Analytic Geometry (Coordinate Geometry) in Three-dimensional Space
Line and Plane

Intersection point of a line and a plane
How determine two planes of which a given line is their intersection line
 Through a given line lay two planes each of which is orthogonal to one of the
coordinate planes.
That way, given line will be determined by any of the following pairs of equations,
as the intersection line of the corresponding planes (each of which is perpendicular to one of the three coordinate planes).
 Example:   Given is a line find two planes of which given line is the intersection line.
Solution:  By eliminating the part of equation which contains x-coordinates determined is a plane which
passes through the given line orthogonal to the
yz coordinate plane that is
Therefore,   P1 ::  y - 18z + 4 = 0 with  N1 = j  - 18k   and  P2 ::  x - 11z + 3 = 0 with  N2 = i - 11k.
The obtained result is correct if the vector product of the corresponding normal vectors, N1 and N2 of the planes equals the direction vector s of the given line, so
By calculating the coordinates of the intersection point, of the line of intersection represented by obtained two planes, and the coordinate planes, by plugging successively, z = 0, y = 0 and x = 0 into their equations, we can prove that the result coincides with that in the first example of this section, because it is the same line.
Intersection point of a line and a plane
The point of intersection is a common point of a line and a plane. Therefore, coordinates of intersection must satisfy both equations, of the line and the plane.
The parametric equation of a line,
x = x0 + at,      y = y0 + bt     and      z = z0 + ct
represents coordinates of any point of the line expressed as the function of a variable parameter t which makes possible to determine any point of the line according to given condition.
Therefore, by plugging these variable coordinates of a point of the line into the given plane determine what value must have the parameter t, this point to be the common point of the line and the plane.
 Example:   Given is a line and a plane 4x - 13y + 23z - 45 = 0, find the
intersection point of the line and the plane.
Solution:  Transition from the symmetric to the parametric form of the line
by plugging these variable coordinates into the given plane we will find the value of the parameter t such that these coordinates represent common point of the line and the plane, thus
x = -t + 4y = 4t - 3 and  z = 4t - 2   =>    4x - 13y + 23z - 45 = 0
which gives,    4 · (-t + 4) - 13 · (4t - 3) + 23 · (4t - 2) - 45 = 0   Þ     t = 1.
Thus, for  t = 1  the point belongs to the line and the plane, so
x = -t + 4 = - 1+ 4 = 3,    y = 4t - 3 = 4 · 1 - 3 = 1 and  z = 4t - 2 = 4 · 1 - 2 = 2.
Therefore, the intersection point A(3, 1, 2) is the point which is at the same time on the line and the plane.
Points, Lines and planes relations in 3D space, examples
 Example:   Through the line lay a plane to be parallel to the y-axis, determine
the equation of the plane.
Solution:  Write the line as the intersection of two planes to be able to form the sheaf of planes, therefore
P1 and P2 are the planes of which the given line is intersection and which are perpendicular to the coordinate planes, xy and yz respectively.
Form the equation of a sheaf to determine the parameter l according to the given condition
P1 + lP2 = 0  or  (5x - 3y - 1) + l(4y - 5z + 3) = 0
then,   5x + (- 3 + 4l) ·  y - 5lz + (3l - 1) = 0   and   Ns = 5i + (- 3 + 4l) j - 5lk
is the equation of planes that all intersect at the given line and the normal vectors of which are perpendicular to the direction vector s of the line.
As the plane to be found should be parallel with the y-axis, then the normal vector of the sheaf Ns must be perpendicular to the unit vector j of the y-axis that is, the scalar product of the normal vector of the plane and the unit vector  j must be zero, thus
Ns ^  j    =>     Ns ·  j = 0.
Since a scalar product of two vectors expressed by their coordinates is
a · b = axbx + ayby + azbz   and since   j = 0 · i + 1 ·  j + 0 · k
then,       Ns ·  j = 5 · 0 + (- 3 + 4l) · 1 + (-5l) · 0 = 0   =>   l = 3/4.
By plugging   l = 3/4  into  (5x - 3y - 1) + l(4y - 5z + 3) = 0  obtained is   4x - 3z + 1 = 0
the plane through the given line that is parallel to the y-axis.
 Example:   Through the line (the same as in the previous example) lay a plane
orthogonal to the plane 4x - 2y + z - 4 = 0 and determine its equation.
Solution:  Write given line as the intersection of two planes (like in the above example) to form the equation
of a sheaf,       5x + (- 3 + 4l) · y - 5lz + (3l - 1) = 0   and   Ns = 5i + (- 3 + 4l) j - 5lk
that is the plane parameter l of which should be determined so the plane to be perpendicular to the given plane.
In this case, the normal vector of the sheaf Ns and the normal vector of the given plane N = 4i  - 2 j + k, are orthogonal to each other, therefore
Ns ^ N    =>     Ns · N = 0  that is,     5 · 4 + (- 3 + 4l) · (-2) + (-5l) · 1 = 0    =>   l = 2.
Then, by plugging   l = 2  into  5x + (- 3 + 4l) · y - 5lz + (3l - 1) = 0
obtained is   5x + 5y - 10z + 5 = 0   or   x + y - 2z + 1 = 0
the equation of the plane which passes through the given line perpendicular to the given plane.
Pre-calculus contents J