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M O D U L A R S Y S T E M
Class 9
GEOMETRY
?
w w w . z a m b a k . c o m
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Copyright Zambak Yaynclk veEitim Gereleri A..All rights reserved.
No part of this book may be reproduced,stored in a retrieval system or
transmitted in any form without the prior
written permission of the publisher.Digital
Assembly
Zambak Typesetting & DesignPage
Design
Serdar YILDIRIMAydn ETN
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Publisher
Zambak Yaynclk veEitim Gereleri A..
Printed
by
alayan A..
Gaziemir / zmir, August 2012
Tel: +90-232 252 22 85
+90-232 522 20 96 / 97
ISBN: 978-605
-112-4
67-4Printed in Turkey
DISTRIBUTION
Zambak Yaynclk veEitim Gereleri A..
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Baclar / STANBUL_______________________Tel.: +90-212 604 21 00Fax: +90-212 604 21 12
http://book.zambak.com
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To
the
Teacher
Analytic Analysis of Lines and Circles is designed to provide students with the
analytic geometry background needed for further college-level geometry
courses. Analytic geometry can be defined as algebraic analysis applied to
geometrical concepts and figures, or the use of geometrical
concepts and figures to illustrate algebraic forms.
Analytic geometry has many applications in different
branches of science and makes it easier to solve a wide
variety of problems. The goal of this text is to help students
develop the skills necessary for solving analytic geometry
problems, and then help students apply these skills. By the
end of the book, students will have a good understanding
of the analytic approach to solving problems. In addition,
we have provided many systematic explanations throughout
the text that will help instructors to reach the goals that
they have set for their students. As always, we have taken
particular care to create a book that students can read,
understand, and enjoy, and that will help students gain
confidence in their ability to use analytic geometry.
To
the
Student
This book consists of two chapters, which cover analytical analysis of lines andcircles respectively. Each chapter begins with basic definitions, theorems, and
explanations which are necessary for understanding the subsequent chapter
material. In addition, each chapter is divided into subsections so that students
can follow the material easily.
Every subsection includes self-test heckYourself problem sections followed by basic
examples illustrating the relevant definition, theorem, rule, or property. Teachers
should encourage their students to solve Check Yourself problems themselves
because these problems are fundemental to understanding and learning the related
subjects or sections. The answers to most Check Yourself problems are given directly
after the problems, so that students have immediate feedback on their progress.
Answers to some Check Yourself problems are not included in the answer key, as they
are basic problems which are covered in detail in the preceding text or examples.
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Giving answers to such problems would effectively make the problems redundant,
so we have chosen to omit them, and leave students to find the basic answers
themselves.
At the end of every section there are exercises categorized according to the
structure and subject matter of the section. Exercises are graded in order,
from easy (at the beginning) to difficult (at the end).
Exercises which involve more ability and effort are
denoted by one or two stars. In addition, exercises which
deal with more than one subject are included in a
separate bank of mixed problems at the end of the
section. This organization allows the instructor to deal
with only part of a section if necessary and to easily determine which exercises
are appropriate to assign.
Every chapter ends with three important sections.
The hapterSummary is a list of important concepts and
formulas covered in the chapter that students can use
easily to get direct information whenever needed.
A oncept heck section contains questions about the
main concepts of the subjects
covered, especially about the definitions, theorems or
derived formulas.
Finally, a hapterReview Testsection consists of three tests, each with sixteen
carefully-selected problems. The first test covers
primitive and basic problems. The second and third tests
include more complex problems. These tests help
students assess their ability in understanding the
coverage of the chapter.
The answers to the exercises and the tests are given at the end of the book so
that students can compare their solution with the correct answer.
Each chapter also includes some subjects which are denoted as optional. These
subjects complement the topic and give some additional
information. However, completion of optional sections is
left to the discretion of the teacher, who can take into
account regional curriculum requirements.
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CHAPTER 1
SECTION 1:ANALYSIS OF VECTORS
GEOMETRICALLY
A. BASIC VECTOR CONCEPTS . . . . . . . . . .10
1. Directed Line Segment . . . . . . . . . . . . . . . . . . . .10
2. Definition of a Vector . . . . . . . . . . . . . . . . . . . . . .11
3. Equal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .11
B. VECTOR OPERATIONS . . . . . . . . . . . . . .12
1. Addition of Vectors . . . . . . . . . . . . . . . . . . . . . . . .12
2. Subtraction of Vectors . . . . . . . . . . . . . . . . . . . . .16
3. Multiplication of a Vector by a Scalar . . . . . . . . . .17
C. PARALLEL VECTORS . . . . . . . . . . . . . . . .18
1. Parallel Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . .18
2. Non-Parallel Vectors . . . . . . . . . . . . . . . . . . . . . . .19
SECTION 2:ANALYSIS OF VECTORS
ANALYTICALLY
A. BASIC CONCEPTS OF VECTORS IN THE
ANALYTIC PLANE . . . . . . . . . . . . . . . . . . . .24
1. Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24
2. The Rectangular Coordinate System . . . . . . . . . .24
3. Position Vector . . . . . . . . . . . . . . . . . . . . . . . . . . .25
4. Components of a Vector . . . . . . . . . . . . . . . . . . .26
5. Equal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .27
B. VECTOR OPERATIONS . . . . . . . . . . . . . .28
1. Addition of Vectors . . . . . . . . . . . . . . . . . . . . . . . .28
2. Subtraction of Vectors . . . . . . . . . . . . . . . . . . . . .29
3. Multiplication of a Vector by a Scalar . . . . . . . . . .30
4. Standard Base Vectors . . . . . . . . . . . . . . . . . . . . .31
C. PARALLEL VECTORS . . . . . . . . . . . . . . . .33
SECTION 3: THE DOT PRODUCT OF
TWO VECTORS
A. DOT PRODUCT . . . . . . . . . . . . . . . . . . .38
1. Properties of the Dot Product . . . . . . . . . . . . . . . .38
B. ANGLE BETWEEN TWO VECTORS . . . . .39
1. Angle Between Two Vectors . . . . . . . . . . . . . . . . .39
2. Perpendicular and Parallel Vectors . . . . . . . . . . .41
CHAPTER 2
SECTION 1: CONGRUENCE
A. THE CONCEPT OF CONGRUENCE . . . .56
1. Congruent Figures and Polygons . . . . . . . . . . . .56
2. Congruent Triangles . . . . . . . . . . . . . . . . . . . . . . .57
B. THE TRIANGLE ANGLE BISECTORTHEOREM . . . . . . . . . . . . . . . . . . . . . . . . .64
SECTION 2: THE CONCEPT OF
SIMILARITY
INTRODUCTION TO SIMILARITY . . . . . . . .67
1. Similar Figures . . . . . . . . . . . . . . . . . . . . . . . . . . .67
2. Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . .67
SECTION 3: THE ANGLE - ANGLE
SIMILARITY POSTULATETHE ANGLE-ANGLE (AA) SIMILARITYPOSTULATE . . . . . . . . . . . . . . . . . . . . . . . . . .73
SECTION 4: WORKING WITH SIMILARITY
TRIANGLES
A. THE SIDE-ANGLE-SIDE (SAS)SIMILARITY THEOREM . . . . . . . . . . . . . . . .83
B. THE SIDE-SIDE-SIDE (SSS)SIMILARITY THEOREM . . . . . . . . . . . . . . . .85
C. THE TRIANGLE PROPORTIONALITYTHEOREM AND THALES THEOREM . . . . . .90
1. The Triangle Proportionality Theorem . . . . . . . . .90
2. Thales Theorem of Parallel Lines . . . . . . . . . . . .93
D. FURTHER APPLICATIONS . . . . . . . . . . . .95
1. Menelaus Theorem . . . . . . . . . . . . . . . . . . . . . . .952. Cevas Theorem . . . . . . . . . . . . . . . . . . . . . . . . . .96
SECTION 5: FURTHER STUDIES
A. EUCLIDEAN RELATIONS . . . . . . . . . . .103
B. MEDIAN RELATIONS . . . . . . . . . . . . . .106
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SECTION 3: SOLIDS WITH CURVED
SURFACES
A. Some Important Polyhedrons . . . . . . . .300
1. Prisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .300
SECTION 4:VOLUMES OF SOLIDS
A. Circular Cylinder . . . . . . . . . . . . . . . . .320
B. Areas of Cones . . . . . . . . . . . . . . . . . .323
C. Spheres . . . . . . . . . . . . . . . . . . . . . . .325
1. Fundamental Definitions . . . . . . . . . . . . . . . . . . .325
SECTION 5: SECTIONS AND
COMBINATIONS OF SOLIDS
1. Volume of a Right Prism . . . . . . . . . . . . . . . . . . .330
2. Volume of a Pyramid . . . . . . . . . . . . . . . . . . . . .331
Volumes of Cones . . . . . . . . . . . . . . . . . .336
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Geometry 9
Some of the quantities we measure in our daily lives are completely determined by their
magnitudes, for example, length, mass, area, temperature, and energy. When we speak of a
length of 3 cm or an area of 5 cm2, we only need one number to describe each of these
quantities. We call such quantities scalarquantities.
On the other hand, to describe a force, we need to record its direction as well as its size. For
example, to describe the velocity of a moving object, we must specify both the speed and the
direction of travel. Quantities such as displacement, velocity, acceleration, and other forces
that have magnitude as well as direction are called vectorquantities. We usually show a
vector quantity as an arrow that points in the direction of the action, with length that shows
the magnitude of the action in terms of a suitable unit. The way to represent such quantities
mathematically is through the use of vectors.
1. Directed Line Segment
When we move from Antalya to Berlin
by bus, we have two quantities: the
direction from Antalya to Berlin, and
the length of the displacement between
these cities.
We can sketch a line segment AB as shown in the figure with starting
pointA and finishing point B to represent the movement from Antalya to
Berlin. The line segmentAB with an arrow has direction and length. The
arrow head specifies the direction, and the length of the arrow specifies
the magnitude, at a suitable scale.A and B are the endpoints of the segment.
