3 APPLY
ASSIGNMENT GUIDE
BASIC
Day 1: pp. 576–577 Exs. 10–30
Day 2: pp. 578–580 Exs. 31–45,
48, 53–60, Quiz 3 Exs. 1–17
AVERAGE
Day 1: pp. 576–577 Exs. 10–30
Day 2: pp. 578–580 Exs. 31–48,
53–60, Quiz 3 Exs. 1–17
GUIDED PRACTICE
✓
Concept Check ✓
Vocabulary Check
1. The magnitude of a
vector is the distance
from its initial point to
its terminal point, and
the direction is the
angle the vector makes
with a horizontal line.
Skill Check
ADVANCED
Day 1: pp. 576–577 Exs. 10–30
Day 2: pp. 578–580 Exs. 31–60
BLOCK SCHEDULE
pp. 576–580 Exs. 10–48, 53–60,
Quiz 3 Exs. 1–17
✓
1. What is meant by the magnitude of a vector and the direction of a vector?
In Exercises 2–4, use the diagram.
See margin.
2. Write the component form of each vector.
Æ„ Æ„ Æ„ Æ„
HOMEWORK CHECK
To quickly check student understanding of key concepts, go over
the following exercises: Exs. 10,
16, 22, 28, 30, 32, 36, 44. See also
the Daily Homework Quiz:
• Blackline Master (Chapter 10
Resource Book, p. 11)
•
Transparency (p. 71)
B
3. Identify any parallel vectors. AB , PQ ; MN , UV
Æ„
V
Æ„
4. Vectors PQ and ST are equal vectors.
Æ„
2
M
1
x
U
N
Write the vector in component form. Find the magnitude of the vector.
Round your answer to the nearest tenth.
Æ„
2. AB = 具º2, º2典;
Æ„
MN = 具0, º3典;
Æ„
PQ = 具3, 3典;
Æ„
UV = 具0, 2典
5.
y
6.
B
7.
y
y
M
P
1
2
A
1
œ
1 x
1
x
N
x
具4, 5典; 6.4
具º4, º2典; 4.5
具2, º5典; 5.4
8. Use the vector in Exercise 5. Find the direction of the vector relative to east.
about 51° north of east
9. Find the sum of the vectors in Exercises 5 and 6.
具0, 3典
PRACTICE AND APPLICATIONS
STUDENT HELP
Extra Practice
to help you master
skills is on p. 820.
FINDING MAGNITUDE Write the vector in component form. Find the
magnitude of the vector. Round your answer to the nearest tenth.
10.
11.
y
12.
y
K
y
1
1 x
S
1
1
R
1
E
2
x
x
J
STUDENT HELP
HOMEWORK HELP
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
576
576
œ
P
Although ST is not shown, the coordinates
of its initial point are (º1, º1). Give the
coordinates of its terminal point. (2, 2)
1
EXERCISE LEVELS
Level A: Easier
10–12
Level B: More Difficult
13–48
Level C: Most Difficult
49–52
A y
Exs. 10–20
Exs. 21–24
Exs. 25–29
Exs. 31–40
Exs. 41–45
F
具4, 1典; 4.1
具º4, º4典; 5.7
具º3, 6典; 6.7
Æ„
FINDING MAGNITUDE Draw vector PQ in a coordinate plane. Write the
component form of the vector and find its magnitude. Round your answer
to the nearest tenth.
13. P(0, 0), Q(2, 7) 具2, 7典; 7.3
14. P(5, 1), Q(2, 6)
15. P(º3, 2), Q(7, 6) 具10, 4典; 10.8
16. P(º4, º3), Q(2, º7) 具6, º4典; 7.2
17. P(5, 0), Q(º1, º4) 具º6, º4典; 7.2
18. P(6, 3), Q(º2, 1) 具º8, º2典; 8.2
19. P(º6, 0), Q(º5, º4) 具1, º4典; 4.1
20. P(0, 5), Q(3, 5) 具3, 0典; 3
Chapter 9 Right Triangles and Trigonometry
具º3, 5典; 5.8
NAVIGATION The given vector represents the velocity of a ship at sea.
Find the ship’s speed, rounded to the nearest mile per hour. Then find the
direction the ship is traveling relative to the given direction.
21. Find direction relative to east.
about 61 mi/h; about 9° north of east
22. Find direction relative to east.
about 64 mi/h; about 51° south of east
y
y
10
A
T
x
20
E
TEACHING TIPS
EXERCISE 16 While direction is
not important in finding magnitude,
the component form depends on
the direction. Make sure the terminal point is Q in this diagram.
S
10
x
10
E
B
23. Find direction relative to west.
about 57 mi/h; 45° north of west
24. Find direction relative to west.
about 64 mi/h; about 39° south of west
y
M
y
10
W
30. In Round 2, since the
vectors have the same
magnitude and
opposite directions,
the rope was pulled in
opposite directions
with equal force, and
the match was a tie.
In Round 1, Team A
won since the
magnitude of their
force vector is greater
than the magnitude of
the force vector for
Team B.
