Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
Line and Plane Points, Lines and planes relations in 3D space, examples
Example:   Through a line which is written as the intersection of two planes, P1 ::  x - 2y + 3z - 4 = 0 and P2 ::  3x + y - z + 1 = 0, lay a plane which passes through the point A(- 1, 2, 1).
Determine the equation of the plane.
Solution:  Write the equation of a sheaf,  P1 + lP2 = 0  or  (x - 2y + 3z - 4) + l(3x + y - z + 1) = 0.
Plug the point A into the equation to determine l such that the plane passes through the given point, so
A(-1, 2, 1 =>   P1 + lP2 = 0 gives (-1 - 2 · 2 + 3 · 1 - 4) + l[3 · (-1) + 2 - 1 + 1] = 0,   l = - 6.
Therefore,    (x - 2y + 3z - 4) - 6 · (3x + y - z + 1) = 0    or   - 17x - 8y + 9z - 10 = 0
is the plane passing through the given line and the point A.
 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.
The angle between line and plane
The angle j between a line and a plane is the angle subtended by the line and its orthogonal projection onto the plane.
Since the normal vector N = Ai + Bj + Ck of the plane forms with the direction vector s = ai + bj + ck of the line the angle y = 90° - j, the angle j between a line and a plane we calculate indirectly, that is    If a line is perpendicular to a plane, its projection onto the plane is a point and therefore its direction vector and the normal vector of the plane are collinear, i.e.,
 s = lN    =>   j = 90°.
 If a line is parallel to a plane, then s ^ N   =>   s · N = 0,   j = 0°.
 Example:   Given is a line and a plane x - 3y + 2z - 8 = 0, find the angle between
the line and the plane.
Solution:  From the equations of the line and the plane,  s = 3i - j - 2k  and  N = i - 3j + 2k  so that    Coordinate geometry contents 