Question

I would like to write a function that takes a 3d triangle (as 3 points (vector3ds)) and returns a 2d triangle (as 3 points (vector2ds)):

When given a 3d triangle, it should return the 2 dimensional coordinates of its points as they lie on its plane. (By ‘its plane’ I mean the plane that all three points lie on).

I can think a long winded way todo this:

• rotate the triangle until its normal is equal to +z (0,0,1), then construct a triangle from the (x, y) coords of each point.

I cant help but think there must be an easier way to achieve the same thing.

If posting code examples please try not to use Greek alphabet. Some pseudo code in a C/java style language would be ideal.

Answer 1

From your comments I infer that you can choose the coordinate system of the plane in an arbitrary way, as long as the Euclidean metric of this coordinate system is the same as the metric induced by the Euclidean metric of your three-dimensional coordinate system. (That is, Euclidean distances will stay the same.)

One possible solution:

x0' = 0
y0' = 0

x1' = sqrt((x1 - x0)^2 + (y1 - y0)^2 + (z1 - z0)^2)
y1' = 0

x2' = ((x1 - x0) * (x2 - x0) + 
       (y1 - y0) * (y2 - y0) + 
       (z1 - z0) * (z2 - z0)) / x1'
y2' = sqrt((x2 - x0)^2 + (y2 - y0)^2 + (z2 - z0)^2 - x2'^2)