PointA is called the initialpoint and point B is called the terminalpoint
of the line segment. The resulting segmentAB is called a directedlinesegment.
Definition directedlinesegment
A line segment with direction is called a directedlinesegment.
We write
B to denote a directed line segment from pointA to point B.
Directed line segments are used in daily life. For example, some
traffic signs for drivers use directed line segments.
In technology we also use directed line segments.
A. BASIC VECTOR CONCEPTS
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Analysis of Vectors Geometrically
PointsM, N, P and K on line d are given. Write all the
directed line segments with endpointsM, N, P, or K.
The directed line segments with endpoints M, N, P, or K are
MN,
MP,
MK,
NP,
NK,NM,
PK,
PN,
PM,
KP,
KN, and
KM.
Notice that
MN is not the same as
NM, and
MP is not the same as
PM. This is because the
line segments have direction. Pairs such as
MN and
NM have the same magnitude but
opposite direction.
Solution
EXAMPLE 1
Definition vector
A directed line segment in the plane is called a vector.
The length of the directed line segment is the length of the
vector.
The direction of the directed line segment is the direction of
the vector.
We write
B to mean a vector with initial pointA and terminal point B. Alternatively, we can
name a vector with a lower-case letter such as u or p.
3. Equal Vectors
Definition equalvectors
Two vectors that have the same direction and length arecalled equalvectors. We show that two vectors
u and
v are
equal by writing u =
v.
In the figure, D, E, and F are the midpoints ofAB,AC and
BC respectively, and DE || BC, EF ||AB, DF ||AC.
Name all the equal vectors.
In triangle ABC,|DE| = |BF| = |FC|
|EF| = |AD| = |DB|
|DF| = |AE| = |EC|.
So
DE =
BF =
FC
EF =
AD =
DB
DF =
AE =
EC
ED =
FB =
CF
FE =
DA =
BD
FD =
EA =
CE.
Solution
EXAMPLE 2
2. Definition of a Vector
and
For example, consider a line segment
AB with length 2 cm.
We can say the length of vector
AB is 2 cm, and write |
AB| = 2 cm.
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2
Geometry 9
Ch e ck Yo u r se l f
ABCD is a parallelogram in the figure.
1 How many pairs of equivalent directed line segments are there?
2 How many pairs of equal vectors are there?
A n sw e r s
1 4 2 4
1. Addition of Vectors
Let
PQ and
QRbe two vectors in a plane.
PQ +
QR denotes the sum of the vectors
PQ and
QR. There are two ways to find the sum of two or more vectors.
a. The Polygon Method
Imagine we want to add n vectors together. Using the polygon method, we draw the first
vector. Then we place the initial point of the second vector at the terminal point of the first
vector, the initial point of the third vector at the terminal point of the second vector, and so
on until we place the initial point of the nth vector at the terminal point of the (n 1)th
vector. The sum is the vector whose initial point is the initial point of the first vector and
whose terminal point is the terminal point of the last vector.
Let us look at an example.Let
AB and
CDbe two vectors in a plane, as in the
diagram. We place the initial point of
AB at the
terminal point of
CD to make
DE (
AB =
DE).
Using the polygon method,
CD +
AB =
CD +
DE =
CE.
B. VECTOR OPERATIONS
Definition oppositevectors
Two vectors are called opposite vectors if and only if their
magnitudes (lengths) are the same but their directions are opposite.
Definition zerovector
A vector whose initial and terminal points are the same is called a zerovector.
We write a zero vector as0.
A zero vector has no direction and no size.
For example, in the figure,
AB and
BA are opposite vectors.
CD and
DC are also opposite vectors. We can write
AB =
BA and
CD =
DC.
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3
Analysis of Vectors Geometrically
b. The Parallelogram Method
To add one vector to another using the parallelogram method, we draw the first vector, and
then we draw the second vector with its initial point at the initial point of the first vector. We
make a parallelogram by drawing two additional sides, each passing through the terminal
point of one of the vectors and parallel to the other vector. We find the sum by drawing a
vector along the diagonal from the common initial point to the intersection of the two lines.
Look at the example of addingu and
v using the parallelogram method:
Now look at an example of adding more than two vectors using the polygon method.
As shown in the figure,u +
v +
w +
x =
AE.
Findu +
v +
w in the figure on the right.EXAMPLE 3
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4
Geometry 9
Let us choose A as a fixed point. We can use the polygon method or the parallelogram
method to add the given vectorsu +
v +
w .
Solution
The velocity of a boat is 25 m/min north and the velocity of a river current is 3 m/min east.
Draw a scale diagram to show the velocities as vectors and find the sum.
First we choose a starting point A and
write
AN = velocity of the boat due north.
AE = velocity of the current due east.
AN and
AE are perpendicular, and
AK is
the sum of
AN and
AE:
|
AK| =
This is the sum of the vectors.
2 225 + 3 = 634.
Solution
EXAMPLE 4
c. Properties of Vector Addition
Letu ,v, and
w be three vectors in a plane P.
1 The sum of any two vectors in P is
also a vector in P (closure property).
2 The sum of any two vectors in P is
commutative (commutative property).
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5
Analysis of Vectors Geometrically
4 The sum of the zero vector and a vector
in P is the vector itself (identity
element).
3 The sum of any three vectors in P is associative (associative property).
u +(
v +
w ) = (
u +
v ) +
w
5 The additive inverse of any vectoru is
u:u + (
u) =
0 (additive inverse).
In a triangleABC, P is the midpoint of
AB. Express
CP in terms of
CA and
CB.
CP = CA + AP
+
CP =
CB +
BP
2
CP =
CA +
CB +
AP +
BP
CP = (
CA +
CB)1
2
Solution
EXAMPLE
0
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6
Geometry 9
Ch e ck Yo u r se l f
1 Find the following using the vectors in the figure.
a
v u b u + w c w + v u
2 In a triangleABC, D [BC] and |BD| = 2 |DC|.
Express
AD in terms of
AB and
AC.
A n sw e r s
1 use the polygon method 2
AD =
AC +
AB1
3
2
3
2. Subtraction of Vectors
Since subtraction is the inverse of addition, we can find the difference of two vectorsu and
v by adding the vectors
u and
v (opposite of
v) using either the parallelogram method
(u v = u + (v )) or the polygon method.
In a triangleABC, G is the centroid. Find
GA +
GB +
GC.
Let us label a point G on the extension of CG which
satisfies |CG| = |GG|. Since G is the centroid of
ABC, |CG| = 2|GK|. Therefore |GG| = 2|GK|,which means that K is the midpoint of GG. We con-
clude thatAGBG is a parallelogram because K is the
midpoint of both diagonalsAB and GG. So we have
AG =
GB which gives us
GA +
GB =
GG.
On the other hand, we have
CG =
GG =
GC. Using
this result in
GA +
GB =
GG, we get
GA +
GB =
GC which gives us
GA +
GB +
CG =
0.
Solution
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7
Analysis of Vectors Geometrically
3. Multiplication of a Vector by a Scalar
Multiplying a vector by a scalar makes the vector longer or shorter depending on the value of
the scalar. If the scalar is greater than 1 or less than 1, multiplying makes a longer vector. If
the scalar is between 1 and 1 and non-zero, it makes a shorter vector.If the scalar is positive, multiplying does not change the direction.
If the scalar is negative, multiplying will make the vectors direction opposite.
For a real number a and a vectoru,
1 if a > 0 then vector au has the same direction as
u and the length |a
u| = a|
u|.
2 if a < 0 then vector au has the opposite direction to
u and the length |a
u| = |a||
u|.
3 if a = 0 then a
u =
0.
Using
AB as shown in the figure, draw vector
diagrams to show 2
AB, 4
AB, and
AB.1
2
Since 2 and are positive, 2
AB and
AB have the
same direction as
AB. However, 2
AB is twice as long
as
AB and
AB is half as long.
On the other hand, 4
AB has opposite direction to
AB (since 4 is a negative scalar) and it is four times
as long as
AB.
1
2
1
2
1
2Solution
EXAMPLE 5
a. Properties of the Multiplication of a Vector by a Scalar
For any vectorsu,v, and
w and real numbers a and b, the following properties are satisfied.
1 a
u is a vector in the plane2 (ab)
u = a(b
u )
3 (a + b)u = a
u + b
u
4 a(u +
v ) = a
u + b
v
5 1u =
u
6 a
0 =
0
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8
Geometry 9
Ch e ck Yo u r se l f
1 Multiply the vector
u by the scalars 2, 3, 0.5 andand draw a vector diagram to show them.
2 PointsA, B, C, andM are on the same line.M is between
A and C.
AB = 2
AC. Express the vector
MC in terms of
the vectors
MA and
MB.
A n sw e r s
2
MC = (
MA +
MB)1
2
1
3
1. Parallel Vectors
C. PARALLEL VECTORS
Definition parallelvectors
Leta and
bbe two vectors.
a and
b are called parallelvectors if and only if
a = k
b where
k 0 and k . We write
a
||
b to show that two vectors are parallel.
For example, in the diagram, |a| = 2 cm,
|b| = 1 cm and |
c| = 4 cm.
We can express vectora as
a =
c and
a = 2
b.
Therefore the vectorsa,b, and
c are parallel, i.e.
a||b||c.
1
2
PointsA, B, C, andM are on the same line.M is between C and B.
AB = 3
AC. Express the
vector
MC in terms of vectors
MA and
MB.
AB = 3
AC so
CB = 2
AC (1)
MA +
AC =
MC (2)
CM +
MB = 2
AC (3)
AC =
MC +
MB (4)
MA
MC +
MB =
MCby (2) and (4).
MA +
MB =
MC +
MC
MA +
MB =
MC
So
MC =
MA +
MB.1
3
2
3
3
2
1
2
12
12
1
2
1
2
1
2
1
2
Solution
EXAMPLE 6
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9
Analysis of Vectors Geometrically
In a triangleABC, D and E are the midpoints of sidesAB andAC respectively.
Show that
DE ||
BC.
EXAMPLE 7
BA +
AC =
BC and
DA +
AE =
DEby the addition of vectors.
