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Area in multidimensional space using Wolfram|Alpha?

Tim Roberts
Posted 2 years ago

Can someone please tell me how to use Wolfram|Alpha to calculate area in a multidimensional space, given the coordinates of the vertices? E.g., the area of a triangle in 5D with vertices (0,0,0,1,2), (1,2,1,3,2), (2,0,1,1,1) ?
POSTED BY: Tim Roberts
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Another variation, using Orthogonalize to find an orthonormal basis of the plane containing the triangle:

a = {1, 2, 1, 3, 2};
b = {0, 0, 0, 1, 2};
c = {2, 0, 1, 1, 1};
{a0, b0, c0} = {a - a, b - a, c - a};
basis = Orthogonalize[{b0, c0}];
triangleInNewBasis = 
 Triangle[{basis . a0, basis . b0, basis . c0}] // Simplify
Area[triangleInNewBasis]
POSTED BY: Gianluca Gorni

I don't know about Wolfram|Alpha, but Mathematica gives an area this way:

b = {0, 0, 0, 1, 2};
a = {1, 2, 1, 3, 2};
c = {2, 0, 1, 1, 1};
Area[Triangle[{a, b, c}]]

I have not checked if the result is correct.
POSTED BY: Gianluca Gorni

ERROR of mine! If you want the area of the triangle you have to divide the foregoing results by 2.
POSTED BY: Hans Dolhaine

And there is another method using the non - commutative algebra of the 2-vectors made of the base-vectors of R5:

1st method was

b = {0, 0, 0, 1, 2}
a = {1, 2, 1, 3, 2}
c = {2, 0, 1, 1, 1}
Norm[b - a] Norm[(c - a) - ((c - a).(b - a)/(b - a).(b - a)) (b - a)]

the 2nd method then is

x1 = b - a
x2 = c - a
jj = Subsets[Range[5], {2}]
vv = Table[
Det[
{Part[x1, jj[[k]]], Part[x2, jj[[k]]]}
],
{k, 1, Length[jj]}
]
Sqrt[vv.vv]
POSTED BY: Hans Dolhaine

Dear Hans,

your second method 

Norm@Table[Det[{x1[[jj[[k]]]], x2[[jj[[k]]]]}], {k, 1, Length[jj]}]/2

seems to me a really remarkable approach! Excellent -I need to remember this!

Best regards -- Henrik
POSTED BY: Henrik Schachner

I just modified my formula.

a=(0,0,0,1,2)
b= (1,2,1,3,2)
c= (2,0,1,1,1)

or any permutaion
POSTED BY: Hans Dolhaine
Tim Roberts
Tim Roberts
Posted 2 years ago

Thanks! Could you tell me how I can calculate a, b, and c, please?
POSTED BY: Tim Roberts

Your vectors (call them a, b and c) span a 2-dimensional subspace of R5.

So I guess your sought area is given by

Norm[b - a] Norm[(c - a) - ((c - a).(b - a)/(b - a).(b - a)) (b - a)]

You could check this in R3.
POSTED BY: Hans Dolhaine
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