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## Homework Statement

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

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

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bump

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First, what do you mean by the area of such an object? Do you mean its n-dimensional volume?## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

Second, please show some attempt or thoughts on the matter. Anything relevant you've been taught? Can you do it for 3 dimensions?

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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

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mfb

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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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mfb

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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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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.

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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I think take it as a standard triangle in n-dim space

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Ray Vickson

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

I think take it as a standard triangle in n-dim space

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.

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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.

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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.

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

HallsofIvy

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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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