Triangle area
This topic is somewhat old I'm sure:
Computing the area of a triangle in planar space (2 dimensional) is simple:
for a triangle (x1,y1) - (x2, y2) - (x3, y3)
V = 1/2((x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1))
//It is half the perp dot product of two edges
The best I've come up with for 3 dimensional is this:
//First get vectors for two edges of the triangle
U.x=x2-x1: U.y=y2-y1: U.z=z2-z1
V.x=x3-x1: V.y=y3-y1: V.z=z3-z1
D.x = U.y*V.z-V.y*U.z: D.y = V.x*U.z-U.x*V.z: D.z = U.x*V.y-V.x*U.y
Area = sqrt(D.x^2 + D.y^2 + D.z^2)/2;
in notation A=|UxV|/2
Essentialy this is half the magnitude of the cross product of the two vectors.
So my wonder is, why should we need to use a square root operation to obtain the area of a triangle in 3D.
If you use the 3D form for a triangle in the XY plane, you will see that the result is:
Area = sqrt( Dz2 )/2 = Dz/2,
which is the same as the 2D form.
Area = sqrt( Dz2 )/2 = Dz/2,
which is the same as the 2D form.
I don't understand... the square root is of the sum of three seperate monomials... once there is adddition in the radicand square roots cannot be factored out. Is that what you meant?
JohnBolton was assuming your original triangle was restricted to the XY plane, in which case the cross product vector would only have a non-zero Z component. Thus you can eliminate the square root because X and Y are automatically 0.
For the general case 3D, you indeed need to use a square root to find the length of the cross product vector, and divide it by 2.
For the general case 3D, you indeed need to use a square root to find the length of the cross product vector, and divide it by 2.
I interpreted your question as "Why is a square root needed in 3D but not in 2D?" I was trying to show that the 2D case does use square root, but since it reduces to the square root of a squared value, the two cancel each other out.
You can change a 3D problem into a 2D problem if you fix one of the axes to a constant value. In the XY plane (z = 0),
You can change a 3D problem into a 2D problem if you fix one of the axes to a constant value. In the XY plane (z = 0),
D.x = U.y*V.z-V.y*U.z = 0 D.y = V.x*U.z-U.x*V.z = 0 D.z = U.x*V.y-V.x*U.y Area = sqrt(D.x^2 + D.y^2 + D.z^2)/2 = sqrt(0 + 0 + D.z^2)/2 = sqrt(D.z^2)/2 = D.z/2
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