DA =
BA,
AE =
AC
DE =
BA +
AC = (
BA +
AC) =
BC
Now
DE =
BC, so
DE ||
BCby the definition of parallel vectors.1
2
1
2
1
2
1
2
1
2
1
2
1
2
Solution
BA +
AC =
BC (1) and
EA +
AF =
EF (2).
EA =
BA
+
AF =
AC
EA +
AF = (
BA +
AC)
EF =
BCby (1) and (2). Therefore,
EF ||
BC.1
2
1
2
1
2
1
2
Solution
EXAMPLE8
In a quadrilateral ABCD, points E and F are the midpoints of side AB and diagonal AC,respectively. Show that EF || BC.
2. Non-Parallel Vectors
By the definiton of parallel vectors we can conclude that ifa and
b are non-zero,
non-parallel vectors, then h
a = k
b when h =k = 0. Look at the proof:Suppose that h k 0,
a =
b.
Thena||b. This is a contradiction, since
a and
b are non-parallel. As a result, h = k = 0.
k
h
EXAMPLE 9 Prove that the diagonals of a parallelogram intersect at their midpoints by using vectors.
Look at the diagram. Let
AB =
a and
BC =
b, so
AC =a +
b
DB =a b
AE =m(a +
b)
EB =n(a b )
Solution
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2
Geometry 9
Ch e ck Yo u r se l f
1 Name all the pairs of parallel
vectors in the figure.
2 Add the vector pairs
u1 and
u5,u2 and
u6, and
u3 and
u7.
3 Find the additive inverse ofu7
andu1.
4 Subtractu8 from
u4.
5 In a quadrilateralABCD, P and
K are the midpoints of sidesAB
and DC respectively. Express
PKin terms of
DA and
CB.
A n sw e r s
1 look at the directions and lengths 2 use the polygon method 3 use the polygon method
4 use the polygon method 5
PK = (
DA +
CB)1
2
Let us draw the triangleABC as in the figure.
If D, E and F are midpoints then
AD = (
AB +
AC) by the result of Example 5.
BF = (
BA +
BC)
+
CE = (
CA +
CB)
AD +
BF +
CE = (
AB +
BA +
AC +
CA +
BC +
CB )
Therefore,
AD +
BF +
CE =0.
1
2
1
2
1
2
1
2
Solution
EXAMPLE 10 In a triangleABC, D, E and F are the midpoints of sides BC, BA andAC respectively. Find thesum
AD +
BF +
CE.
0
0
0
AE +
EB =m(a +
b ) +n(
a b )
a =m(
a +
b) +n(
a b)
(m +n 1)a =
b(n m).
Since
a and
b are non-zero and non-parallel, we have (m +n 1) = (n m) = 0. Therefore
m +n = 1 andm =n, and som =n = .1
2
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2
Analysis of Vectors Geometrically
Project:Describe some other areas in which we use directed line segments
(for example: flowcharts, keyboards, ...).
A. Basic Vector Concepts
1. Draw any two parallel directed line segments withthe same length but opposite direction. Express
one of the line segments in terms of the other.
EXERCISES 1
2. How many equal directed line segments can we
find on two parallel lines?
3. Make a scale diagram showing the vectors in each
statement and find their sum.
a A 6 km trip east is followed by a 3 km trip
southeast.
b The velocity of a swimmer is 5 m/min west
and the velocity of a river current is 2 m/min
north.
4. Using the vectors given on the right,sketch the following vectors.
a
u +
v b
w (
u +
v)
c w
v
u d
u + (
w +
v)
e
u 3
v + 2
w
B. Vector Operations
5. In a plane, [AB] is given. Point K is the midpoint
of [AB] and point O is any point in the same
plane. Express
OK in terms of
OA and
OB.
6. In a triangleABC, points D and E lie on [BC] and
|BD| = |DE| = |EC|. Express the vector
AD +
AE
in terms of
AB and
AC.
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Analysis of Vectors Geometrically
12. In a six-sided polygonABCDEF,
AB =
ED,
BC =
FE,
and
CD =
AF. Show that FBCE is a parallelogram.
In a parallelogram ABCD, K is the midpoint of
side DC, and the intersection point of diagonals
AC and BK is T. Show that
AC = 3
TC.
18.
In a quadrilateralABCD, E and F are the midpoints
of the diagonalsAC and BD respectively.
Show that
AB +
AD +
CB +
CD = 4
EF.
15.
14. In a triangle ABC, |BD| = |DE| = |EC|, and
E, D [BC]. If |
AD+
AE| = 9 cm, find |
AB +
AC|.
Show that the centroid of a triangle divides a
median in the ratio 1:2 using vectors.
17.
13. A trapezoid is a four-sided figure with only two
parallel sides. A line segment which joins the
midpoints of the non-parallel sides is called the
median of the trapezoid. Prove that the median of
a trapezoid is parallel to the two parallel sides,and has magnitude equal to half of their sum.
In the figure,
Tis the midpoint of BC,
ABC is a triangle,
2|AK| = |KB|, and
2|AM| = |MC|.
Use vectors to show that
|
AL| = |
LT|.1
2
16.
11. ABCD is a quadrilateral and M, N, P, Q are the
midpoints of AB, BC, CD, and DA respectively.
Show thatMNPQ is a parallelogram.
C. Parallel Vectors
10. In a triangleABC, G is the point of intersection of the
medians and P is the midpoint of
BG. Show thatPA +
PC = 4
PG.
Consider any two points A and B in a plane.For
any point P in the same plane, the symmetry of
point P with respect to point A is Q and the
symmetry of point Q with respect to point B is
point R. Show that
PR is always 2
AB. (Hint: letA,
B, and Cbe collinear. If |AB| = |BC| then A is
the symmetry of C with respect to B.)
9.
8. Point O is in the plane of a triangle ABC. Point G
is the centroid of triangle ABC. Show that
OA +
OB +
OC = 3
OG.
7. Show that (
AB +
BC) +
CD =
AB + (
BC +
CD)
by using the parallelogram method.
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Geometry 9
We have studied vectors geometrically. Now let us look at a method for describing vectorsanalytically.
We will begin this section by looking at some important axioms.
1. Axioms1 For each pair of points P and Q there exists a unique vector
v such that
PQ =v.
2 For each point P and vectorv, there is a
unique point Q such thatv =
PQ.
A. BASIC CONCEPTS OF VECTORS IN THE ANALYTIC
PLANE
Con c l u s i o n
1 Two points in a plane determine two opposite vectors.
2 In a plane, if one point is fixed as an initial point then all the other points in the plane
can be chosen as the terminal point of any vector.
2. The Rectangular Coordinate System
The rectangularcoordinatesystem is formed by
two perpendicular intersecting number lines, as
shown in the diagram opposite.
1 The horizontal number line is called the x-axis.
2 The vertical number line is called the y-axis.
The origin is the point of intersection. At this
intersection, both number lines are 0. The
rectangular coordinate system is split into four
quadrants, which are marked in the diagram
with roman numerals.
Each point in the coordinate system is associated with a pairofrealnumbers. In an x, y
system, the x-coordinate always comes first and the y-coordinate always comes second in the
pair (x, y). The first coordinate is called the abscissa of the point and the second coordinate
is called the ordinate of the point.
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Analysis of Vectors Analytically
A(2, 3) lies in quadrant I.
B(1, 2) lies in quadrant II.
C(3, 4) lies in quadrant III.
D(2, 0) lies on the x-axis.
E(0, 5) lies on the y-axis.
Solution
EXAMPLE 11 Plot each pair of coordinates and name the quadrant or axis in which the point lies.A(2, 3), B(1, 2), C(3, 4), D(2, 0), E(0, 5)
Ch e c k Y o u r s e l f
Plot the pairs of real numbers and name the quadrant or axis in which the point lies.
A(3, 2), B(2, 1), C(4, 3), D(0, 2), E(5, 0)
From this definition we can conclude the following:
1 For every vector in the plane there exists a position
vector
OP which is determined by a pair (x, y), that is,
OP = (x, y).
2 If
OP =
AB, then OPBA is a parallelogram.Look at the diagram. We can calculate that x = x2 x1 and
y = y2 y1. Therefore the vector
AB determined by the
pointsA(x1, y1) and B(x2, y2) has position vector
OP =
AB =
OB
OA = (x, y) = (x2 x1, y2 y1).
OP =
KL =
OL
OK = (4 2,5 1)
OP = (2, 4)
Solution
EXAMPLE
12Find the position vector of
KL with endpoints
K(2, 1) and L(4, 5).
3. Position Vector
Definition positionvector
Avector
OP whose initial point is at the origin of the rectangular coordinate plane and which
is parallel to a vector
AB is called the positionvector of
AB in the plane. In other words, ifOP is the position vector of
AB, then
OP ||
AB, |
OP| = |
AB|, and
OP =
AB.
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Geometry 9
4. Components of a VectorWe have seen how to describe a vector in the plane with
reference to its unique position vector.
We can also express a vector in the plane as the sum of
two vectors, one of which is parallel to the x-axis and the
other parallel to the y-axis. These two vectors are called
the components of the original vector. The component
parallel to the x-axis is called the horizontalcomponent
of the vector and the component parallel to y-axis is
called the verticalcomponent. Expressing a vector as the
sum of its components like this is called resolving the vector. For example, in the figure
opposite, the vector
u is the sum of the two components
ux and
uy.We can represent the vector
u as an ordered pair of real numbers:
u = (u1, u2) or
u = ,
where u1 is the horizontal scalar component ofu, and
u2 is the vertical scalar component ofu.
Now,ux = (u1, 0) and
uy = (0, u2).
