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

Line and Plane
The line of intersection of two planes
Two planes are either parallel or they intersect in a line. If planes are parallel, their coefficients of coordinates x, y and z are proportional, that is
 and then, the vector product of their normal vectors is zero N1 ´ N2 = 0. When two planes intersect, the vector product of their normal vectors equals the direction vector s of their line of intersection, N1 ´ N2 = s. To be able to write the equation of a line of intersection of two planes we still need any point of that line.
We can use the intersection point of the line of intersection of two planes with any of coordinate planes (xy, xz or yz plane) as that point.
Example:   Given are planes, P1 ::  -3x + 2y - 3z - 1 = 0 and P2 ::  2x - y - 4z + 2 = 0, find the line of intersection of the two planes.
Solution:  Since the equation of the plane and its normal,
P º Ax + By + Cz + D = 0 and  N = Ai + Bj + Ck  then,  N1 = -3i + 2j  - 3k  and  N2 = 2i - j - 4k,
To write the equation of the line of intersection, i.e.,
we still need the coordinates of any of its point P(x0, y0, z0).
Let this point be the intersection of the intersection line and the xy coordinate plane.
Then, coordinates of the point of intersection (x, y, 0) must satisfy equations of the given planes.
Therefore, by plugging  z = 0  into P1 and P2 we get,
so, the line of intersection is
Using the same method we can check validity of obtained equation by calculating coordinates of another intersection point of the intersection line and the yz coordinate plane, and plug them into mentioned equation.
So, coordinates of the intersection (0, y, z) must satisfy equations of both given planes
Now, plug the coordinates of the intersection point Pyz into the equation of the line of intersection
so, obtained equation is valid.
Projection of a line onto coordinate planes
In the equation of a line,
represents the orthogonal projection of the given line onto the xy coordinate plane.
At the same time, it is spatial equation of the plane, which passes through the given line, and is perpendicular to the xy coordinate plane. Similarly,
represent the equations of the projections of the given line onto the xz and the yz coordinate plane respectively.
While, in relation to space, they are the equations of planes passing through the given line orthogonal to the xz and the yz coordinate planes respectively.
 Example:   Find the (orthogonal) projections of the line onto coordinate planes and
find the coordinates of the intersection points of the line with the coordinate planes.
 Solution:  Equation is projection of the given line onto the xy
coordinate plane, i.e., it is the equation of the plane passing through the given line orthogonal to the xy plane, or it is the trace of that plane in the xy plane.
While the equations,
are the projections of the given line onto the xz plane and yz plane respectively, i.e., they are the equations of planes which are orthogonal onto these coordinate planes passing through the given line.
The coordinates of the intersection point of the given line and the xy coordinate plane we calculate by plugging z = 0 into the equation of the given line that is,
Similarly, the intersection point of the given line and the xz coordinate plane we calculate by plugging y = 0,
The intersection of the given line and the yz coordinate plane we calculate by plugging x = 0 into the equation of the given line that is
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.
Pre-calculus contents J