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Coordinate geometry (or Analytic geometry) in three-dimensional space
 :: 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 Hessian normal form of the equation of a plane
Position of a plane in space can also be defined by the length of the normal, drawn from the origin to the plane and by angles, a, b and g, that the normal forms with the coordinate axes.
If  N°  is the unit vector of the normal then its components are,  cosa, cosb and cosg.
Then the projection of the position vector r, of any point P(x, y, z) of the plane, onto the normal has the length p.
Since the projection is determined by expression
that is,  r · N° = p or written in the coordinates
x · cosa + y · cosb + z · cosg - p = 0
represents the equation of a plane in the Hessian normal form.

The intercept form of the equation of a plane

If, l, m and n are the intercepts of x, y and z axes and a plane respectively, then projections of these segments in direction of the normal drawn from the origin to the plane are all equal to the length of the normal, that is

By plugging these values of cosines into Hessian normal form of the equation of plane, obtained is

the intercept form of the equation of plane.

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
Comparison of general form and the Hessian normal form of equations of a plane
Coefficients, A, B and C of coordinates x, y and z, of the general form of equation of a plane
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 equation of the plane through three points, A(-1, 1, 4), B(2, 4, 3) and C(-3, 4, 1).
Solution:  Equation of a plane
Ax + By + Cz + D = 0
is determined by the components (the direction cosines) of the normal vector N = Ai + Bj + Ck and the coordinates of any point of the plane.
The normal vector is perpendicular to the plane determined by given points, and as the vector
AB = rB - rA = (2i + 4j + 3k) - (-i  + j + 4k)
rB - rA = 3i + 3j - k, and the vector
  AC = rC - rA = -2i + 3j - 3k
then,

By plugging the coordinates, of any given point, into equation of the plane determines the value of the coefficient D.  Thus, plugging

A(-1, 1, 4) into  - 6x + 11y + 15z  + D = 0  gives  - 6 · (-1) + 11 · 1 + 15 · 1 + D = 0,   D = -77.

Therefore,  - 6x + 11y + 15z - 77 = 0  is the equation of the plane the given points lie on.

The coordinate of the remaining two points B and C must also satisfy the obtained equation, prove.

Example:  Segments that a plane cuts on the axes, x and y, are l = -1 and m = -2 respectively, find the standard or general equation of the plane if it passes through the point A(3, 4, 6).

Solution:  As the given point A(3, 4, 6) and the segments, l = -1 and m = -2 must satisfy the intercept form of the equation of plane, then

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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