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Triangle area

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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.

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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?

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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.

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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),
    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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