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Title: Chapter 7: Vectors and the Geometry of Space


1
Chapter 7 Vectors and the Geometry of
Space Section 7.5 Lines and Planes in Space
Written by Karen Overman Instructor of
Mathematics Tidewater Community College, Virginia
Beach Campus Virginia Beach, VA With Assistance
from a VCCS LearningWare Grant
2
  • In this lesson you will learn
  • Lines in Space
  • Parametric equations for a line
  • Symmetric equations for a line
  • Relationships between lines in space
  • Planes in Space
  • Standard form and General form of a plane
  • Sketching planes using traces
  • The line of intersection of two planes
  • Distances in Space
  • The distance between a point and a plane
  • The distance between a point and a line

3
Lines in Space
Previously you have studied lines in a
two-dimensional coordinate system. These lines
were determined by a point and a direction or the
slope. In 3-dimensional space, a line will also
be determined by a point and a direction, except
in 3-dimensional space the direction will be
given by a parallel vector, called the direction
vector.
4
Lines in Space
To determine the equation of the line passing
through the point P and parallel to the
direction vector, , we will use our knowledge
that parallel vectors are scalar multiples. Thus,
the vector through P and any other point Q
on the line is a scalar multiple of the
direction vector, .
In other words,
5
Equations of Lines in Space
Equate the respective components and there are
three equations.
These equations are called the parametric
equations of the line.
If the components of the direction vector are all
nonzero, each equation can be solved for the
parameter t and then the three can be set equal.
These equations are called the symmetric
equations of the line.
6
Equations of Lines in Space
A line passing through the point P
and parallel to the vector, is
represented by the parametric equations A
nd if all three components of the direction
vector are nonzero, the line is also represented
by the symmetric equations
7
Example 1 Find the parametric and symmetric
equations of the line passing through the point
(2, 3, -4) and parallel to the vector, lt-1, 2, 5gt
.
Solution Simply use the parametric and symmetric
equations for any line given a point on the line
and the direction vector.
Parametric Equations
Symmetric Equations
8
Example 2 Find the parametric and symmetric
equations of the line passing through the points
(1, 2, -2) and (3, -2, 5).
Solution First you must find the direction
vector which is just finding the vector from one
point on the line to the other. Then simply use
the parametric and symmetric equations and either
point.
  • Notes
  • For a quick check, when t 0 the parametric
    equations give the point (1, 2, -2) and
    when t 1 the parametric equations give the
    point (3, -2, 5).
  • The equations describing the line are not unique.
    You may have used the other point or the vector
    going from the second point to the first point.

9
Relationships Between Lines
In a 2-dimensional coordinate system, there were
three possibilities when considering two lines
intersecting lines, parallel lines and the two
were actually the same line. In 3-dimensional
space, there is one more possibility. Two lines
may be skew, which means the do not intersect,
but are not parallel. For an example see the
picture and description below.
If the red line is down in the xy-plane and the
blue line is above the xy-plane, but parallel to
the xy-plane the two lines never intersect and
are not parallel.
10
Example 3 Determine if the lines are parallel or
identical.
Solution First look at the direction vectors
Since , the lines are parallel.
Now we must determine if they are identical. So
we need to determine if they pass through the
same points. So we need to determine if the two
sets of parametric equations produce the same
points for different values of t. Let t0 for
Line 1, the point produced is (3, 2, 4). Set the
x from Line 2 equal to the x-coordinate produced
by Line 1 and solve for t.
Now let t1 for Line 2 and the point (3, 2, -1)
is produced. Since the z-coordinates are not
equal, the lines are not identical.
11
Example 4 Determine if the lines intersect. If
so, find the point of intersection and the cosine
of the angle of intersection.
Solution Direction vectors Since , the
lines are not parallel. Thus they either
intersect or they are skew lines.
Keep in mind that the lines may have a point of
intersection or a common point, but not
necessarily for the same value of t. So equate
each coordinate, but replace the t in Line 2 with
an s.
System of 3 equations with 2 unknowns Solve the
first 2 and check with the 3rd equation.
