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 intercept form of the equation of a plane
      A plane in a 3D coordinate system, examples
         Equation of the plane through three points, example
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
 
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