Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
     Line and Plane
     Point, Line and Plane - orthogonal projections, distances, perpendicularity of line and plane
      Through a given point pass a line perpendicular to a given plane
      Given a line and a point, through the point lay a plane perpendicular to the line
Point, line and plane orthogonal projections, distances, perpendicularity of line and plane
Through a given point pass a line perpendicular to  a given plane
In this case, the normal vector N of a plane is collinear or coincide with the direction vector
s = ai + bj + ck  of a line,
that is,   s = N = Ai + Bj + Ck.
The point A(x0, y0, z0) plugged into rewritten equation of the line gives
 
Example:   Determine the equation of a line which passes through the point A(-3, 5, -1) perpendicular to the plane 2x - y + 4z - 3 = 0.
Solution:  By plugging the point A(-3, 5, -1) and the components of the normal vector
                N = s = 2i  - j + 4k   of the given plane into the above equation of the line obtained is 
the equation of the line perpendicular to the given plane that passes through the given point.
Given a line and a point, through the point lay a plane perpendicular to the line
The direction vector s of a line is now collinear or coincide with the normal vector N of a plane so that
N = s = ai + bj + ck.
Coordinates of the given point A(x0, y0, z0) is plugged into the rewritten equation of a plane 
P ::  a  x0 + b  y0 + c  z0 + D = 0
to determine the parameter D.
 
Example:   Given is a line  and a point A(-2, 1, 4), through the point lay a plane
perpendicular to the line.
Solution:  In the above equation of the line, the zero in the denominator denotes that the direction vector's component c = 0, it does not mean division by zero. Consider this as symbolic notation.
It means that the given line is parallel with the xy coordinate plane on the distance z = 3 that is, the coordinate z of each point of the line has the value 3.
Since  N = s   then,   N = -i  + 3j    or    N = Ai + Bj + Ck.
The coordinate of the point must satisfy the equation of the plane Ax + By + Cz + D = 0 that is,
A(-2, 1, 4)    =>    Ax + By + Cz + D = 0  gives   -1  (-2) + 3  1 + 0  4 + D = 0,   D = -5.
Therefore,   P ::  -2x + 3y - 5 = 0  is the plane through the given point perpendicular to the given line.
Coordinate geometry contents
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