12
Solution to Example 4 Continued Solving the
system, we get t 1 and s -1. Line 1 t 1
produces the point (5, -2, 3) Line 2 s -1
produces the point (5, -2, 3)
The lines intersect at this point.
Recall from lesson 7.3 on the dot product,
13
Solution to Example 4 Continued
Thus,
14
Planes in Space
In previous sections we have looked at planes in
space. For example, we looked at the xy-plane,
the yz-plane and the xz-plane when we first
introduced 3-dimensional space. Now we are going
to examine the equation for a plane. In the
figure below P, , is a point in the
highlighted plane and is
the vector normal to the highlighted plane.
n
For any point Q, in the plane, the vector from
P to Q , is also in the plane.
Q
P
15
Planes in Space
Since the vector from P to Q is in the plane,
are perpendicular and their dot product
must equal zero.
n
This last equation is the equation of the
highlighted plane. So the equation of any plane
can be found from a point in the plane and a
vector normal to the plane.
Q
P
16
Standard Equation of a Plane
The standard equation of a plane containing the
point and having normal
vector, is
Note The equation can be simplified by using
the distributive property and collecting like
terms. This results in the general form
17
Example 5 Given the normal vector, lt3, 1, -2gt to
the plane containing the point (2, 3, -1), write
the equation of the plane in both standard form
and general form.
Solution Standard Form
To obtain General Form, simplify.
18
Example 6 Given the points (1, 2, -1), (4, 0,3)
and (2, -1, 5) in a plane, find the equation of
the plane in general form.
Solution To write the equation of the plane we
need a point (we have three) and a vector normal
to the plane. So we need to find a vector normal
to the plane. First find two vectors in the
plane, then recall that their cross product will
be a vector normal to both those vectors and thus
normal to the plane.
Two vectors From (1, 2, -1) to (4, 0, 3) lt
4-1, 0-2, 31 gt lt3,-2,4gt From (1, 2,
-1) to (2, -1, 5) lt 2-1, -1-2, 51 gt lt1,-3,6gt
Their cross product
Equation of the plane
19
Sketching Planes in Space
If a plane intersects all three coordinate
planes (xy-plane, yz-plane and the xz-plane),
part of the plane can be sketched by finding the
intercepts and connecting them to form the
plane. For example, lets sketch the part of the
plane, x 3y 4z 12 0 that appears in the
first octant. The x-intercept (where the plane
intersects the x-axis) occurs when both y and z
equal 0, so the x-intercept is (12, 0, 0).
Similarly the y-intercept is (0, 4, 0) and the
z-intercept is (0, 0, 3). Plot the three points
on the coordinate system and then connect each
pair with a straight line in each coordinate
plane. Each of these lines is called a
trace. The sketch is shown on the next slide.
20
Sketch of the plane x 3y 4z 12 0 with
intercepts, (12, 0, 0), (0, 4, 0) and (0, 0, 3).
z
y
Now you can see the triangular part of the plane
that appears in the first octant.
x
21
Another way to graph the plane x 3y 4z 12
0 is by using the traces. The traces are the
lines of intersection the plane has with each of
the coordinate planes. The xy-trace is found by
letting z 0, x 3y 12 is a line the the
xy-plane. Graph this line.
z
y
x
22
Similarly, the yz-trace is 3y 4z 12, and the
xz-trace is x 4z 12. Graph each of these in
their respective coordinate planes.
z
y
x
23
Example 7 Sketch a graph of the plane 2x 4y
4z 12 0.
Solution The intercepts are (6, 0, 0), (0, -3,
0) and (0, 0, 3). Plot each of these and connect
each pair with a straight line.
24
Example 7 Sketch a graph of the plane 2x 4y
4z 12 0.
Solution The intercepts are (6, 0, 0), (0, -3,
0) and (0, 0, 3). Plot each of these and connect
each pair with a straight line.
z
y
Hopefully you can see the part of the plane we
have sketched appears on the negative side of the
y-axis.
x
25
More on Sketching Planes
Not all planes have x, y and z intercepts. Any
plane whose equation is missing one variable is
parallel to the axis of the missing variable. For
example, 2x 3y 6 0 is parallel to the
z-axis. The xy trace is 2x 3y 6, the yz trace
is y 2 and the xz trace is x 3.
