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Analytic Geometry (Coordinate Geometry) in Three-dimensional Space |
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Line
and Plane
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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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| 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, |
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P1
+ lP2 = 0 |
or |
(A1x + B1
y
+ C1 z + D1) + l(A2
x + B2 y + C2 z
+ D2) = 0 |
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| 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. |
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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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| Example:
Through the line |
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lay
a plane to be parallel to the
y-axis, determine
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| 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
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| 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.
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| 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 |
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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.
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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
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Ns
^
j
=>
Ns
·
j = 0.
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Since a scalar product of two vectors expressed by their coordinates is
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a · b = axbx +
ayby + azbz
and since j
= 0 · i + 1 ·
j + 0 · k
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then,
Ns
·
j = 5 · 0 + (-
3
+ 4l) · 1 + (-5l)
· 0 = 0
=>
l
= 3/4.
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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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| Pre-calculus contents
J |
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