# Thread: Dip angle and direction of a plane described by 3 points

1. ## Dip angle and direction of a plane described by 3 points

I'm trying to make a model in excel which calculates the dip angle
and direction (azimuth or bearing) of a plane descriped by 3 points in a coordinate
system.
The planned use is a quick-look geomodel where I would like to
calculate formation dip based on the observation of a specific
geological feature in three separate wells. The points where the
well intersects the formation in each well is described by a UTM
latitude/longitude and depth below a fixed datum. Point 1 (x1, y1,
z1), point 2 (x2, y2, z2), point 3 (x3,y3, z3).
By definition, the dip angle is perpendicular to the strike (a
horizontal line through the plane) 0 degrees meaning a horizontal
plane/90 degrees meaning a vertical plane, and the dip azimuth is
the compass direction/bearing of the dip angle.

2. Originally Posted by prevaricator
I'm trying to make a model in excel which calculates the dip angle
and direction (azimuth or bearing) of a plane descriped by 3 points in a coordinate
system.
The planned use is a quick-look geomodel where I would like to
calculate formation dip based on the observation of a specific
geological feature in three separate wells. The points where the
well intersects the formation in each well is described by a UTM
latitude/longitude and depth below a fixed datum. Point 1 (x1, y1,
z1), point 2 (x2, y2, z2), point 3 (x3,y3, z3).
By definition, the dip angle is perpendicular to the strike (a
horizontal line through the plane) 0 degrees meaning a horizontal
plane/90 degrees meaning a vertical plane, and the dip azimuth is
the compass direction/bearing of the dip angle.
Given the three points, you can construct two vectors $\displaystyle <x_2-x_1,y_2-y_1,z_2-z_1>$ and $\displaystyle <x_3-x_1,y_3-y_1,z_3-z_1>$. The cross product of those two vectors is normal to the plane. I'm not going to do the calculation myself, but call that $\displaystyle <v_x,v_y,v_z>$. The "dip angle", the angle between that normal vector and the horizontal plane which is the angle between $\displaystyle <v_x,v_y,v_z>$ and $\displaystyle <v_x,v_y,0>$. The angle, $\displaystyle \theta$, is given by $\displaystyle cos(\theta)= \frac{u\cdot v}{|u||v|}$.