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3D geometry parallelepiped problem

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data

    Given a rectangular parallelepiped ABCDEFGH, the diagonal [AG] crosses planes BDE and CFH in K and L. Show K and L are BDE's and CFH's centres of gravity.

    I think I have understood the problem, could you verify my demo please ? Thanks

    2. Relevant equations

    3. The attempt at a solution

    Let us call ##G_1## and ##G_2## BDE's and CFH's center of gravity.

    ## \vec{AG} = \vec{AB} + \vec{AD} + \vec{AE} = 3 \vec{AG_1}##

    So ## A,G_1, G## are aligned and ##G_1## belongs to plane BDE so we must have ##K = G_1##

    Then

    ## 3 \vec{AG_2} = \vec{AC} + \vec{AF} + \vec{AH} = \vec{AB} + \vec{BC} + \vec{AE}+\vec{EF} + \vec{AD} + \vec{DH} = \vec{AG} + \vec{BC} + \vec{EF}+\vec{DH} = 2\vec{AG}##

    So, similarly, ## A,G_2, G## are aligned and ##G_2## belongs to plane CFH so we must have ##L = G_2##

    Furthermore, ##K## and ##L## are at the third and two third of the diagonal.
     
  2. jcsd
  3. Dec 14, 2016 #2

    jedishrfu

    Staff: Mentor

    It would help if you provided a picture of your regular parallelepiped with vertices labeled.
     
  4. Dec 14, 2016 #3
    Hi, here is the picture
     

    Attached Files:

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