Chapter 7: Vectors and the Geometry of Space

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Chapter 7 Vectors and the Geometry of Space

• Section 7.2
• Space Coordinates and Vectors 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
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• In this lesson you will learn
• 3 Space - The three-dimensional coordinate
  system
• Points in space, ordered triples
• The distance between two points in space
• The midpoint between two points in space
• The standard form for the equation of a sphere

• Vectors in 3 Space
• Different forms of vectors
• Vector operations
• Parallel vectors
• Applications of vectors

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Three-Dimensional Space
Previously you studied vectors in the Cartesian
plane or 2-dimensions, now we are going to
expand our knowledge of vectors to 3-dimensions.
Before we discuss vectors, lets look at
3-dimensional space. To construct a
3-dimensional system, start with a yz plane flat
on the paper (or screen).
Next, the x-axis is perpendicular through the
origin. (Think of the x-axis as coming out of
the screen towards you.) For each axis drawn
the arrow represents the positive end.
z
y
x
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z
This is considered a right-handed system. To
recognize a right-handed system, imagine your
right thumb pointing up the positive z-axis, your
fingers curl from the positive x-axis to the
positive y-axis.
y
x
In a left-handed system, if your left thumb is
pointing up the positive z-axis, your fingers
will still curl from the positive x-axis to the
positive y-axis. Below is an example of a
left-handed system.
z
x
Throughout this lesson, we will use right-handed
systems.
y
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The 3-dimensional coordinate system is divided
into eight octants. Three planes shown below
separate 3 space into the eight octants.
The three planes are the yz plane which is
perpendicular to the x-axis, the xy plane which
is perpendicular to the z-axis and the xz plane
which is perpendicular to the y-axis. Think
about 4 octants sitting on top of the xy plane
and the other 4 octants sitting below the xy
plane.
z
y
yz plane
x
z
z
xy plane
y
y
xz plane
x
x
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Plotting Points in Space
Every position or point in 3-dimensional space is
identified by an ordered triple, (x, y, z).
Here is one example of plotting points in
3-dimensional space
z
P (3, 4, 2)
y
The point is 3 units in front of the yz plane, 4
points in front of the xz plane and 2 units up
from the xy plane.
x
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Here is another example of plotting points in
space. In plotting the point Q (-3,4,-5) you
will need to go back from the yz plane 3 units,
out from the xz plane 4 units and down from the
xy plane 5 units.
z
y
Q (-3, 4, -5)
x
As you can see it is more difficult to visualize
points in 3 dimensions.
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Distance Between Two Points in Space
The distance between two points in space
is given by the formula
Take a look at the next two slides to see how we
come up with this formula.
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Consider finding the distance between the two
points,
. It is helpful to think of a
rectangular solid with P in the bottom back
corner and Q in the upper front corner with R
below it at .
Using two letters to represent the distance
between the points, we know from the Pythagorean
Theorem that PQ² PR² RQ²
Q
P
R
Using the Pythagorean Theorem again we can show
that PR²
Note that RQ is .
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Starting with PQ² PR²
RQ² Make the substitutions PR²
and RQ
Thus, PQ² Or the distance from P to Q, PQ
Q
P
R
Thats how we get the formula for the distance
between any two points in space.
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We will look at example problems related to the
three-dimensional coordinate system as we look at
the different topics.
Example 1
Find the distance between the points P(2, 3, 1)
and Q(-3,4,2).
Solution Plugging into the distance formula
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Example 2
Find the lengths of the sides of triangle with
vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3).
Then determine if the triangle is a right
triangle, an isosceles triangle or neither.
Solution First find the length of each side of
the triangle by finding the distance between each
pair of vertices.
(0, 0, 0) and (5, 4, 1)
(0, 0, 0) and (4, -2, 3)
(5, 4, 1) and (4, -2, 3)
These are the lengths of the sides of the
triangle. Since none of them are equal we know
that it is not an isosceles triangle and since
we know it is not a right triangle. Thus it is
neither.
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The Midpoint Between Two Points in Space
The midpoint between two points,
is given by
Each coordinate in the midpoint is simply the
average of the coordinates in P and Q.
Example 3 Find the midpoint of the points P(2,
3, 0) and Q(-4,4,2).
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Equation of a Sphere
A sphere is the collection of all points equal
distance from a center point. To come up with
the equation of a sphere, keep in mind that the
distance from any point (x, y, z) on the sphere
to the center of the sphere, is the constant r
which is the radius of the sphere. Using the
two points (x, y, z), and r, the
radius in the distance formula, we get
If we square both sides of this equation we get
The standard equation of a sphere is where r is
the radius and is the center.
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Example 4
Find the equation of the sphere with radius, r
5 and center, (2, -3, 1).
Solution Just plugging into the standard
equation of a sphere we get
Example 5
Find the equation of the sphere with endpoints of
a diameter (4, 3, 1) and (-2, 5, 7).
Solution Using the midpoint formula we can find
the center and using the distance formula we can
find the radius.
Thus the equation is
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Example 6
Find the center and radius of the sphere,
.
Solution To find the center and the radius we
simply need to write the equation of the sphere
in standard form, . Then we can
easily identify the center, and
the radius, r. To do this we will need to
complete the square on each variable.
Thus the center is (2, -3, -4) and the radius is
6.
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Vectors in Three-Dimensional Space
Now that we have an understanding of the
three-dimensional system, we are ready to discuss
vectors in the three-dimensional system. All the
information you learned about vectors in the
previous lesson will apply, only now we will add
in the third component.
Vectors in component form in three dimensions are
written as ordered triples, in other words, now a
vector in component form is
.
In three dimensions the zero vector is O lt 0,
0, 0gt and the standard unit vectors are
.
z
Each of the unit vectors represents one unit of
change in the direction of each of their
respective positive axes.
y
x
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Given the initial point, and
the terminal point, , the
component form of the vector can be found the
same way it was on the Cartesian Plane.
Component form of a vector
Be sure to subtract the initial points
coordinates from the terminal points coordinates.
The same vector can be written as a combination
of the unit vectors.
Standard Unit Vector Notation
We will look at examples using both forms.
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More on Vectors in Three-Dimensions
Let and let c be a
scalar.