Look at the diagram opposite. By applying thePythagorean theorem for triangle PQR in the figure, we
can see that the length of the vectoru = (u1, u2) is
|u| =
2 2
1 2+ .u u
1
2
u
u
a |u| =
b |v| =
c |
w| =
2 23 4 9 16 9 16 25
( ) +( ) = = = =15 5 25 25 25 25
2 23 + 0 = 9 = 3
2 22 +(3) = 4+9= 13Solution
EXAMPLE 13 Find the length of each vector.
a
u = (2, 3) b
v = (3, 0) c
w =3 4
( , )5 5
u = (3 1, 6 2) = (2, 4)
|u| =
2 22 + 4 = 4+16 = 20
Solution
EXAMPLE 14 Find the length of the vectoru with initial point (1, 2) and terminal point (3, 6).
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Analysis of Vectors Analytically
|
AB| =
(1 a)
2
+ 4 = 20(1 a)2 = 16
1 a = 4 or 1 a = 4
a = 3 or a = 5
2 2 2(1 ) +(4 2) = (1 ) + 4 =2 5a aSolution
5. Equal Vectors
Definition equalvectors
Two vectors are equal if and only if their corresponding scalar components are equal.
In other words, the vectorsu = (u1, u2) and
v = (v1,v2) are equal if and only if u1 =v1 and
u2 =v2.
u =
v , so a + b = 1 and 3 = b a.Solution
EXAMPLE 16u and
v are equal vectors with
u = (a + b, 3),
v = (1, b a). Find a and b.
Let the terminal point ofvbe (x, y), so
(x 2, y 4) = (3, 7)
x 2 = 3 and y 4 = 7 by the equality of vectors.
So x = 5 and y = 11.
Solution
EXAMPLE 17 The vectorv = (3, 7) has initial point (2, 4). What is its terminal point?
Ch e c k Y o u r s e l f
1 Find the position vector of
ML with endpointsM(3, 2) and L(2, 3).
2 Find the length of each vector.
a
u = (3, 1) b v = (0, 3) c AB with endpointsA(2, 1) and B(5, 4)
A n s w e r s
1
ML = (1, 1) 2 a 10 b 3 c 32
a + b = 1
b a = 3, so b = 1 and a = 2.
EXAMPLE 15 Find the possible values of a given |
AB| = 25 and the endpointsA(a, 2) and B(1, 4).
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Analysis of Vectors Analytically
EXAMPLE 22 Given
w = (1, 3), K(3, 2) and P(1, 4), find
w +
KP and
w
KP.
2. Subtraction of Vectors
Ifu = (u1, u2) and
v = (v1,v2) then
u
v = (u1 v1, u2 v2).
The figure opposite shows how the analytic definition
of vector subtraction corresponds to the geometric
one.
u
v = (2 3, 4 2) = (1, 2)Solution
EXAMPLE 21 Subtractv = (3, 2) from
u = (2, 4).
a. Properties of Vector Subtraction
Letu = (u1, u2),
v = (v1, v2), and
w = (w1, w2) be vectors in a plane. Then the following
properties hold.
1 The difference of any two vectors in a plane is a vector.
2
u
v
v
u. Therefore, vector subtraction is not commutative.3
u (
v
w ) (
u
v )
w . Therefore, vector subtraction is not associative.
4
u
0
0
u. Therefore there is no identity element for subtraction.
a. Properties of Vector Addition
Letu = (u1, u2),
v = (v1,v2), and
w = (w1,w2) be vectors in a plane. Then the following
properties hold.
1 The sum of any two vectors in a plane is a vector. (closure property)
2
u +
v =
v +
u (commutative property)
3
u + (
v +
w ) = (
u +
v ) +
w (associative property)
4
0 is the identity element:u +
0 = (u1 + 0, u2 + 0) = (u1, u2) =u.
5 u is additive inverse of
u:
u + (u) = (u1 u1, u2 u2).
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Geometry 9
3. Multiplication of a Vector by a Scalar
Letv = (v1,v2) and c , then c
v = (cv1, cv2).
2u = (2(3), (2)2) = (6, 4)
3v = (32, 3(1)) = (6, 3)
3u + 2
v = 3(3, 2) + 2(2, 1) = (9, 6) + (4, 2) = (5, 4)
Solution
EXAMPLE24
u = (3, 2) and
v = (2, 1) are given. Find 2
u, 3
v , and 3
u + 2
v .
a. Properties of the Multiplication of a Vector by a Scalar
Let
u = (u1, u2),
v = (v1,v2) and c, d . Then the following properties hold:1 c(
u +
v) = c
u + c
v. Look at the proof:
c(u +
v ) = c(u1 +v1, u2 +v2)
= (cu1 + cv1, cu2 + cv2)
= c(u1, u2) + c(v1,v2)
= cu + c
v.
2 (c + d)u = c
u + d
u
3 (cd)u = c(d
u ) = d(c
u )
4 1u =
u
5 0u =
0
6 c0 =
0
7 |cu| = |c||
u|.
KP = (1 3, 4 2) = (4, 2)w +
KP = (1, 3) + (4, 2) = (1 4, 3 + 2) = (3, 1)w
KP = (1, 3) (4, 2) = (1 + 4, 3 2) = (5, 5)
Solution
MN +
NK =
MK
|
MN +
NK| = |
MK| = 13
(m 1)2 + 25 = 169
(m 1)2 = 144
m 1 = 12 orm 1 = 12
m = 13 orm = 11
2 2( 1) +( 1 4) =13m
Solution
EXAMPLE 23 M(1, 4), N(3, 2m), K(m, 1), and |
MN +
NK| = 13 are given. Findm.
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Analysis of Vectors Analytically
4. Standard Base Vectors
a u 3v = (3, 1) 3(2, 2) = (3, 1) + (6, 6) = (3 6, 1 + 6) = (9, 7)
b 3u + 2
v = 3(3, 1) + 2(2, 2) = (9, 3) + (4, 4) = (9 + 4, 3 4) = (5, 1)
c 4u +
v = 4(3, 1) + (2, 2) = (12, 4) + (2, 2) = (12 + 2, 4 2) = (10, 2)
Solution
EXAMPLE 26 Find each vector, givenu = (3, 1) and
v = (2, 2).
a
u 3
v b 3
u + 2
v c 4
u +
v
2u 3
v = 2(2, 3) 3(1, 1) = (4, 6) (3, 3) = (4 3, 6 + 3) = (7, 9)
Therefore, |2
u 3
v | = 2 2
( 7) +9 = 49+81= 130.
Solution
EXAMPLE 25 Find |2u 3
v| given
u = (2, 3) and
v = (1, 1).
Ch e c k Y o u r s e l f
1 Find |w| given
u = (3, 2),
v = (1, 4),
w = (a, b) and
v
w = 3
u.
2 Findx given 2
x +
y = (1, 2), and
x
y = (4, 4).
A n s w e r s
1 |w| = 102 2
x = (1, 2)
u
|u|
1
|u|
= u 21
2 2 2 2
1 2 1 2
= , .+ +
uu
u u u u
u
|
u|
22 2 2 22
2 2 1 21 1
2 2 2 2 2 22 2 2 2
1 2 1 2 1 21 2 1 2
+= + = + = =1.
+ + ++ +
u u u uu u
u u u u u uu u u u
So
We sometimes use to find the direction ofu.
u
|u|
There are two important unit vectors,i and
j, defined as
i = (1, 0) and
j = (0, 1).
These vectors are special because we can use them to express any vector.
We call these vectors standardbasevectors.
Definition unitvector
A vector of length 1 is called a unitvector.
For any non-zero vectoru = (u1, u2), is a unit vector because and
1
|u|
2 21 2
1=
+u u
u
|u|
For example, the vectorw = is a unit vector.
3 4( , )5 5
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Geometry 9
b The properties of addition and scalar multiplication of vectors show that we can
manipulate vectors in the same way we manipulate algebraic expressions. Therefore,
2u + 3
v = 2(3
i + 2
j ) + 3(
i + 6
j )
= (6i + 4
j ) + (3
i + 18
j )
= (6 3)i + (4 + 18)
j
= 3i + 22
j.
a
u = 5
i + (8)
j = 5
i 8
j.Solution
EXAMPLE 27 a Write the vectoru = (5, 8) in terms of
i and
j.
b Ifu = 3
i + 2
j and
v =
i + 6
j , find 2
u + 3
v.
For example, let us express the vectorv = (v1,v2) in terms of
i and
j:
v = (v1,v2) =v1
i +v2
j.
In this expression,
v1
i is the horizontal component,v2
j is the vertical component,
v1 is the horizontal scalar component, and
v2 is the vertical scalar component.
We can prove the proposition above by using algebraic operations on vectors and the
properties of real numbers:v = (v1,v2) = (1v1 + 0, 0 + 1v2) = (1v1, 0) + (0, 1v2)
=v1(1, 0) +v2(0, 1)=v1
i +v2
j.
Letv be a vector in the plane with its initial point at the
origin. Let be the positive angle between the positive
x-axis andv (see the figure). If we know the length and
direction ofv , then we can resolve the vector into
horizontal and vertical components in terms of :v has length |
v|, and
v = (v1,v2) =v1
i +v2
j.
Sov1 = |v | cos and v2 = |
v | sin.
EXAMPLE 28 a |v| = 4 cm and the angle between
v and the positive x-axis is 60. Find the horizontal and
vertical components ofv and express
v in terms of
i and
j.
b Find the angle between the vectoru = 3
i +
j and the positive x-axis.
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Analysis of Vectors Analytically
Ch e c k Y o u r s e l f
1 Expressv = (3, 2) in terms of standard base vectors.
2 Resolve the vector u with length 3 cm into its components if the angle between the
positive x-axis andu is 120.
3 Find the angle between the vectorv = 3
i 33
j and the positive x-axis.
A n s w e r s
1
v = 3
i + 2
j 2
ux = ,
uy = 3 = 240
3 3(0, )
2
3( , 0)
2
a We havev = (v1,v2), where the scalar components are given by v1 = 4cos60 = 2 and
v2 = 4sin60 = 23. Therefore, the horizontal component is 2i and the vertical
component is 23j . Therefore,
v = 2
i + 23
j.
b From the figure we see that
has the property that
tan(180 ) = .
Thus 180 = 30, and so = 150.
1 3=
33
Solution
, so they are parallel.2 1
= = 2112
Solution
EXAMPLE 29 Show thatu = (2, 1) and
v = (1, ) are parallel.
