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Analytic Geometry (Coordinate Geometry) in Three-dimensional Space |
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Line
and 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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The angle between line and 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 |
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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Example:
Through the line |
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(the
same as in the previous example) lay a plane
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orthogonal to the
plane
4x
-
2y + z -
4 = 0 and determine its equation.
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Solution: Write given line as the intersection of two planes (like in
the above example) to form the equation
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of a
sheaf,
5x
+ (-
3
+ 4l) ·
y -
5lz +
(3l
-
1) = 0
and Ns
= 5i + (-
3
+ 4l)
j -
5lk
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that is the plane parameter l
of which should be determined so the plane to be perpendicular to the given plane.
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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
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Ns
^ N
=>
Ns
· N = 0 that
is, 5 · 4 + (-
3
+ 4l) · (-2)
+ (-5l) · 1 = 0
=> l =
2.
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Then, by plugging l
= 2 into 5x
+ (-
3
+ 4l) ·
y -
5lz +
(3l
-
1) = 0
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obtained is 5x
+ 5y -
10z + 5 = 0
or x
+ y -
2z + 1 = 0
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the equation of the plane which passes through the given line perpendicular to the given plane.
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The angle between line and plane
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The angle j
between a line and a plane is the angle subtended by the line and its orthogonal projection onto
the plane.
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Since the normal vector
N =
Ai + Bj + Ck
of the plane form 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
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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.,
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If
a line is parallel to a plane, then
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s
^ N
=>
s
· N = 0,
j =
0°.
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Example:
Given is a line |
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and
a plane x
-
3y + 2z -
8 = 0, find the angle between
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the line
and the plane.
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Solution: From the equations of the line and the plane,
s = 3i
-
j -
2k and N =
i -
3j + 2k
so that
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Pre-calculus contents
J |
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