Coordinate Geometry (Analytic Geometry) in Three-dimensional Space
     Line and Plane
      How determine two planes of which, a given line is their intersection line
      Intersection point of a line and a plane
      Sheaf or pencil of planes
         Points, Lines and planes relations in 3D space, examples
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.
Sheaf or pencil of planes
A sheaf of planes is a family of planes having a common line of intersection. If given are two planes
P1 ::  A1x + B1 y + C1 z + D1 = 0  and  P2 ::  A2 x + B2 y + C2 z + D2 = 0
of which we form a linear system,
        P1 + lP2 = 0   or    (A1x + B1 y + C1 z + D1) + l(A2 x + B2 y + C2 z + D2) = 0
then this expression, for every value of the parameter l represents one plane which passes through the common line of intersection.
Thus, by changing the value of the parameter l, each time determined is another plane of the sheaf of which, the axis of the sheaf is the line of intersection of the given planes, P1 and P2.
For example, to find equation of a plane of a sheaf which passes through a given point A(x1, y1, z1), plug the coordinates of the point into the above equation of the sheaf, to determine the parameter l so that the plane contains the point A.
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.
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