1
2
EXAMPLE 30 Find the relation between x and y givenA(3, 1), B(2, 3), C(5, 4), D(x, y), and
CD ||
AB.
We know from the geometrical analysis of vectors that two non-zero vectors are parallel if and
only if multiplying one of them by a suitable scalar equals the other, that is,
for any c 0,u 0, and
v 0,
u ||
v if and only if
u = c
v.
It follows that ifu = (u1, u2) and
v = (v1,v2), then (u1, u2) = (cv1, cv2).
So
u || v ifandonlyif 21
2 2
v v
C. PARALLEL VECTORS
S l ti
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Geometry 9
Ch e c k Y o u r s e l f
1 Show that
u = (2, 3) and
v = (4, 6) are parallel.
2 Find the value of k ifu = (1, k) and
v = (3, 6) are parallel.
A n s w e r s
1 check the scalar components 2 k = 2
AB = (2 3, 3 (1)) = (1, 4)
CD = (x 5, y + 4)
CD ||
AB so
4x 20 = y 4, so 4x + y 16 = 0.
5 +4= .
1 4
x y
Solution
Let us write E(x, y) and F(m,n).
BE =
EA
BE = (x + 3, y + 2)EA = (3 x, 2 y)
x + 3 = 3 x
2x = 0
x = 0
y + 2 = 2 y
2y = 0
y = 0
and
CF =
FA
CF = (m 2,n + 3)FA = (3 m, 2 n)
m 2 = 3 m
2m = 5
m =
n + 3 = 2 n
2n = 1
n =1
2
5
2
Solution
EXAMPLE 31 A triangleABC has verticesA(3, 2), B(3, 2), and C(2, 3). E and F are the midpoints ofsidesAB andAC respectively. Find the coordinates of E and F.
emember
If two parallel vectorsa
andb have at least one
point in common, thena and
b lie on the same
straight line (they are
collinear).
For instance, if
AB = k
BC for some
k \ {0}, then A, B,
and C are collinear.
Therefore the coordinates are E(0, 0) and5 1
( , ).2 2
F
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Analysis of Vekctors Analytically
EXERCISES 1.2
A. Basic Concepts of Vectors in the
Analytic Plane
1. Plot the points A(1, 1), B(2, 1), C(3, 1), andD(3, 1) in the plane.
3. Find the length of
MN givenM(1, 4) and N(2, 1).
2. Sketch the position vector of the vector with the
given endpoints.a b c
EF
E(0, 3)
F(4, 2)
CD
C(1, 5)
D(0, 2)
AB
A(2, 3)
B(4, 1)
5. Describe the vector with initial point P and
terminal point Q.
4. The figure shows the
vectorsu and
v. Sketch
the following vectors.
a 2v b
u
c
u +
v d
u 2
v
e 2u +
v
c P(3, 2), Q(8, 9) d P(1, 3), Q(1, 0)
a b
Project:
Use The Geometers Sketchpad,
Cabri Geometry, or Javascript
sketchpad to sketch the vectors
2u,
v,
u +
v,
u 3
v and
PQ, QP if P(3, 4) and Q(4, 3).
We can use computer applications such as TheGeometers Sketchpad, Cabri Geometry, or
Javascript sketchpad to sketch vectors and solveproblems. We can use an application to aproblem, and then change certain values to seetheir effect. We can also use a computer
application to add and subtract vectors, and tomultiply a vector by a scalar.
The screen opposite shows a simple problem illustrated usingThe Geometers Sketchpad. Suppose a current flows at acertain velocity w downstream. A boat moves at a constantspeed v. Which direction the boat take in order to reach theother side of the river in the shortest possible time?
The Geometers Sketchpad calculates the time as we move thepoint H on the screen to set the direction of the boat. Using theSketchad we can move H to find the shortest possible time inthe problem.
6 Find the vectorsa and
b if 2
a 3
b (4 2) and
11 Given
AB 5i + 6
j and B( 4 8) find the
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Geometry 9
6. Find the vectors a and b if 2a 3b = (4, 2) anda +
b = (2, 1).
B. Vector Operations
7. Find u + v, u v, 2u, 2u v, u + 3v, and
4u + 3
v for the given vectors
u and
v.
a
u = (2, 6),
v = (1, 3)
b
u = (2, 3),
v = (8, 2)
c
u = (1, 0),
v = (0, 2)
d
u =
i
v =
j
e
u = 3
+
j
v =
i j
f
u = 7
i + 5
j
v =
j
i
8. Find |u|, |
v|, |2
v|, |
v|, |
u +
v|, and |
u
v|
for the given vectorsu and
v.
a
u = 3i + j v = i + 2j
b
u = 2
i
j
v =
i
j
c
u = (2, 3),
v = (0, 1)
d
u = (3, 4)
v = (2, 5)
1
3
9. Find the horizontal and vertical scalar components
of the vector with the given length and angle with
the positive x-axis. Express the vector in terms of
standard base vectors.
a |v | = 20, = 30
b |v | = 30, = 120
c |v | = 1, = 225
d |u | = 80, = 135
e |v | = 4, = 10
f |u | = 3, = 300
10. Given
AB = 7i + 2
j and B(3, 11), find the
coordinates of pointA.
11. Given AB = 5 i + 6j and B(4, 8), find the
coordinates of pointA.
12.u = 3
i + 4
j and
v = 4
i +
j are given. Which
vector is the longest?
13. Givenu = 3
i + 4
j, calculate |
u
i|.
1
2
14. Show thatu = (a, b) and
v = (2a, 2b) are
parallel.
16. In a triangle ABC, the vertices are A(2, 3),
B(0, 1) and C(4, 1). Points D(1, 2) and E(1, 2)
are on the sidesAB andAC respectively.
Show that
DE ||
BC.
17. In a triangle ABC, the vertices are A(1, 3),
B(2, 1), and C(3, 2). E( , 2) is on the sideAB.
Find the coordinates of F if F is on AC and
EF ||
BC.
3
2
15.u ||
v,u = (1, k 3), and
v = (k, k 4) are given.
Find the value of k.
C. Parallel Vectors
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Geometry 9
We have seen how to add and subtract vectors, and how to multiply vectors by a scalar.In this section we will introduce another operation on vectors, called the dotproduct.
The dot product is sometimes called the scalar product because the resulting product is a
number and not a vector. It has applications in mathematics, as well as in engineering and
physics.
A. DOT PRODUCT
Definitiondotproduct scalarproduct,orEuclideaninnerproduct)
Letu = (u1, u2) and
v = (v1,v2) be two vectors in the plane.
The dotproduct ofu and
v, denoted by
u
v, is defined by
= 11 + 22.
Thus, to find the dot product of two vectors we multiply the corresponding scalar components
and then add them together.
a.
u
v = 25 + 34 = 10 + 12 = 2
b.
u
v = 15 + 11 = 5 + 1 = 4
Solution
EXAMPLE 32 Find the dot product of the given vectors.a.
u = (2, 3) and
v = (5, 4) b.
u =
i +
j and
v = 5
i +
j
Proof1.
u
v = u1v1 + u2v2 =v1u1 +v2u2 =
v
u , by the commutative property of real numbers.
2.
u(
v +
w ) = (u1, u2)(v1 +w1, v2 +w2) = u1(v1 +w1) + u2(v2 +w2)
= u1v1 + u1w1 + u2v2 + u2w2
= u1v1 + u2v2 + u1w1 + u2w2
=u
v +
u
w
1. Properties of the Dot ProductThe definition of the dot product gives us the following properties.
1.
u
v =
v
u (commutative property)
2.
u(
v +
w ) =
u
v +
u
w (associative property)
3. c(u
v ) = (c
u)
v
4.
u
u = |
u|2
5.
u
v 0, and
u
u = 0 if and only if
u =
0.
3. c(u
v ) = c(u1v1 + u2v2) = cu1v1 + cu2v2 = (c
u )
v
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The Dot Product of Two Vectors
u
u = 22 + (2) (2) = 4 + 4 = 8. Therefore, |
u| = 8.Solution
EXAMPLE 33 Find the length of the vectoru = (2, 2) by using the dot product.
Ch e c k Y o u r s e l f
1. Find the dot product of
u = 3
j and
v =
i +
j.2. Find the quantity (
u +
v)(
u 3
v ) given
u =
j ,
v =
i.
A n s w e r s
1. 3 2. 2
Definition anglebetweentwovectors
Letu =
OE andv =
OFbe two non-zero vectors. The angle EOF
is called the angle between and . We use to represent the
smaller angle betweenu and
v when their initial points coincide.
1. Angle Between Two Vectors
Proof The proof is a nice application of the law of cosines.
As we know, |v
u|2 = (
v
u)(
v
u) =
v 2 +
u 2 2
v
u (1).
Applying the cosine law to triangle EOF in the figure,
|v
u|2 = |
u|2 + |
v|2 2|
u||
v| cos (2) (0 < < )
(v
u)(
v +
u) = |
u|2 + |
v|2 2|
u|2|
v|2cos
Theorem
Let be the angle measure between two non-zero vectorsu and
v. Then
u
v = |
u||
v|cos.
dotproducttheorem
B. ANGLE BETWEEN TWO VECTORS
cosinelaw:
a2 = b2 + c2 2bccos A
( ) ( 1 1 2 2) 1 1 2 2 ( )
4.
u
u = u1u1 + u2u2 = u
2
1 + u2
2 = |u|2, so
5. This proof is left as an exercise for you.
= |u|.u u
By (1) and (2),
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Geometry 9
EXAMPLE 34 Givenu = (2, 2),
v = (5, 8) and
w = (4, 3), find the following.
a.
u
v b. (
u
v )
w c.
u (2
v ) d. |
w|2
a. By the definition of the dot product we haveu
v = 2 5 + (2)8 = 6.
b. Using the result from (a) we have (u
v )
w = 6
w = 6( 4, 3) = (24, 18).
c. By property 3 of the dot product we have u (2v) = 2 (u v) = 2(6) = 12.
d. By property 4 of the dot product we have |
w|2 =
w
w = (4)2 + 32 = 25.
