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Coordinate
Geometry (Analytic Geometry) in Three-dimensional Space |
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Line
and Plane
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Point,
Line and Plane
- orthogonal
projections, distances, perpendicularity of line and plane
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Through a given point pass a line perpendicular to a given plane
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Given a line and a point, through
the point lay a plane perpendicular to the line
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Point,
line and plane – orthogonal
projections, distances, perpendicularity of line and plane
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Through a given point pass a line perpendicular to
a given plane |
In this case, the normal vector
N
of a plane is
collinear or coincide
with the direction vector |
s =
ai + bj + ck
of a line, |
that is,
s =
N =
Ai + Bj + Ck. |
The point
A(x0,
y0,
z0)
plugged into rewritten
equation of the line gives |
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Example:
Determine the equation of a line which passes through
the point A(-3, 5,
-1) perpendicular to the
plane
2x -
y + 4z - 3 =
0.
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Solution: By
plugging the point A(-3, 5,
-1)
and the components of the normal vector
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N =
s
= 2i -
j + 4k
of the given plane into
the above equation of the line obtained is
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the equation of the line perpendicular to the given plane that
passes through the given point.
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Given a line and a point, through
the point lay a plane perpendicular to the line
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The direction vector
s
of a line is now collinear or coincide
with
the normal vector N of a plane
so that |
N =
s =
ai + bj + ck. |
Coordinates of the given point
A(x0,
y0,
z0)
is plugged into the
rewritten equation
of a plane |
P
::
a · x0 + b · y0 + c · z0 + D = 0 |
to determine the parameter
D. |
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Example:
Given is a line |
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and a point
A(-2, 1,
4), through the point lay a plane
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perpendicular to the line. |
Solution: In the above equation of the line, the zero in the denominator denotes that the direction
vector's
component c =
0, it does not mean division by zero. Consider this as symbolic
notation.
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It means that the given line is parallel with the
xy coordinate plane on the distance
z = 3 that is, the coordinate
z
of each point of the line has the value 3.
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Since N =
s
then, N =
-i
+ 3j
or
N =
Ai + Bj + Ck.
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The coordinate of the point must satisfy the equation of the plane
Ax + By + Cz +
D = 0
that is,
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A(-2, 1,
4) => Ax + By + Cz + D = 0
gives -1 · (-2)
+ 3 · 1
+ 0 · 4 +
D = 0, D =
-5.
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Therefore, P
:: -2x +
3y -
5 = 0 is the plane through the given point perpendicular to the given line.
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Coordinate
geometry contents |
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