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| Coordinate
Geometry (Analytic Geometry) in Three-dimensional Space |
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| Point,
Line and Plane
- orthogonal
projections, distances, perpendicularity of line and plane
|
Through a given point lay a line perpendicular to a given line,
example
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| Through a given point lay a line perpendicular to a given line
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| A line which will pass through a given point perpendicular to a given line will lie in a plane that is perpendicular to the given line and which passes through the given point. |
| The equation of that line is then
determined by two points, the given point and by its projection onto the given line or the intersection with the
plane. |
| Example: Through the point
A(-6,
-3, 4)
pass a line perpendicular to the line |
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| Solution:
Through the given point lay a plane perpendicular to the given line,
therefore |
| N =
s = 3i -
4j -
3k. |
|
Plug the point A
into the plane,
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| A(-6,
-3, 4)
=>
3x
-
4y -
3z + D = 0,
D =
18. |
| So, the equation of the plane |
| P
::
3x -
4y -
3z + 18 = 0. |
| The intersection
AŽ, as the common point of the given line
and the plane P, can be determined by plugging its
parametrically expressed coordinates
into the equation of the plane, that is |
|
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|
 |
| plug
these variable coordinates of a point that moves along the line AAŽ
into the plane P |
| 3 · (3t + 2) - 4 · (-4t -
1) -
3 · (-3t -
2) + 18 = 0,
t = -1. |
| Thus,
determined is parameter t
such that the point AŽ
lies onto the lines l
and l1
and the plane P,
so that |
| x =
3t + 2 = 3 · (-1)
+ 2 = -1, y =
-4t -
1 = -4
· (-1)
-
1 = 3 and z =
-3t
-
2 = -3(-1)
-
2 = 1 |
| the
coordinates of the intersection AŽ(-1,
3, 1). |
| To
write the equation of the line l1
through the points, A(-6,
-3, 4)
and AŽ(-1,
3, 1), that is perpendicular to
the
given line l,
we should calculate its direction vector and include one of its points. |
| The
direction vector s1
we get as the difference between positions vectors of the points, A
and AŽ, |
| s1
= rA -
rA′ = (-6i
-
3 j
+ 4k) -
(-i
+ 3 j
+ k) = -5i
-
6j
+ 3k |
| then,
we plug any of its points, A
or AŽ
to form the equation of the line l1.
So, let plug the point A |
 |
| Let
check the result by calculating the scalar product of the vectors,
s
and s1
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| s · s1 =
(3i -
4j -
3k) · (-5i
-
6j + 3k) = 3 · (-5)
+ (-4) · (-6)
+ (-3) · 3
= 0, since s
^ s1. |
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| Coordinate
geometry contents |
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