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### 3D Projection Onto a 2D Plane

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Date: 7/9/96 at 15:22:57
From: Anonymous
Subject: 3D Projection Onto a 2D Plane

I'm dealing with a theoretical geological problem where I consider a
roughly planar orebody as being 2D.  I have drill-hole values and 3D
coordinates for a number of points.  Fitting a plane to these points
is simple.  What I want to do is project these points onto the plane,
estimate values in 2D and then project the estimates back onto the
original orebody.

Many thanks.
```

```
Date: 7/10/96 at 13:34:48
From: Doctor Jerry
Subject: Re: 3D Projection Onto a 2D Plane

I'm not sure I understand what you want.  Here's what I understand.

1. You have several points (x1,y1,z1), (x2,y2,z2), . . . ,(xn,yn,zn).

2. You have fitted a plane, say z = Ax+By+C, to these points.

3. You want to "project these points onto the plane...".  I take this
to mean that you want to find where a vertical line through, say,
(xi,yi,zi), passes through the plane.  If this is what you want, then
all you have to do is substitute xi and yi into the equation of the
plane.  The projection of (xi,yi,zi) onto the plane would be
(xi,yi,Axi+Byi+C).

4. I don't know what you mean by estimate values in 2D and then
project the estimates back onto the original orebody.

I'd appreciate a little help with what you want.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 7/14/96 at 22:43:28
From: Anonymous
Subject: Re: 3D Projection Onto a 2D Plane

Thank you for your reply.  I really haven't been clear at all have I?
The problem is as follows.

I am attempting to deal with an almost flat 3D ore-body in 2D. Given
a set of sample values and thicknesses throughout the body I want to
fit the best plane to the points. We assume that the 3D locations are
placed at the centre of lines drawn perpendicular to the plane.
Fitting the plane z = Ax+By+C to these points I can do.  In order to
reduce the problem to a 2D one I intend to work with accumulations
(grade * thickness). For each point (x,y,z) I need a transformation
to the fitted plane so that the new coordinates would be (x'=f(x,y,z),
y'=g(x,y,z) z'=0). Each point should be projected onto the plane
along a line passing through the point and perpendicular to the fitted
plane.

Once I have (x',y') I can use 2D geostatistics to evaluate the ore-
body.  Through the process of semi-variogram modeling and kriging I
should be able mark off the plane into rectangles and estimate values
and statistical variances for each of them.  Once I have done this I
will need to transform the centres of each rectangle back into the
original 3D coordinates.

The reason for all this is that geostatistics is much easier in 2D.
3D geostatistics tends to take the form of evaluating multiple
horizontal slices, somewhat messy and each slice is  evaluated
independently from the next slice.  Since this is almost a planar
situation 2D evaluation would be the best option.

```

```
Date: 7/29/96 at 13:32:0
From: Doctor Jerry
Subject: Re: 3D Projection Onto a 2D Plane

Perhaps I'm beginning to understand what you want.  Let me state my
understanding and a reply.  If you haven't given up on me and this
still isn't what you want, please feel free to try again.

It appears to me that the (x',y') values are coordinates of the
projected points, but relative to a system of coordinates in the
plane.  This requires that we somehow choose a set of coordinate axes
in the  fitted plane.

In my earlier comments, I stated how to obtain the coordinates (x,y,z)
of the projection of a sample point into the fitted plane.  With this
as a given, so that z = Ax+By+C, we need to obtain (x',y'), which
gives the position of this point relative  to a set of axes in the
plane.

Here's an example.  Suppose the plane is z=2x-3y+5 and we are
considering the point (4,7,-8) in this plane.

A normal vector to this plane is (-2,3,1).  So, by inspection, a
vector perpendicular to the plane is (3,2,0).  We may cross (-2,3,1)
and (3,2,0) to get the vector (-2,3,-13).  Let e1=(1/sqrt(13))(3,2,0)
and e2=(1/sqrt(182))(-2,3,-13).  These are orthogonal unit vectors in
the fitted plane.  Now choose a fixed point in the plane as origin,
say, (0,0,5).  To calculate x' and y' for (4,7,-8), we calculate the
vector from (4,7,-8) to (0,0,5), which is (4,7,-8)-(0,0,5) =
(4,7,-13).  This vector lies in the plane and, hence, can be written
as a linear combination of e1 and e2, which also lie in this plane.
So, we solve

(4,7,-13)=x'*e1+y'e2,

obtaining x'=2 sqrt(13) and y'=sqrt(182).  These are the 2D
coordinates of the point (4,7,-8).

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry

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