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

1. Aug 16, 2014

### LFCFAN

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. Aug 16, 2014

### LFCFAN

bump

3. Aug 16, 2014

### haruspex

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. Aug 16, 2014

### LFCFAN

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.

5. Aug 16, 2014

### LFCFAN

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

6. Aug 16, 2014

### Staff: 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.

7. Aug 16, 2014

### LFCFAN

Some elaboration on that would help, is possible?

8. Aug 16, 2014

### Staff: Mentor

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

9. Aug 16, 2014

### Ray Vickson

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. Aug 16, 2014

### LFCFAN

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. Aug 16, 2014

### Ray Vickson

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. Aug 16, 2014

### LFCFAN

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. Aug 16, 2014

### Ray Vickson

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. Aug 16, 2014

### HallsofIvy

Staff Emeritus
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 $sin(arcos(u))= \sqrt{1- cos^2(\theta)}= \sqrt{1- cos^2(arcos(\theta))}=\sqrt{1- u^2}$