Part of the plane is outlined in red.
Any plane whose equation is missing two variables
is parallel to the coordinate plane of the
missing variables. For example, 2x 6 0 or x
3 is parallel to the yz-plane.
The plane is outlined in blue and is at the x
value of 3.
26
Intersecting Planes
Any two planes that are not parallel or identical
will intersect in a line and to find the line,
solve the equations simultaneously.
For example in the figure above, the white plane
and the yellow plane intersect along the blue
line.
27
Example 8 Find the line of intersection for the
planes x 3y 4z 0 and x 3y 2z 0.
Solution To find the common intersection, solve
the equations simultaneously. Multiply the
first equation by 1 and add the two to eliminate
x.
Back substitute y into one of the first equations
and solve for x.
Finally if you let z t, the parametric
equations for the line are
28
Distance Between a Point and a Plane
Let P be a point in the plane and let Q be a
point not in the plane. We are interested in
finding the distance from the point Q to the
plane that contains the point P. We can find
the distance between the point, Q, and the plane
by projecting the vector from P to Q onto the
normal to the plane and then finding its
magnitude or length.
n, normal
Q
Projection of PQ onto the normal to the plane
P
Thus the distance from Q to the plane is the
length or the magnitude of the projection of the
vector PQ onto the normal.
29
Distance Between a Point and a Plane
If the distance from Q to the plane is the length
or the magnitude of the projection of the vector
PQ onto the normal, we can write that
mathematically
Now, recall from section 7.3,
So taking the magnitude of this vector, we get
30
Distance Between a Point and a Plane
The distance from a plane containing the point P
to a point Q not in the plane is
where n is a normal to the plane.
31
Example 9 Find the distance between the point Q
(3, 1, -5) to the plane 4x 2y z 8.
Solution We know the normal to the plane is lt4,
2, -1gt from the general form of a plane. We can
find a point in the plane simply by letting x and
y equal 0 and solving for z P (0, 0, -8) is a
point in the plane. Thus the vector, PQ lt3-0,
1-0, -5-(-8)gt lt3, 1, 3gt
Now that we have the vector PQ and the normal, we
simply use the formula for the distance between a
point and a plane.
32
Lets look at another way to write the distance
from a point to a plane. If the equation of the
plane is ax by cz d 0, then we know the
normal to the plane is the vector, lta, b, cgt .
Let P be a point in the plane, P
and Q be the point not in the plane, Q
. Then the vector,
So now the dot product of PQ and n becomes
Note that since P is a point on the plane it will
satisfy the equation of the plane, so
and the dot product can be rewritten
33
Thus the formula for the distance can be written
another way The Distance Between a Point and a
Plane The distance between a plane, ax by cz
d 0 and a point Q is
Now that you have two formulas for the distance
between a point and a plane, lets consider the
second case, the distance between a point and a
line.
34
Distance Between a Point and a Line
In the picture below, Q is a point not on the
line , P is a point on the line, u is a
direction vector for the line and is the
angle between u and PQ.
Q
D Distance from Q to the line
P
u
Obviously,
35
We know from Section 7.4 on cross products that
Thus,
So if, then from above, .
36
Distance Between a Point and a Line
The distance, D, between a line and a point Q
not on the line is given by
where u is the direction vector of the line and P
is a point on the line.
37
Example 10 Find the distance between the point Q
(1, 3, -2) and the line given by the parametric
equations
Solution From the parametric equations we know
the direction vector, u is lt 1, -1, 2 gt and if
we let t 0, a point P on the line is P (2, -1,
3). Thus PQ lt 2-1, -1-3, 3-(-2) gt lt 1, -4,
5 gt
Find the cross product
Using the distance formula
38
You have three sets of practice problems for this
lesson in Blackboard under Chapter 7, Lesson 7.5
Parts A, B and C.
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