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

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Homework Statement



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

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The Attempt at a Solution

 

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  • #3
haruspex
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Homework Statement



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

Homework Equations





The Attempt at a Solution

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?
 
  • #4
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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.
 
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? Let me know if I've explained the problem sufficiently
 
  • #6
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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.
 
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Some elaboration on that would help, is possible?
 
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Did you draw a sketch? What you try so far to follow that idea?
 
  • #9
Ray Vickson
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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.
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.
 
  • #10
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I've literally typed out the question as it has been given.

I think take it as a standard triangle in n-dim space
 
  • #11
Ray Vickson
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I've literally typed out the question as it has been given.

I think take it as a standard triangle in n-dim space
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.
 
  • #12
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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.
 
  • #13
Ray Vickson
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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.
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.
 
  • #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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