Plane equation for 3D vectors.

Question

Hi, all.I want to find a 3D plane equation given 3 points.I have got the normal calculated after applying the cross product. But the equation of a plane is known to be the normal multiply by another vector which what i am taught to be as P.OP. I substitute my main reference point as OP and i want P to be in (x, y, z) form. So that i can get something like e.g, OP = (1, 2, 3) I want to get something like that: (x-1) (y-2) (z-3) May i know how? Below is my reference code.(Note: plane_point_1_x(), plane_point_1_y(), plane_point_1_z() are all functions asking for the user input of the respective points)

"""
I used Point P as my reference point so i will make use of it in this section
"""

vector_pop_x = int('x') - int(plane_point_1_x())
vector_pop_y = int('y') - int(plane_point_1_y())
vector_pop_z = int('z') - int(plane_point_1_z())

print vector_pop_x, vector_pop_y, vector_pop_z

All the above is what i did, but for some reason it did not work. I think the problem lies in the x, y , z part. Hope you guys can help. Thanks!

Answer 1

Plane implicit Eqn:

All points P = (x, y, z) satisfying

<n, QP> = 0

where

• n is the plane normal vector,
• Q is some point on the plane (any will do)
• QP is the vector from Q to P
• <a, b> is the scalar (dot) product operator.

(Remember that QP can be computed as P - Q)

Answer 2

One good way is:

| x1 y1 z2 1 |
| x2 y2 z2 1 |
| x3 y3 z3 1 | = 0
| x  y  z  1 |

Where the vertical pipes mean the determinant of the matrix, and (x1 y1 z1), (x2 y2 z2), and (x3 y3 z3) are your given points.