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How to find the area of an n-dimensional triangle?

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data

    How to find area of an n-dimensional triangle using vectors?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 16, 2014 #2
    bump
     
  4. Aug 16, 2014 #3

    haruspex

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    First, what do you mean by the area of such an object? Do you mean its n-dimensional volume?
    Second, please show some attempt or thoughts on the matter. Anything relevant you've been taught? Can you do it for 3 dimensions?
     
  5. Aug 16, 2014 #4
    Find the area of the triangle with sides

    A = (a1 ... an)
    B = (b1 ... bn)
    and A-B = (a1-b1 ... an-bn)

    I don't even know where to start. I know how to do it in 3D with the cross product, but that obviously won't work for higher dimensions.

    So I need help generalizing for Rn.
     
  6. Aug 16, 2014 #5
    ? Let me know if I've explained the problem sufficiently
     
  7. Aug 16, 2014 #6

    mfb

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    There are n-dimensional generalisations of the cross-product, one of them will work.

    Alternatively, use the dot product and some vector additions in the same way you can do it in two dimensions.
     
  8. Aug 16, 2014 #7
    Some elaboration on that would help, is possible?
     
  9. Aug 16, 2014 #8

    mfb

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    Did you draw a sketch? What you try so far to follow that idea?
     
  10. Aug 16, 2014 #9

    Ray Vickson

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    Your terminology is incorrect and highly misleading. You are not talking about an "n-dimensional triangle"; you are talking about a standard triangle located in an n-dimensional space. Your triangle is still a "2-dimensional" object. Recognizing that is very important for solving the problem.
     
  11. Aug 16, 2014 #10
    I've literally typed out the question as it has been given.

    I think take it as a standard triangle in n-dim space
     
  12. Aug 16, 2014 #11

    Ray Vickson

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    OK, so we are agreed on that. Now what would you do next?

    If I were doing this problem I would look for formulas for a triangle that involve only norms and inner products of the vectors forming the triangle's sides.
     
  13. Aug 16, 2014 #12
    A = {2,1,2,4}
    B = {4,1,6,2}

    angle = arccos( (A.B)/(norm(A)*norm(B)) )

    area = (1/2)A.B sin(angle)

    Is this correct? if yes, it is easy to generalize for Rn.
     
  14. Aug 16, 2014 #13

    Ray Vickson

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    You can make it a lot neater by expressing ##\sin(\arccos(u))## in a simpler way---just think about what ##\arccos(u)## actually means, and what that says about its sine.
     
  15. Aug 16, 2014 #14

    HallsofIvy

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    One method: draw a triangle with angle [itex]\theta= arcos(u)[/itex]. Then, since cosine is "near side over hypotenuse", we can label the hypotenuse 1 and the near side u. By the Pythagorean theorem, the opposite side has length [itex]\sqrt{1- u^2}[/itex] so that [itex]sin(arcos(u))= sin(\theta)= \sqrt{1- u^2}/1= \sqrt{1- u^2}[/itex].


    Or: since [itex]sin^(\theta)+ cos^2(\theta)= 1[/itex], [itex]sin(\theta)= \sqrt{1- cos^2(\theta)}[/itex]. So [itex]sin(arcos(u))= \sqrt{1- cos^2(\theta)}= \sqrt{1- cos^2(arcos(\theta))}=\sqrt{1- u^2}[/itex]
     
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