Solution
v 2 2
v
u +
u2 = |
u|2 + |
v|2 2|
u||
v|cos
2 v
u = 2|
u||
v|cos
Therefore,
u
v = |
u||
v|cos
.
One of the most important uses of the dot product is to find the angle between two vectors if
the scalar components of the vectors are given. We simply calculate the angle by solving the
equation given by the dot product theorem for cos . Let us state this important result clearly.
If is the angle measure between two non-zero vectorsu and
v then .
u
v
cos = ---------------------|
u||
v|
FINDING THE ANGLE BETWEEN TWO NON-ZERO VECTORS
a.
u
v = 2 5 + 5 2 = 20
|u| =
|v| =
by the formula, and so
b.
u
v = 1 2 + 2 22 = 2 + 4 = 6
|u| =
|v| =
Therefore, = 0.
6 6
cos = = =13 12 36
4+8 = 12
21 +2 = 3
20
cos = .29
20
cos =29 29
2 25 + 2 = 29
2 22 + 5 = 29
Solution
EXAMPLE 35 a. Find the cosine of the angle between the vectorsu = (2, 5) and
v = (5, 2).
b. Find the angle between the vectorsu = (1, 2) and v = (2, 22).
Ch e c k Y o u r s e l f
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The Dot Product of Two Vectors
Definition perpendicular orthogonal)vectors
Two non-zero vectorsu and
v are perpendicular if and only if the
angle measure between them is 90.
By the dot product thorem,u
v = |
u||
v|cos. We can conclude that for two non-zero
vectors
u and
v ,
u
v is zero if and only if equals 90.
2.Perpendicular and Parallel Vectors
Theorem
Two non-zero vectorsu and
v are perpendicular if and only if
u
v = 0.
u
v = 61 + (2)3 = 0So the vectors are perpendicular.
Solution
EXAMPLE 36 Are the vectorsu = (6, 2) and
v = (1, 3) perpendicular?
u
v = 32 + 5(6) = 6 30 = 24. So
u and
v are not perpendicular.Solution
EXAMPLE 37 Are the vectorsu = (3, 5) and
v = (2, 6) perpendicular?
1. Find the dot product ofu = (0, 3) and
v = (1, 2).
2. Find the length ofu = (1, 3) using the dot product.
3. Find (u
v )
w given
u = (0, 3),
v = (1, 2), and
w = (2, 1).
4. Find the angle betweenu = (2, 1) and
v = (1, 3).
A n s w e r s
1. 6 2. 10 3. (12, 6) 4. = 45
From the definition of parallel vectors we know thatu ||
v if and only if
u = k
v . We can
write,
|u| = |k||
v| (1)
u
v = k
v
v =k|
v|2 (2)
u
v = |
u||
v|cos (3).
From (1), (2) and (3),
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Geometry 9
|k||v||
v|cos = k|
v|2
cos =
cos = 1 or cos = 1 = 0 or = 180.
From these results we can conclude that two vectors are parallel if and only if the angle
measure between them is 0 or 180.
| |
k
k
Con c l u s i o n
Let be the angle measure between nonzero vectors
u and
v.
Then
|| ifandonlyif
=|
|
|
|or
=|
|
|
|.
Letv = (v1,v2) be perpendicular to
u.
u
v = 4v1 + 2v2 = 0
v2 = 2v1
Ifv1 = t R,v2 = 2t, thenv = (t, 2t).
Let t = 1 or 3:v = (1, 2) and
v = (3, 6) are both perpendicular to
u.
Solution
EXAMPLE 38 Determine two vectors in the plane which are perpendicular tou = (4, 2).
SoAD
BC and
AD
BC = 0.
Solution
EXAMPLE 39 In an equilateral triangleABC, D is the midpoint of BC. Find
AD
BC.
In an equilateral triangle, the median is also the
altitude, as show in the diagram.
EXAMPLE 40 In a squareABCD, E is the midpoint of side BC and |BC| = 4 cm. Find
AE
AB.
AB =a
b
Solution
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The Dot Product of Two Vectors
A(ABC) =
AH = (x0 2, y0 3)
BC = (3, 1)
AH
BC = 3(x0 2) + (y0 3) = 0
3x0 + y0 6 3 = 0
3x0 + y0 = 9
| | | |2
AH BC
AH = (0.4, 1.2)
|
AH| = 1
0.16+1.44 = 1.6 = 410
BC = (3, 1)
|
BC| = 9+1 = 10
2x0 x0y0 = 3y0 3 x0y0 + x0
x0 = 3y0 3
x0 3y0 = 3
3x0 + y0 = 9 (1)
x0
3y0
= 3 (2)
9x0 + 3y0 = 27
x0 3y0 = 3
10x0 = 24
x0 = 2.4 and y0 = 1.8.
BH = k
HC
BH = (x0, y0 1)
HC = (3 x0, 2 y0)
Solution
0 0
0 0
1=
3 2
x y
x y
A(ABC) = = 1 4
10 22 10
AE =a +
AE
AB =
a(
a + ) =
a 2 + =
a 2
Sincea
b,
a 2 = |
a|2 = 42 = 16.
2
a b
2
b
2
b
EXAMPLE 41 Find the area of the triangle with vertices A(2, 3), B(0, 1), C(3, 2).
+
Let us multiply (1)by 3. Then,
Theorem triangleinequality
If
d
i h l h Thi i ll d h
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Geometry 9
If u and v are vectors in the plane, then |
+ | |
|+|
|. This property is called the
triangleinequality.
By the property of the dot product, |
u +
v|2
= (
u +
v )2
=
u2
+ 2
u
v +
v2
.2
u
v 2|
u||
v|
{|u|2 + |
v|2 + 2
u
v } {|
u|2 + |
v|2 + 2|
u||
v|}
|u +
v|2 (|
u| + |
v|)2, since both |
u| + |
v| and |
u +
v| are non-negative.
Therefore, |u +
v| |
u| + |
v|.
Proof
Theoremu and
v are perpendicular in the plane if and only if |
u +
v|2 = |
u|2 + |
v|2.
Proof |u +
v|2 = (
u +
v )2 =
u 2 + 2
u
v +
v 2 = |
u |2 + 2
u
v + |
v |2
= |u |2 + |
v |2, since
u
v = 0.
|2u 3
v|2 = (2
u 3
v)2 = 4
u2 + 9
v2 12
u
v
= 4|u|2 + 9|
v|2 12|
v||
v|cos
= 432 + 942 1234
= 36 + 144 72
= 108.
Now |2u 3v|2 = 108, and so|2u 3v| = 108 = 63.
1
2
Solution
EXAMPLE42 |
u|= 3, |
v|= 4, and the angle between
u and
v is 60. Find |2
u 3
v|.
Let us choose avariable point B(x, y):
AB = (x + 1, y 3).
AB and
n are perpendicular, so
ABn = 0.
3(x + 1) + 5(y 3) = 0
3x + 3 + 5y 15 = 0
3x + 5y 12 = 0 is the required equation.
Solution
EXAMPLE 43 Find an equation for the line passing throughA(1, 3) which is perpendicular ton = (3, 5).
Ch e c k Y o u r s e l f
Fi d th l f if
( 1) i di l t
(3 4)
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The Dot Product of Two Vectors
1. Find the value of a if u = (a, 1) is perpendicular to v = (3, 4).
2. In the right triangleABC,AC is the hypotenuse, BHAC, and H lies onAC.
Find (
HC +
CB)(
AB +
BH).
3. Find an equation for the line passing through P(3, 1) which is perpendicular ton = (3, 1).
A n s w e r s
1. 2. 0 3. 3x + y 8 = 0 4
3a
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The Dot Product of Two Vectors
ACROSS
1. A vector with length zero.
5. An undefined concept in geometry which has noshape or size.
7. Being the main or most important part.
10. Being parallel.
12. The coordinate representing the position of a pointalong a line perpendicular to the y-axis in a plane.
13.A thin straight stick with a point, sometimes used toshoot things.
15.A mathematical statement that establishes theequality of two expressions.
16.Any of the four areas into which a plane is dividedby the reference axes in a rectangular coordinatesystem, designated first, second, third, and fourth,counting counterclockwise from the area in which
both coordinates are positive.
17. One of the two horizontal or vertical vectors whosesum is equal to a given vector.
18. In a triangle, the formula cosA = (b2 + c2 a2)/2bc.
21. Vectors which have the same direction and length.
24. To arrange in sets of two.
25. The act or process of adding.
26. The direction 90 clockwise from north.
27. Either of two points marking the end of a linesegment.
DOWN
2. A single vector that is the equivalent of a set ofvectors.
3. The measure of heat of a body or environment.
4. The principal structural member of a ship.
6. The solution of an equation in which every variableis equal to zero.
7. An acute angle measured from due north or duesouth.
8. The size of a flat surface, calculated mathematically.
9. The property that states: if a, b A, then a b Afor an operation.
11.A point whose position is constant.
14. The direction 270 clockwise from due north anddirectly opposite east.
19.A picture such as a pie chart or bar graph, used toillustrate quantitative relationships.
20. To give the meaning of (a word or idea).
22.An undefined concept in geometry that describes aset of points along a path.
23.A relatively small, usually open craft used byfishermen.
EXERCISES 1.3
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Geometry 9
A. Dot Product
1. Find the dot product of each pair of vectors.
a.
u = (2, 1) v = (3, 6)
b.
u = 7
i + 24
j,
v = 3
i
j
c.
u = 3i + j, v = 3i j
d.
u =
i
j,
v = 2
i
e.
u = 19
j,
v = 16
j
2. Find the following quantities given
u =
i + 2
j,v = 3
i +
j , and
w = 4
i + 3
j.
a.
u
w +
u
w
b.
w(
u +
v )
c. (u +
w )(
u
w)
d. 2(u
v )
e. (
w
v )(
w +
u )f. (3
u 2
v )(
u + 2
v )
d.
u = 2
i 8
j ,
v = 12
i + 3
j
e.
u = 3
i 4
j ,
v = 8
i 6
j
f.
u = 4i , v = 3i
8. Find two perpendicular vectors to each given
vector.
a.
a = 3
i
j b.
b = 8
i 6
j c.
c =
i + 2
j
9. For what values of t areu = 4
i 5
j t and
v = 3
i 2
j perpendicular?
