Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
Plane in a three-dimensional (3D) coordinate system

The intercept form of the equation of a plane

The distance between a point and a plane - plane given in Hessian normal form
Equations of a plane in a coordinate space
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.

A plane in a 3D coordinate system, examples
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:   Given are points, A(-1, 1, 1) and B(3, -2, 6), find the equation of a plane which is normal to the vector AB and passes through the point A.
Solution:   According to the given conditions the vector AB = N so,
N = rB - rA = (3i - 2j + 6k) - (-i  + j + k) = 4i - 3j + 5k
As the plane should pass through the point A
plug  A(-1, 1, 1)  into    4x - 3y + 5z  + D = 0    =>   4 · (-1) - 3 · 1 + 5 · 1 + D = 0,    D = 2
Therefore, the equation of the plane   P ::  4x - 3y + 5z  + 2 = 0.

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

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.
Pre-calculus contents J