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| Coordinate
Geometry (Analytic Geometry) in Three-dimensional Space |
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Line
and Plane
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How determine two planes of which, a given line is their
intersection line
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Intersection point of a line and a plane
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Sheaf or
pencil of planes
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Points, Lines and planes relations in 3D space, examples
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| How determine two planes of which, a given line is their
intersection line
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| Through a given line |
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lay two planes each of which is orthogonal to one of the |
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| coordinate planes. |
| That way, given line will be determined by any of
the following pairs of
equations, |
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| as the intersection line of the corresponding planes
(each of which is perpendicular to one of the three coordinate
planes). |
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| Example:
Given is a line |
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find two planes of which given line is the intersection line. |
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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 |
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| 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 |
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| 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. |
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| Intersection point of a line and a plane
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| 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. |
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| Example:
Given is a line |
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and a plane
4x -
13y + 23z -
45 = 0, find the
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| intersection
point of the line and the plane. |
| Solution: Transition from the symmetric to the parametric form
of the line
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| 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 +
4, y
= 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. |
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| Sheaf or
pencil of planes
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| 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. |
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| 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. |
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| Points, Lines and planes relations
in 3D space, examples
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| Example:
Through a line which is written as the intersection of two planes, P1
::
x -
2y
+ 3z -
4 = 0 and
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| 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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| Coordinate
geometry contents |
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