10. |u| = 5 and |
v| = 3 are given. For what values
of t areu + t
v and
u t
v perpendicular?
12. Show that the equality |
u +
v| = |
u
v| holdswhen
u and
v are orthogonal.
11. For what values of t doa = t
i +
j and
b =
i + t
j have angle measure 120 between
them?
7. Find the measure of the interior angles of the triangle
ABC with verticesA(1, 3), B(1, 2), and C(2, 2).
4. Evaluate the following quantities if the angle measure
betweenu and
v is 60 and |
u| = 4, |
v| = 3.
a.
u
v b. |
u +
v|2
c. |u
v|2 d. (3
u + 2
v ) (
u + 2
v )
5. Evaluate |u +
v| given |
u| = 13, |
v| = 5, and
|u
v| = 12.
3.u,
v and
w are unit vectors such that
u +
v +
w = 0. Find
u
v +
v
w +
w
u = 0.
13. Verify the equality
|u +
v|2 + |
u
v|2 = 2(|
u|2 + |
v|2).
B. Angle Between Two Vectors
6. Determine whether the given vectors are
perpendicular or not.
a.
u = (4, 6),
v = (3, 2)
b.
u = (5, 0),
v = (0, 4)
c.
u = 3
i ,
v =
j
14. In a rhombusABCD, one side measures 6 cm and
E, F are the midpoints of sides AD and DC
l Th l b d AD
Find the area of a triangle with vertices
A(2, 2), B(0, 2), and C(1, 4).
21.
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The Dot Product of Two Vectors
respectively. The angle measure between sidesAD
and DC is 120. Find the quantity
BE
AF.
15. In a rectangleABCD,
DC = 3
AD and point E is on
DC. Find the quantity
AE
BE given
DE = 2
EC
and |
AD| = 3 cm.
16. Show that the inequality |u +
v| > |
u
v| holds
when the angle measure betweenu and
v is less
than 90.
17. Find two unit vectors that make angles of 45
with i +
j.
19. Write the equation of the line passing through
A(1, 1) which is perpendicular to
u = (3, 4).
18. Letu and
v be vectors and let be a scalar. Verify
the given properties.a.
u
v =
v
u
b. (u )
v = (
u
v ) =
u(
v )
Show that the diagonals of a rhombus are
perpendicular using vectors.
20.
Find the area of a rhombus with vertices A(2, 0),
B(3, 3), C(8, 0), and D(3, 3).
22.
Find the area of a rectangle with verticesA(3, 2),
B(9, 2), C(9, 5), and D(3, 5).
23.
Show that the altitudes of an acute-angled triangle
are concurrent using vectors.
25.
Find the distance of P(x0, y0) from the line
ax + by + c = 0 using vectors.
26.
Find the area of a parallelogram with vertices
A(2, 1), B(6, 0), C(8, 3), and D(4, 2).
27.
For any vectors
u,
v, and
w prove that(
u
w )
v
u(
v
w) is perpendicular to
w.
24.
CHAPTER SUMMARY
A li g t ith di ti i ll d di t d li
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Geometry 9
A line segment with direction is called a directed line
segment.
A vector in the plane is a directed line segment.
AdditionofTwoVectorsGeometrically
For two vectors
PO and
QR in a plane,
PQ +
QR is the
sum ofPQ and
QR. There are two ways to add vectors
geometrically: the polygon method and the parallelogram
method.
ThePolygonMethod
In this method we draw the first vector. Then we place
the initial point of the second vector at the terminal point
of the first, the initial point of the third vector at theterminal point of the second, and so on until we place the
initial point of the nthvector at the terminal point of the
(n 1)th vector. The vector whose initial point is the
initial point of the first vector and whose terminal point
is the terminal point of the last vector is the resultant
(sum) vector.
TheParallelogramMethod
In this method we draw the first vector, and then draw
the second vector with its initial point at the initial pointof the first vector. We make a parallelogram by drawing
two additional sides, each passing through the terminal
point of one of the vectors and parallel to the other
vector. The sum is drawn along the diagonal from the
common initial point to the intersection of the two lines.
MultiplicationofaVectorbyaScalar
For a real number a and a vectoru:
If a > 0, then the vector au has the same direction
to u and the length |au| = a|u|.
If a < 0, then the vector au has the opposite
direction tou and the length |a
u| = |a||
u|.
If a = 0, then au =
0.
ParallelVectors
Leta and
bbe two vectors.
a and
b are parallel if and
only ifa = k
b where k 0.
Non-ParallelVectorIf
a is not parallel to
b, then h
a =k
b when h = k = 0.
ComponentsofaVectors
A vector in a plane has two components, called the
horizontal and vertical components of the vector.
Let |a| = (a1, a2) be a vector, then |
a| = 2 2
1 1+ .a a
EqualVectors
Two vectors are equal if and only if their correspondingcomponent vectors are equal.
In other words, for a = (a1, a2) and b = (b1, b2), thena =
b if and only if a1 = b1 and a2 = b2.
StandardBaseVectorsi = (1, 0) and
j = (0, 1) are called the standard base
vectors.
LinearCombinationofVectors
Letu1,
u2, ... ,
ukbe vectors in a plane and let c1, c2, ... ck
be scalars.v = c1
u1 + c2
u2 + ... + ck.
uk is called a
linear combination of vectors.
DotProduct
The dot product ofu = (u1, u2) and let
v = (v1,v2) is the
scalar quantityu
v = u1v1 + u2v2.
AngleBetweenTwoVectors
Let be the angle between two non-zero vectorsu and
v.
1. u ||
v if and only if
u
v = |
u||
v| or
u
v = |
u||
v| because
= 0 or
= 180.2.
u and
v are perpendicular if and only if
u
v = 0.
1. Can we use directed line segments in traffic? Give anexample.
2. What is the difference between a vector and a scalar?
3. Is it possible to add a hundred vectors in a plane using thepolygon method or the parallelogram method?
4. How do you change the direction of a vector using a realnumber?
5. a. Draw a diagram to show how to add two vectors.
b. Draw a diagram to show how to subtract two vectors.
6. Can you equalize two non-parallel vectors usingmultiplication by two real numbers?
7. In a plane, how many components do you need to resolvea vector?
8. How many standard base vectors are there in the plane?
9. What is the difference between the inner product and dotproduct of two vectors?
10.How do you use the dot product to find the angle measurebetween two vectors?
11.How do you use the dot product to determine whether twovectors are parallel or not?
12.How do you use the dot product to determine whether twovectors are perpendicular or not?
Concept Check
CHAPTER REVIEW TEST 1A
1 I t i l ABC G i th t id D ib
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Chapter Review Test 1A
5. (1, 4) +v = (4, 5) is given. Find
v.
A) (3, 9) B) (5, 1) C) (5, 1)
D) (1, 5) E) (9, 3)
8. Find the unit vector with opposite direction to
AB givenA(6, 3), and B(2, 6).
A) B) C)
D) E) 4 3
( , )5 5
4 3
( , )7 7
4 3
( , )7 7
4 3
( , )5 5
4 3
( , )5 5
3. In the figure,
|AB| = |BC| = |CD|
and the vectorsa,
b,
c,
d
are given. Which one of
the following is the linearcombination of
c in terms of
a and
d?
A) B) C)
D) E)
a + 2
d
-------------------2
a + 2
d
-------------------3
2a +
d
-------------------3
2a +
d
-------------------2
a +
d
----------------2
7.u = 2
i + 3
j and
v =
i 2
j are given.
Find 3u 2
v.
A) 4i + 5
j B) 8
i + 5
j
C) 8i + 13
j D) 4
i + 13
j
E) 4i + 13
j
1. In triangle ABC, G is the centroid. Describe
AG +
GC +
CA.
A)0 B)
AG C)
BG D) 2
CA E)
AC
10. Describe the unit vectora in the
figure.
A) i + j B) i + j
C)
i j D)
i +
j
E)
i3
2
3
2
1
2
3
2
1
2
12
3
2
12
3
2
4.
AD =
AB and
BE =
BC are given. Express
DE in terms of
AB and
BC.
A)
AB +
BC B)
AB +
BC
C)
AB
BC D)
AB
BC
E)
BC
AB
3
4
1
2
1
2
1
4
1
2
3
4
1
2
1
4
1
2
3
4
12
14
2. In the figure, K, L,M, N, P, R are the midpoints of
the sidesABCDEF
respectively. Describe
KB+
LC+
MD+
NE+
PF.
A)
AR B)
RA C)
FK D)
FA E)
PA
6. Find the coordinates of B if
AB = (7, 3) and
A(2, 1).
A) (9, 4) B) (9, 2) C) (9, 4)
D) (9, 4) E) (9, 2)
9.u =
i 5
j and
v = 2
i + 3
j are given. Find
3u + 2
v.
A) 3i 2
j B) 5
i
j C) 7
i 9
j
D) 2i + 3
j E) 8
i
j
16. ABCD is the rectangle in
the figure.
11. For how many values ofm > 0 area = (2, 1 m)
andb = (m + 1, 4) parallel?
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Geometry 9
15. Which one of the following is perpendicular tov = (3, 4)?
A) B) (4, 3) C)
D) E) 4 3
( , )5 5
3 4
( , )5 5
3 4( , )5 3
4 3
( , )5 5
FindABAC if
|AB| = 4 and
|BC| = 3.
A) 9 B) 12 C) 15 D) 16 E) 20
18. G is the centroid of a triangle with vertices
A(1, 1). B(4, 2), and C(0, 6). Find the scalar
product of
AB
BG.
A) 8 B) 14 C) 0 D) 1 E) 12
20. ABC is a triangle with
|AB| = 4,|AC| = 5,
|BC| = 6. Find
AC (
AB +
BC).
