# Thread: Help Calculating 3d Coordinates on a Plane Given a Line

1. ## Help Calculating 3d Coordinates on a Plane Given a Line

Hi Everyone,

I have a problem that I'm trying to solve. I'm using a program that simulates fractures (as planes with finite dimensions) and I need to plot these fractures. The program outputs the following:

1. Fracture Centroid: x,y,z coordinates of the center of the plane
2. Line normal to the Centroid: given as a dip direction and a dip. Two angular measurements. Dip direction is a 0 - 359 number and dip is a 0 to 90 number.
3. Fracture height and length. These are numbers in meters used to define how far the plane extends in each direction.

I need to solve for the corner points of these planes so I can plot them in 3d.

For example, my first fracture that I'm trying to solve for looks like this:

x = 1.6
y = 8.9
z = 0.3

dip direction = 88.9
dip = 86.5

length = 40.1
height = 32.5

Given this information, I need to calculate the 3d coordinates of this plane. I'm really struggling with this. By the way, I'm using Matlab 2013 so I have all those built in functions available.

Thanks!

2. ## Re: Help Calculating 3d Coordinates on a Plane Given a Line

Well, you don't mean "plane". A plane has no "corners" but extends infinitely. You mean some rectangle in a plane. In order that your explanation of "dip direction" and "dip" make sense, I think we have to assume that the "dip direction" is measure, parallel to the xy-plane, clockwise from the positive x direction. If that is the case then, with "dip direction" $\displaystyle \theta$, a normal vector to this plane will have x and y components $\displaystyle a cos(\theta)$ and $\displaystyle a sin(\theta)$, respectively. The "dip" then is below the xy-plane so if the "dip" is angle $\displaystyle \phi$ then the z component of the normal vector will be $\displaystyle -a sin(\phi)$. That is, a normal vector is of the form $\displaystyle (a cos(\theta), a sin(\theta), -a sin(\phi))$ and so the equation of the plane is $\displaystyle a cos(\theta)(x- x_0)+ a sin(\theta)(y- y_0)- a sin(\phi)(z- z_0)= 0$.

You say that "fracture height and length", measure "how far the plane extends in each direction" but you do not say what those directions are. Is one the "dip direction" and the other perpendicular to it?

3. ## Re: Help Calculating 3d Coordinates on a Plane Given a Line

HallsofIvy,

You're right, I do mean a rectangle on a plane.

The dip and dip direction actually describe a pole that is normal to the plane. I'm not sure why the program provides this information. The input measurements (as taken by a geologist) would be the actual strike and dip of the vein (or plane). The program requires these be converted into a pole to that plane. It's not a big deal really; just a minor annoyance.

Also, if it's easier, I can set the length and height to be the same. These are actually unknown parameters since we cannot make these measurements from the drill core samples that we collect.

The length would be the distance in the "strike direction" which would be perpendicular to the dip direction. Height would be perpendicular to the length.

4. ## Re: Help Calculating 3d Coordinates on a Plane Given a Line

I've been working on this a bit more. I've solved for the normal vector using the following:

nvec = < a b c >

a = sin(dip direction) * cos(dip)
b = cos(dip direction) * cos(dip)
c = sin(dip)

Using the normal vector and the centroid point (x,y,z), I solved for the equation of the plane using:

ax + by + cz + d = 0

Now, given this equation, how can I solve for the corner points that I need?