant is indeed not equal to 0, then the intersection point is justified?
     
  5. Mar 15, 2010 #4
    Well if we use those equations then are we would have x+0y+0z=3; 0x+y+0z=4; 0x+0y+z=5. Which would make our determinant 1.
     
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