Anybody for Some C/C++ Math?

[AustinJ]

So simply put, I have three points in 3d space and I know all the x, y, and z's. Assuming that they are all on the same plane how can I find another point that would also be on this plane if I know just the x and the z part of the point?

[vm'N]

You could in a sense connect these 3 points to a triangle, then calculate the triangle center (which should be on the same plane).

[ItsDenn]

have you had any vector math in school? If you did you can pretty much set up (I am using some translate site now because idk how its called in english sorry) a parametric description of the plane. And because you already know the x and z, it shouldnt be too hard to calculate the y. I recommend writing it out on paper first and then translate it into some code. I could provide pictures of my calculation if you want but I have to go to school first now so it would take a few hours before I can give them to you.

[fizzlesticks]

If you know the 3 points A, B and C the equation for a plane can be written as:

 

P(u,v) = A + (B - A)u + (C - A)v

 

which can be broken down into:

 

x = Ax + (Bx - Ax)u + (Cx - Ax)v

y = Ay + (By - Ay)u + (Cy - Ay)v

z = Az + (Bz - Az)u + (Cz - Az)v

 

in those equations, plug in everything you know then use the equations for x and z (system of equations) to calculate the values for u and v, plug them into the equation for y and solve for the answer.

[AustinJ]

 

If you know the 3 points A, B and C the equation for a plane can be written as:

 

P(u,v) = A + (B - A)u + (C - A)v

 

which can be broken down into:

 

x = Ax + (Bx - Ax)u + (Cx - Ax)v

y = Ay + (By - Ay)u + (Cy - Ay)v

z = Az + (Bz - Az)u + (Cz - Az)v

 

in those equations, plug in everything you know then use the equations for x and z (system of equations) to calculate the values for u and v, plug them into the equation for y and solve for the answer.

 

 

The problem with substition in programming is that there is so many variables that you can't simplify at all so it quite the mess. So if I do take the system of equations and say that the variables that I am solving for are part of the point D. I can solve Dx for 'u' and then plug 'u' into the equation to get Dz and I get something like this: "Dz = Az+((Bz-Az)*(-((Ax+((Cx-Ax)*v)-dX)/(Bx-Ax))))+((Cz-Az)*v)" which it quite the mess. that would need to be solved for 'v' yet and when I did get 'v'. Then I'd have to plug that back into the other eqation to get 'u' which would also be quite the mess.

[fizzlesticks]

The problem with substition in programming is that there is so many variables that you can't simplify at all so it quite the mess. So if I do take the system of equations and say that the variables that I am solving for are part of the point D. I can solve Dx for 'u' and then plug 'u' into the equation to get Dz and I get something like this: "Dz = Az+((Bz-Az)*(-((Ax+((Cx-Ax)*v)-dX)/(Bx-Ax))))+((Cz-Az)*v)" which it quite the mess. that would need to be solved for 'v' yet and when I did get 'v'. Then I'd have to plug that back into the other eqation to get 'u' which would also be quite the mess.

Instead of substituting you can use linear algebra to solve for u and v after transforming the equations so matrices

  

|(Bx - Ax), (Cx - Ax) |    |  u  |      | x - Ax |

|(Bz - Az), (Cz - Az) | x |  v   |  = | z - Az |

 

Doing it that way you could at least find some library to do the work for you.

[ItsDenn]

I made a basic calculation on my phone as an example I hope it is clear enough:

 

http://www.mediafire.com/view/81xovlo77x9ecr5/Screenshot_2015-02-06-17-37-09.png

 

with that equation you can calculate the "remaining" x if you already have y and z.  

[Satlen]

Wouldn't you want to solve this using matrices? If I read OP correctly you have [ x1 y1 z1 ] in row 1 and you have [ x2 y2      ] in row 2

 

This is the first thing that came to my head.