Find the top vertex coordinate of a regular tetrahedron

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SUMMARY

The discussion focuses on calculating the top vertex coordinate of a regular tetrahedron with base vertices A(1,1,0), B(3,1,0), and C(2,1+(3^(1/2)),0). The user attempts to find the height and z-coordinate using the apothem formula a = [sqrt(3)/6]*s, where s is the edge length of 2. The calculations lead to a height of 4/sqrt(3) for the vertex S, but the user expresses uncertainty about the correctness of this solution compared to an external source. The discussion emphasizes the importance of vector analysis in determining the coordinates.

PREREQUISITES
  • Understanding of regular tetrahedrons and their properties
  • Familiarity with the distance formula in three-dimensional space
  • Knowledge of vector operations and geometric center calculations
  • Ability to apply the apothem formula in geometric contexts
NEXT STEPS
  • Study the derivation and application of the apothem formula in regular tetrahedrons
  • Learn about vector representation and calculations in three-dimensional geometry
  • Explore methods for calculating heights and coordinates in polyhedra
  • Investigate solutions to similar geometric problems for better understanding
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Students studying geometry, particularly those focused on polyhedra, as well as educators and tutors looking for problem-solving strategies in three-dimensional shapes.

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Homework Statement



A regular tetrahedron has the vertices of its base A(1,1,0) B(3,1,0) C(2,1+(3^(1/2),0). Find coordinate of vertex S?

Homework Equations



The Attempt at a Solution


If this is a tetrahedron
image004.jpg


Then we know the length by caclulating the distance formula, which gives length of 2.

My game plan is to find the height in order to determine the z-coordinate.
I thought I could just get length of the apothem, which is given by a = [sqrt(3)/6]*s, where s is 2 in this case. this gives us sqrt(3)/3

Then I tried to calculate the length of BH... so (sqrt(3)/3))^2 + (1)^2 = BH^2
and i have BH = 2/sqrt(3), or 2*sqrt(3)/3

anyway. this leads to caclulate the height AOH, and i had [2/sqrt(3)]^2 + (2)^2 = AOH^2, where 2^2 comes from the length of AB...
and AOH is 4/sqrt(3)

I don't have any solution to this problem, but upon googling someone attempted the problem.
http://answers.yahoo.com/question/index?qid=20100903153810AAmVNTY

I am not sure where I did wrong, if that solution is corrected.
Just by looking at the z-value, I would have 4/sqrt(3)... which is different from whatever was solved on yahoo answer.

Can anyone please help me on this? Thanks!
 
Physics news on Phys.org
first find the centre, H in the diagram, of the base, if you think it through you should be able to find something in terms of vectors like:
H = A + AB/2 + (AC + BC)*(1/2)*(1/3)

then you if S = (x,y,z), and now you know x & y

to find z just consider a side length
 

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