Ok so your diagram didn't show what you wanted :D
In that case you have no choice but to do a coordinate transformation as _swx_ suggests. Find the basis vectors and origin point for the plane that the point cloud is based on; this will give you a 4×4 transformation matrix
m of the type:
( ....x.... 0 )( ....y.... 0 )( ....n.... 0 )( ....o.... 1 )
... where
o is the origin point of the plane,
n is the normal and
x and
y are two perpendicular basis vectors within the plane. (If your plane is only a little deviant from the target space, I recommend you make these vectors
x=n×Y and
y=X×n respectively, which will align them with the target's X and Y axes as closely as possible.)
Similarly, find the basis vectors for the plane onto which you which to project the point cloud, which make up a matrix
M:
( ....X.... 0 )( ....Y.... 0 )( ....N.... 0 )( ....O.... 1 )
It is likely that your target axes
X,
Y and
N will be the world axes [1,0,0], [0,1,0] and [0,0,1], making this a simple transformation matrix, but they do not have to be.
Now, to project a point
p out of the coordinate space defined by
m, and into the coordinate space defined by
M, simply multiply by the inverse of
m and then by
M:
P = pm-1
M