Analytic Geometry (Coordinate Geometry) in Three-dimensional Space
Line and Plane
Intersection point of a line and a plane
Sheaf or pencil of planes
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.
 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.

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