Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
      Plane in a three-dimensional (3D) coordinate system
      Equations of a plane in a coordinate space
         The equation of a plane in a 3D coordinate system
      The distance between a point and a plane - plane given in Hessian normal form
         Comparison of the general form and the Hessian normal form of equations of a plane
      The distance of a point to a plane - plane given in general form
         Angle (dihedral angle) between two planes
Equations of a plane in a coordinate space
The equation of a plane in a 3D coordinate system
A plane in space is defined by three points (which don’t all lie on the same line) or by a point and a normal vector to the plane.
Then, the scalar product of the vector P1P = r - r1, drawn from the given point P1(x1, y1, z1) of the plane to any point P(x, y, z) of the plane, and the normal vector N = Ai + Bj + Ck, is zero, that is
   which is called the vector equation of a plane.
       Or,    
Therefore,
(x - x1) · A + (y - y1) · B + (z - z1) · C = 0
giving A · x + B · y + C · z + D = 0
  where,   D - (Ax1+ By1+ Cz1)  
  the general equation of a plane in 3D space.
If plane passes through the origin O of a coordinate system then its coordinates, x = 0, y = 0, and z = 0 plugged into the equation of the plane, give
A · 0 + B · 0 + C · 0 + D = 0   =>   D = 0.
Thus, the condition that plane passes through the origin is   D = 0.
 
The distance between a point and a plane, plane given in Hessian normal form
Distance from a point A0(x0, y0, z0) to a plane is taken to be positive if the given point is on the one side while the origin is on the other side regarding to the plane, as is in the right figure.
Through the point A lay a plane parallel to the given plane. The length of a normal segment from the origin to the plane through A, written in the normal form, is p + d.
As the point A lies in that plane its coordinates must satisfy the equation
x0 · cosa + y0 · cosb + z0 · cosg  = p + d.
Thus, the distance of the point to the given plane is
d = x0 · cosa + y0 · cosb + z0 · cosg - p
 
To express the distance d in terms of the coefficients of the equation of a plane given in the general or standard form, we should examine the relations of the two forms of equations.
Comparison of general form and the Hessian normal form of equations of a plane
Coefficients, A, B and C, beside coordinates x, y and z, of the general form of equation
Ax + By + Cz + D = 0
are the components of the normal vector N = Ai + Bj + Ck, while in the Hessian’s equation
x · cosa + y · cosb + z · cosg - p = 0,
on that place, there are components (the direction cosines) of the unit vector N° of the same vector.
Thus, to convert from general form to the Hessian normal form, divide the general form of equation by
  taking the sign of the square root opposite to the sign of D, where D is not 0
The distance of a point to a plane - plane given in general form
Let replace in the Hessian formula for the distance  d = x0 · cosa + y0 · cosb + z0 · cosg - p, the direction cosines and the length of the normal p, by coefficients of the general form, to obtain
   
the formula for the distance of a point to a plane given in the general form.
Angle (dihedral angle) between two planes
The angle subtended by normals drawn from the origin to the first and second plane, that is, the angle which form the normal vectors of the given planes is
   
the angle between two planes.
If planes are parallel normal vectors are collinear that is,
N1 = lN2    <=>   j = 0°,
and if planes are orthogonal, then their normal vectors are orthogonal too, so their scalar product is zero,
N1 · N2 = 0    <=>   j = 90°
Example:   Find the angle between planes, P1 ::  2x - y + 4z  - 3 = 0 and P2 ::  -x - 3y + 5z + 6 = 0.
Solution:    Thus,
N1 = 2i - j + 4k   and   N2 = -i - 3j + 5k, then
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