Calculus 3: Show Intersection of Tetrahedron Vertices & Centers of Gravity

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SUMMARY

The discussion focuses on demonstrating that the intervals connecting the vertices of a tetrahedron to the centers of gravity of opposite sides intersect at a single point, specifically the center of gravity of the tetrahedron, calculated as (P+Q+R+S)/4. Participants emphasize the necessity of using vectors and symmetry arguments to prove this intersection. The conversation also touches on the implications of uniform density and the use of triple integration, although the primary method discussed involves vector analysis. The conclusion is that the intersection occurs at the center of gravity, supported by vector properties and symmetry.

PREREQUISITES
  • Understanding of vector mathematics
  • Familiarity with the concept of center of gravity
  • Basic knowledge of tetrahedron geometry
  • Introduction to triple integration (optional for deeper understanding)
NEXT STEPS
  • Study vector properties and their applications in geometry
  • Learn about the center of gravity calculations for different shapes
  • Explore the concept of triple integration in calculus
  • Investigate symmetry arguments in geometric proofs
USEFUL FOR

Students in calculus and geometry courses, particularly those studying vector mathematics and the properties of tetrahedrons, as well as educators looking for teaching strategies related to these topics.

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Homework Statement


So we're suppose to show that the intervals connecting vertices of a tetrahedron with centers of gravity of opposite sides interesect at one point, namely the center of gravity of a normal tetrahedron. The center of gravity is (P+Q+R+S)/4.


Homework Equations


All I know is we have to somehow work with vectors so we at the end, we end up with result.


I really have no idea how to start. Please help :(
 
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you have to use triple integration. it's not really a proof, but a calculation.
 
That can't be right, we just barely started this class and we're only learning about vectors and haven't done anything with integration :bugeye:
 
you said center of gravity of a normal tetrahedron. that must involve triple integration, unless the density is uniform throughout the tetrahedron, which you did not state.

if the density is uniform, simply intersect two lines, each of which go from one vertex to the center of the opposite triangle, which is at one third the height of the triangle.
 
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Well we show that the lines going from one vertice to the center of gravity of opposite sides will end up crossing each other right in the middle of the tetrahedron(the very center). The problem says that at that point, we know the answer is (P+Q+R+S)/4 where the letters correspond to vertices. We're suppose to use vectors to show that, at that middle point, we'll get (P+Q+R+S)/4. I think you're interpreting it another way, perhaps this made more sense?
 
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you can think along the lines of use the center of gravity formula for 4 masses:

cg = [m1(P) + m2(Q) +m3(R) + m4(S)]/(m1+m2+m3+m4)

think of what m1,m2,m3,m4 would be in your special case.

but the symmetry argument using the interesection of two lines should also work too.
 
well there would be four lines all crossing at the center of gravity and I'm suppose to show that with vectors though because this is a vector problem. I know the physics concept of it(learned it last semester) but I need to show it with vectors. Do you have any ideas?
 
using nothing but vectors?

how about place the center of gravity at the origin, by symmetry (which we can use due to the simple conditions of the object) the four vertices must be equidistant from the origin. you can figure out the other symmetry properties to argue that the sum of the four vectors must be zero.

i don't like this solution very much, but i haven't taken vector courses since i was 11.
 
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hm...ok...i'll look into it. thanks
 

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