3D geometry parallelepiped problem

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SUMMARY

The discussion centers on proving that points K and L, where the diagonal AG of a rectangular parallelepiped ABCDEFGH intersects planes BDE and CFH, are the centers of gravity for these planes. The user demonstrates the alignment of points A, G1, and G, establishing that K equals G1, and similarly shows that L equals G2. The calculations confirm that K and L are located at one-third and two-thirds along the diagonal AG, respectively, affirming their roles as centers of gravity for the respective planes.

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  • Familiarity with the concept of centers of gravity in geometric shapes
  • Knowledge of the properties of rectangular parallelepipeds
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Homework Statement


[/B]
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

Homework Equations



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.
 
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It would help if you provided a picture of your regular parallelepiped with vertices labeled.
 
Hi, here is the picture
 

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