A) 20 B) 24 C) 25 D) 30 E) 50
17. In the figure, D and E
are the midpoints of
the sides of triangle
ABC.
Find (
AD
AE)
BC.
A) 100 B) 150 C) 0 D) 50 E) 100
19. A triangleABC has verticesA(1, 1), B(4, 2), and
C(0, 6). Find the scalar component of
AB alongBC.
A) B) C) D) E) 3
2
3
4
3
4
4
2
4
2
12. Which one of the following is true for vectorsa = (3, 6),
b = (6, 7) and
c = (9, 13)?
A)a = 4
b + 3
c B)
a = 3
b + 4
c
C)
a = 4
b 3
c D)
a = 3
b 4
c
E)a =
b +
c
A) 1 B) 2 C) 3 D) 4 E) 5
14. A(2, 5), B(1, 3), C(m, 6), and
AB
BC are given.
Findm.
A) 3 B) 2 C) 1 D) 2 E) 3
13. Find the measure of the angle between the unit
vectors
a and
b in degrees if their dot product is
A) 45 B) 60 C) 120 D) 135 E) 150
1.2
CHAPTER REVIEW TEST 1B
1 |
| |
|
6 h h l A C A [AC]
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Chapter Review Test 1A
1. |a| = 3 |b| = 5, and the angle between
a and
b is 60. Find |
a +
b|.
A) 7 B) 8 C) 9 D) 10 E) 11
7. What is the angle betweena = (1, 3) and
b = (4, 43) in degrees?
A) 60 B) 120 C) 135 D) 150 E) 180
8. A squareABCD has side
3 cm, and K, L are the
midpoints of sidesAB and
BC respectively. P and Q
trisectAC as shown in the
figure. FindDP
PQ.
A) 0 B) 1 C) 1 D) 2 E) 3
9.a,
b, and
c are three vectors such that
b =
a 2
c,
b
c, and |
a| = 4|
c|. Find the angle between
a and c in degrees.
A) 30 B) 45 C) 60 D) 75 E) 90
10. Which one of the following is false for two vectorsa,
b and a scalar k ?
A) If a b, ab = 0.
B) Ifa ||
b,
b
a = 0.
C)a
b =
b
a
D) (ka + k
b) = k(
a +
b)
E) (kak
b) = k2
a
b
3. Given |a| = 12, |
b| = 5, and |
a
b| = 8, find
cos(a,
b).
A) B) C) D) E) 12
13
5
13
3
4
7
8
3
8
4. In an equilateral triangleABC, D [AC], |BC| = 6,
and |CD| = 2. Find
BC
BD.
A) 30 B) 24 C) 20 D) 18 E) 12
5. |a| = 7, |
b| = 10, and |
a +
b| =73 are given.
Find |a
b|.
A) 3 B) 8 C) 13 D) 15 E) 17
2. In the figure,
|AB| = |AC| = 4,
mA = 120, and
|AD| = |DC|.
Find
BC
BD.
A) 48 B) 36 C) 30 D) 24 E) 18
6. In the right triangleABC, mA = 90, D [AC],
and |AB| =k. Find
BA
BD.
A)k B)k2 C) D) E) 2k22
k2
2
k
11. In the figure, ABCD is a
parallelogram and
|BC| = 1,
16. In a triangle ABC,
AB = (4, 2a),
AC = (a, 4)
and the length of
BC is 10 cm. Find a possible
value of a
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Geometry 9
| |
|DC| = 2,
|DE| = |EC|. Find(
BC +
CE)(
AD +
DE).
A) 0 B) 1 C) 2 D) 3 E) 4
12. In an equilateral triangle ABC, D and E are the
midpoints of [AC] and [BC] respectively.
Find
DC (
BA +
AE).
A) |
AB|2 B) |
EB|2 C) |
DC|2
D) |
DC|2 E) |
EB|1
2
1
2
1
2
13. In an equilateral triangle ABC with side 1 cm,
what is
BC
CA?
A) B) C) D) 2 E) 3
2
3
2
1
2
1
2
14. For what values of k are the vectorsa = (12k, 9)
and b = (4, 3) linearly dependent?
A) 3 B) 2 C) 1 D) 0 E) 1
value of a.
A) 1 B) 2 C) 3 D) 4 E) 5
18.a +
b = (1, 3) and 2
a +
b = (4, 6) are given.
Find the measure of the angle betweena and
b.
A) 30 B) 45 C) 90 D) 135 E) 150
19. In a triangleABC,
AB = (2, 5) and
AC = (2, 2). Find the length of
BC.
A) 1 B) 2 C) 3 D) 4 E) 5
20.a = (12, 5), and
b = (3, 4) are given. Find
sin(a,
b).
A) B) C) D) E) 4
13
10
13
63
65
12
13
5
13
17. In a squareABCD
with side 2 cm,
|AE| = |ED| and
|DF| = |FC|.Find
EF (
EA +
AB).
A) 1 B) 2 C) 2 D) 1 E) 0
15. In a square ABCD, the
side is 4 cm,
[AB] [EP], [PD] [PC],and |AE| = |EB|.
Find
PE (
PC +
PD).
A) 16 B) 12 C) 10 D) 8 E) 6
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Geometry 9
A. THE CONCEPT OF CONGRUENCE
1. Congruent Figures and Polygons
The world around us is full of objects of various shapes and sizes. If we tried to compare some
of these objects we could put them in three groups:
objects which have a different shape and size,
objects which are the same shape but a different size, and
objects which are the same shape and size.
The tools in the picture at the right have different shape and size.
The pictures below show tools which have the same shape but different size. In geometry,
figures like this are called similarfigures. We will study similar figures in Chapter 3.
The pictures below show objects which are the same size and shape.
In this section, we will study figures which have this property.Factories often need toproduce many parts
with exactly the samesize and shape.
Definition congruentfigures
Figures that have the same size and shape are called congruentfigures. We say A is congruent
to B (or B is congruent toA) ifA and B are congruent figures.
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Congruence
The pictures at the bottom of the previous page show some examples of congruent objects.
The pictures below show two more examples. In these two examples there is only one piece
left to fit in the puzzle. Therefore, without checking anything, we can say that each piece and
its corresponding place are congruent.
Congruence in nature:the petals of this flower
are congruent.
Challenge
Remove five toothpicksto make five congruenttriangles.
In the figure below, ABC and DEF are congruent because their corresponding parts are
congruent. We can write this as follows:
A D AB DE
B E and BC EF
C F AC DF.
We can show this symbolically in a figure as follows:
Definition congruenttriangles
Two triangles are congruent if and only if their corresponding sides and angles are congruent.
We write C DEF to mean that ABC and DEF are congruent.
2. Congruent Triangles
EXAMPLE 1 Given that MNP STK, state the congruent angles and sides in the two triangles withoutdrawing them.
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Geometry 9
Solution M S MN ST
N T and NP TK
P K PM KS.
As we can see, the order of the vertices in congruent triangles is important when we are
considering corresponding elements. Any mistake in the ordering affects the correspondence
between the triangles.If two triangles are congruent then we can write this congruence in six
different ways. For instance, if ABC is congruent to DEF, the following statements are all
true:
ABC DEF
ACB DFE
BAC EDF
BCA EFD
CAB FDE
CBA FED.
A short history of the symbol:
GottfriedWilhelm
Leibniz
(1640-1716)
introduced for congruence
in an unpublished manuscript in
1679.
In 1777,
JohannFriedrich
Hseler
(1372-1797)
used (with the tilde reversed).
In 1824,
Carl randan
Mollweide
(1774-1825)
used the modern symbol for
congruence in Euclids Elements.
If two triangles are congruent then we can write this congruence in six different ways. For
instance, if ABC is congruent to DEF, the following statements are all true:
ABC DEF
ACB
DFE
BAC EDF
BCA EFD
CAB FDE
CBA FED.
EXAMPLE
2 Complete each statement, given that
PRS
KLM.a. PR _____ b. _____ K c. _____ SP
d. S _____ e. ML _____ f. L _____
Solution a. PR KL b. P K c. MK SP
d. S M e. ML SR f. L R
EXAMPLE 3 Decide whether or not the two triangles inthe figure are congruent and give a reason for
your answer.
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Congruence
y
Solution Let us calculate the missing angles:
m(C) = 60 (Triangle Angle-Sum Theorem in ABC)
m(M) = 30 (Triangle Angle-Sum Theorem in KMN)
Now we can write the congruence of corresponding parts:
AB KM (Given)
BC KN (BC = KN = 4)
AC MN (AC =MN = 8)
A M (m(A) =m(M) = 30)
B K (m(B) =m(K) = 90)
C N (m(C) =m(N) = 60)
Therefore, ABC MKNby the definition of congruent triangles.
EXAMPLE 4 ABC EFD is given with AB = 11 cm, BC = 10 cm and EF + ED = 19 cm. Find theperimeter of EFD.
Solution Since ABC EFD,AB = EF, BC = FD and
AC = ED by the definition of congruence.
So by substituting the given values we get
11 = EF, 10 = FD andAC = ED.
Since we are given that EF + ED = 19 cm,
we have 11 + ED = 19 cm; ED = 8 cm.
So P(EFD) =EF + ED + FD = 11 + 8 + 10 = 29 cm.
Ch e c k Y o u r s e l f
1. KLM XYZ is given. State the corresponding congruent angles and sides of the
triangles.
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Geometry 9
2. State the congruence JKM SLX in six different ways.
3. Triangles KLM and DEF are congruent. P(KLM) = 46 cm, the shortest side of KLM
measures 14 cm, and the longest side of the DEF measures 17 cm. Find the lengths of
all the sides of one of the triangles.
4. Triangles DEF and KLM are congruent. If DE = 12.5 cm, EF = 14.4 cm and the perimeter
of the triangle KLM is 34.6 cm, find the length of the side DF.
5. Two line segments KL andABbisect each other at a point T. IfAL = 7 and the lengths of
the segments KL andAB are 22 and 18 respectively, find the pe