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

## Homework Statement

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

## The Attempt at a Solution

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haruspex
Homework Helper
Gold Member

## Homework Statement

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

## 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?

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.

? Let me know if I've explained the problem sufficiently

mfb
Mentor
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.

Some elaboration on that would help, is possible?

mfb
Mentor
Did you draw a sketch? What you try so far to follow that idea?

Ray Vickson
Homework Helper
Dearly Missed
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.

I've literally typed out the question as it has been given.

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

Ray Vickson
Homework Helper
Dearly Missed
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.

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.

Ray Vickson
Homework Helper
Dearly Missed
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.

HallsofIvy
One method: draw a triangle with angle $\theta= arcos(u)$. 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 $\sqrt{1- u^2}$ so that $sin(arcos(u))= sin(\theta)= \sqrt{1- u^2}/1= \sqrt{1- u^2}$.
Or: since $sin^(\theta)+ cos^2(\theta)= 1$, $sin(\theta)= \sqrt{1- cos^2(\theta)}$. So $sin(arcos(u))= \sqrt{1- cos^2(\theta)}= \sqrt{1- cos^2(arcos(\theta))}=\sqrt{1- u^2}$