• Vector Equality
• Magnitude or Length of a Vector
• Vector Addition
• Scalar Multiplication
• Unit Vector in the Direction of

Note This is simply the vector multiplied by the
reciprocal of its magnitude.
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Lets look at some example problems involving
vectors. Example 1 Sketch the vector with
initial point P(2, 1, 0) and terminal point Q(3,
5, 4). Then find the component form of the
vector, the standard unit vector form and a unit
vector in the same direction.
Solution First draw a 3D system and plot P and
Q. The vector connects P to Q.
Q
P
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Example 1 Continued
Second, find the component form of the vector. Do
this by subtracting the initial points
coordinates from the terminal points coordinates.
Component form
Standard Unit Vector Form
Last, find a unit vector in the same direction.
Do this by multiplying the vector by the
reciprocal of the magnitude.
Note You can verify its a unit vector by
finding its magnitude.
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Example 2 Given the vectors find the
following a. b. c.
Solution a.
b.
c.
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Parallel Vectors
You may recall from the previous section that a
nonzero scalar multiple of a vector has the same
direction as the vector (positive scalar) or the
opposite direction as the vector (negative
scalar). Since this is the case, any nonzero
scalar multiple of a vector is considered a
parallel vector. In other words, if two
vectors, , are parallel, then there
exists some scalar, c such that . The
zero vector does not have direction so it cannot
be parallel.
To get the idea, look at these vectors on the
Cartesian Plane.
y
x
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Example 3 Determine if the vector with initial
point, P(3,2,-2) and terminal point, Q(7,5,1-3)
is parallel to the vector
.
Solution First find the component form of the
vector from P to Q.
Second, if the two vectors are parallel, then
there exists some scalar, c, such that
Then 12 4c c -3 And -9 3c
c -3 And 3 -c c
-3 For the two vectors to be parallel, c would
have to be the same for each coordinate. Since it
is, the two vectors are parallel.
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Example 4 Determine whether the points
A(2,3,-1), B(0,1,3) and C(-3,-2,8) are collinear.
Solution We need to find two vector between the
three points and determine if they are parallel.
If the two vectors are parallel and pass through
a common point then the three points must be in
the same line.
The vector from A to B is
Now we need to find the vector from A to C or B
to C. The vector from A to C is
To be parallel -2 -5c c 2/5
-2 -5c c
2/5 4 9c
c 4/9 Since c is not the same in each case,
the vectors are not parallel and the points are
not collinear.
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Example 5 Find a vector parallel to the
vector with magnitude 5.
Solution Be careful. We might quickly assume
that all we need to do is to multiply the vector
by 5. This would be fine if we were dealing with
a unit vector. Since we are not, we need to
multiply by the reciprocal of the magnitude first
to get a unit vector and then multiply by 5.
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Solution to Example 5 Continued You can verify
the new vector is parallel if you look at the
form Obviously the scalar multiple is
.
You can verify the magnitude is 5 by finding the
magnitude of the form
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Example 6 The weight of an 80lb. chandelier
hanging 2.5 feet from the ceiling is distributed
over 3 chains. If the chains are located as shown
below, represent the force exerted on each chain
with a vector.
(-1,-1,0)
1 ft
1 ft
1 ft
(0,1,0)
1 ft
(1,-1,0)
(0,0,-2.5)
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Solution to Example 6 First find the vectors
from the chandelier to the three points on the
ceiling. Each force is a multiple of the vector
since we can find the direction, but we dont
know the magnitude.
(-1,-1,0)
1 ft
1 ft
1 ft
(0,1,0)
1 ft
(1,-1,0)
(0,0,-2.5)
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Solution to Example 6 Continued The sum of the
three forces must negate the downward force of
the chandelier from its weight. So,
This gives us a system of three equations in
three unknowns, a,b and c.
Solving the system, you get a 16, b 8 and c
8. Thus the three forces are
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You can find your practice problems for this
lesson in Blackboard under the Assignments button
under Lesson 7.2.