# Is this possible? [Pythagorean theorem in 3D]

## Main Question or Discussion Point

Imagine the shape you would get if you chopped off the corner of a cube (the *smaller* corner piece with 4 sides and 6 edges).

Can the area of the largest side be expressed in terms of the areas of the 3 smaller sides? Like the Pythagorean theorem in 3D.

mgb_phys
Homework Helper
The distance in 3D is the same Pythagorus theorem = sqrt( x2 + y2 + z2 )

tiny-tim
Homework Helper
Hi Georgepowell!
Imagine the shape you would get if you chopped off the corner of a cube (the *smaller* corner piece with 4 sides and 6 edges).

Can the area of the largest side be expressed in terms of the areas of the 3 smaller sides? Like the Pythagorean theorem in 3D.
You mean a pyramidy thingy?

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?

The distance in 3D is the same Pythagorus theorem = sqrt( x2 + y2 + z2 )
I am wanting to do it for areas, not distances. I already knew that.

Hi Georgepowell!

You mean a pyramidy thingy?

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?
You misunderstood. I never said that the "pyramidy thingy" would be perfectly cut off the corner (so it could, for example, be a very tall thin pyramidy thingy). So the edge would not have to be √2. And the triangle would not have to be equilateral. Sorry if I was unclear. Do you understand?

Ben Niehoff
Gold Member
Yes, it is true that for a right tetrahedron, the square of the area on the diagonal face is equal to the sum of the squares of the areas of the other three faces.

D^2 = A^2 + B^2 + C^2

Try to figure out how to prove it.

An analogous theorem holds in higher dimensions: for a right n-simplex, the square of the (n-1)-volume of the diagonal face is equal to the sum of the squares of the (n-1)-volumes of the other faces.

tiny-tim
Homework Helper
ok … then the area times the normal unit vector is 1/2 (b - a) x (c - a) = 1/2 (b x c + c x a + a x b) …

so the area is the magnitude of that …

which you can find by using Pythagoras.

Yes, it is true that for a right tetrahedron, the square of the area on the diagonal face is equal to the sum of the squares of the areas of the other three faces.

D^2 = A^2 + B^2 + C^2

Try to figure out how to prove it.

An analogous theorem holds in higher dimensions: for a right n-simplex, the square of the (n-1)-volume of the diagonal face is equal to the sum of the squares of the (n-1)-volumes of the other faces.
Ahh cool! Just what I suspected! (although that is easy to say after I have the answer :tongue:) I will have a crack at proving it, although if I struggle I will ask for a clue.

ok … then the area times the normal unit vector is 1/2 (b - a) x (c - a) = 1/2 (b x c + c x a + a x b) …

so the area is the magnitude of that …

which you can find by using Pythagoras.
I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?

tiny-tim
Homework Helper
I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?
Have you done vectors and cross-products?

If not, ignore what I said.

Have you done vectors and cross-products?

If not, ignore what I said.
Vectors, yes a bit. Cross-products no, but I am have just skimmed over the wiki page on it and it is making sense.

Hi Georgepowell!

You mean a pyramidy thingy?

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?
I don't think that's what George means. He never said that the chopped off pyramid has an equilateral base. Pythagoras doesn't only work for an isosceles right triangle, but for any right triangle. George is trying to prove a 3D analog for Pythagoras even when the three right triangles in his pyramid are not mutually congruent. Right?

However, an illustrative first step would be to prove tiny-tim's special case, where all the right triangles are mutually congruent and the base triangle is therefore equilateral.

Heron's formula for the area of a triangle in terms of it's perimeter came in really handy for me here. With that in hand, it was just plug and chug. Now, if my right triangles have area A, B, and C, then I'm really hoping for the area of my "hypotenuse-like" triangle to have an area D like...

$$D = \sqrt{ A^2 + B^2 + C^2}$$

... to be perfectly analogous to the two dimensional Pythagorean. However, that's not what I got. I have an extra factor of $$\sqrt 2$$ in my calculation. It's worth someone double checking this, but I'm convinced that there is a 3D analog of the Pythagorean theorem, with a form similar but not identical to the 2D version. Nice question, George!

Okay, here's a related one. Prove the converse of the 3D Pythagorean theorem. In other words, given a pyramid whose faces obey the certain relation for their areas, must three of those sides meet at a vertex with all right angles?

However, that's not what I got. I have an extra factor of $$\sqrt 2$$ in my calculation. It's worth someone double checking this, but I'm convinced that there is a 3D analog of the Pythagorean theorem, with a form similar but not identical to the 2D version. Nice question, George!
Oh darn. $$2^4 = 16$$, not eight. So my root two factor was indeed spurious, and I'm now in agreement with Ben Niehoff's formula.

tiny-tim
Homework Helper
Hi Georgepowell!

just got up … :zzz:
I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?
Vectors, yes a bit. Cross-products no, but I am have just skimmed over the wiki page on it and it is making sense.
ok, now you've gt the basics:

the area of a triangle is 1/2 bc sinθ, which happens to be the same as the magnitude of the cross product 1/2 b x c, assuming the vertices are at 0 b and c.

b x c, of course, is a vector: its magnitude is twice the area of the triangle, and its direction is normal (perpendicular) to the triangle.

In the more general case where the third vertex is at a (not 0), the sides are b - a and c - a, and so the area times the normal unit vector is 1/2 (b - a) x (c - a) = 1/2 (b x c + c x a + a x b)

Square that, put the origin at the vertex of the cube, so that (b x c) x (c x a) = 0 and so on (can you see why?), and you get the result.

Last edited:
It turned out that today in Maths we learnt about the dot product of two vectors, it looks very similar to the cross product. The way I understand it:

The cross product of two vectors axb =|a|*|b|*sin(θ) [vector, perpendicular to a and b]
The dot product of two vectors a.b =|a|*|b|*cos(θ) [scalor]

The dot product a.a is equal to the area of a square with side lengths a. (because cos(0) = 1)

The size of the cross product axb is equal to the area of a parallelogram made with the vectors a and b.

I was playing around with these two tools and I proved the original Pythagorean theorem (I think). Here goes:

A right angled triangle has two smaller sides represented by vectors a and b.

0.5*[a.a+b.b] is the sum of the squares of the two smaller sides.

0.5*[(a-b).(a-b)]=a.a+b.b+2*a.b a and b are at right angles to each other so 2*a.b = 0

And that is it, the rest is obvious!

The proof of the equation D²=a²+b²+c² for a pyramidy thingy with three sides given as vectors a, b and c:

The area of the larger side is given by 0.5*the area of a parallelogram formed by the two sides.

The area of the larger sides = 0.5*|(b-c)x(a-c)|

Expanding the cross product I got

=0.5*|cxa+cxb+axb| The vectors are all perpendicular to each other, so |cxa|=|c|*|a|.

Then to find the size of this final vector I used Pythagoras' theorem (because each of the individual vectors are perpendicular to each other).

=0.5*√[(|c|*|a|)²+(|b|*|a|)²+(|b|*|b|)²]

The 0.5 on the outside can be turned into 0.25, and brought inside the square root. Then it can be brought into each individual square, to get:

0.5*√[(|c|*|a|)²+(|b|*|a|)²+(|b|*|b|)²]=√[(0.5*|c|*|a|)²+(0.5*|b|*|a|)²+(0.5*|b|*|b|)²]

And it is obvious that 0.5*|c|*|a| is the area of the triangle on the side of the pyramidy thing.

This post looks a bit messy, I think it makes sense though

Last edited: