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Vector/Geometry toughie 2

  1. Mar 29, 2008 #1
    If a, b, and c are the x, y and z intercepts of a plane respectively and d is the distance from the origin to the plane, prove that:

    1/dsquared = 1/a squared + 1/b squared + 1/c squared

    I made three points
    A (a, 0, 0)
    B (0, b, 0)
    C (0, 0, c)

    Then I made vector AB [-a, b, 0] and vector BC [0, -b, c]. Then I found the cross product of AB x BC, which gave me a normal vector of [bc, ac, ab]. If these are my A, B and C for the scalar equation of a plane, then:

    bcx + acy + abz + D= 0
    I solved for D, which is -abc.

    Therefore, the scalar equation for the plane is

    bcx + acy + abz -abc= 0

    Now I am completely stuck as how to use that scalar equation to prove the above equation.
     
  2. jcsd
  3. Mar 29, 2008 #2

    tiny-tim

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    Homework Helper

    Hi emma! :smile:

    Hint: [bc,ca,ab] = abc[1/a,1/b,1/c]. :smile:
     
  4. Apr 1, 2008 #3
    thank you tiny-tim...
     
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