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There is a plane defined by a normal and an origin. For simplicity's sake, the origin is $(0,0,0)$.

And then, there are two coordinates ($x$ and $z$) of a point on this plane.

How can I find the third coordinate ($y$) of the point ?

Notes:

  • The plane is never vertical, so the point always exists.
  • This is exactly like finding the interection between the plane and an infinite vectical vector passing trough $(x,0,z)$
  • In this space, the vertical (up) vector is $(0,1,0)$
  • The computation needs to be cheap/fast.
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  • $\begingroup$ Compute the equation of the plane, for eg. like $ax +by+ cz + d = 0$; here $a, b, c, d$ are known. Substitute $x$ and $z$ to find $y$ $\endgroup$
    – ab123
    Aug 31, 2018 at 8:49

1 Answer 1

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Let $ax+by+cz=0$ be the equation of your plane with $\vec n=(a,b,c)$ being the normal vector. Clearly, the plane passes through the origin. Say you have a point $P(x_0,y_0,z_0)$ then
$$y_0=-(ax_0+cz_0)/b.$$

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