L
W
x
10
10
x
10
P
PARALLEL AND EQUAL VECTORS
In Exercises 25–28, use the diagram
shown at the right.
magnitude but different directions.
y
F
D
H
25. Which vectors are parallel?
Æ
„ Æ„
Æ„
EF , CD , and AB
26. Which vectors have the same direction?
Æ
„
Æ„
EF and CD
27. Which vectors are equal?
Æ
„
Æ„
EF and CD
28. Name two vectors that have the same
Æ„
FOCUS ON
APPLICATIONS
O
1
E
C
A
G
x
1
K
J
B
Æ
„
GH and JK
TUG-OF-WAR GAME In Exercises 29 and 30, use the information below.
The forces applied in a game of tug-of-war can be represented by vectors. The
magnitude of the vector represents the amount of force with which the rope is
pulled. The direction of the vector represents the direction of the pull. The
diagrams below show the forces applied in two different rounds of tug-of-war.
Round 1
Round 2
Team A
A
RE
FE
L
AL I
TUG-OF-WAR
In the game
tug-of-war, two teams pull
on opposite ends of a rope.
The team that succeeds in
pulling a member of the
other team across a center
line wins.
Team B
C
B
A
center line
Æ
„
Team A
Team B
C
B
center line
Æ„
29. In Round 2, are CA and CB parallel vectors? Are they equal vectors? yes; no
30. In which round was the outcome a tie? How do you know? Describe the
outcome in the other round. Explain your reasoning. See margin.
9.7 Vectors
577
577
COMMON ERROR
EXERCISES 31–34 Students
may make mistakes adding vectors
if they do not pay attention to the
location of the initial point and the
terminal point. Remind them to pay
attention to the direction of the
vector when representing it with
components.
APPLICATION NOTE
EXERCISES 41–45 The forces
acting on a skydiver are gravity and
air resistance. The amount of air
resistance depends on the speed of
the skydiver and the cross-sectional
area of the skydiver. That is the
reason some skydivers position
themselves with their bodies
parallel to Earth. It increases their
cross-sectional area, which in turn
increases the air resistance. This
slows their descent.
„
„
PARALLELOGRAM RULE Copy the vectors u
and v
. Write the
„
„
component form of each vector. Then find the sum u
+v
and draw
„
„
the vector u + v .
Ex. 31–34, check students’
drawings.
31.
„
º4典;
6典;
u + v = 具5, 2典
„
1
u
33.
34.
43. about 126 mi/h; the
speed at which the
skydiver is falling,
taking into account the
breeze
FOCUS ON
APPLICATIONS
L
AL I
10
RE
„
v
E
DOWN
45.
UP
FE
W
„
s
„
u
1
„
v
„
x
v
„
1
u
1
A skydiver who has
not yet opened his or her
parachute is in free fall.
During free fall, the skydiver
accelerates at first. Air
resistance eventually stops
this acceleration, and the
skydiver falls at terminal
velocity.
x
„
„
„
ADDING VECTORS Let u
= 〈7, 3〉, v
= 〈1, 4〉, w
= 〈3, 7〉, and „
z = 〈º3, º7〉.
Find the given sum.
35. „
v+„
w 具4, 11典
„
具º2, º3典
36. „
u +„
v 具8, 7典
„
„
39. u + z
37. „
u+„
w
„
具4, º4典
UP
43. Find the magnitude of „
s . What
information does the magnitude give
you about the skydiver’s fall? See margin.
„
u
44. If there were no wind, the skydiver
would fall in a path that was straight
down. At what angle to the ground is
the path of the skydiver when the
skydiver is affected by the 40 mile
per hour wind from the west? about 71.6°
„
v
10
W
10
DOWN
See margin for graph. The new velocity is s = 具º30, º120典.
10
DOWN
578
E
578
具0, 0典
SKYDIVING In Exercises 41–45, use the information and diagram below.
the west at 30 miles per hour. Sketch
a new diagram and find the skydiver’s
new velocity.
10
具10, 10典
A skydiver is falling at a constant downward velocity of 120 miles per hour. In
the diagram, vector „
u represents the skydiver’s velocity. A steady breeze pushes
the skydiver to the east at 40 miles per hour. Vector „
v represents the wind
velocity. The scales on the axes of the graph are in miles per hour.
45. Suppose the skydiver was blown to
W
„
40. w + z
41. Write the vectors „
u and „
v in
„
u = 具0, º120典;
component form. „
v = 具40, 0典
42. Let „
s =„
u +„
v . Copy the diagram
and draw vector „
s . See margin.
SKYDIVING
x
y
1
38. v + z
10
1
x
u + v = 具3, º3典
UP
„
u
u
y
34. „
u = 具2, 3典;
„
v = 具1, º6典;
„
„
1
„
1
„
42.
v
v
u + v = 具6, 5典
33. „
u = 具2,
„
v = 具3,
„
„
y
„
31. „
u = 具4, 1典;
„
v = 具2, 4典;
„
„
32. „
u = 具º6, 2典;
„
v = 具5, 3典;
„
u +„
v = 具º1, 5典
32.
y
Chapter 9 Right Triangles and Trigonometry
E
46.
47.
Writing Write the component form of a vector with the same magnitude as
JK = 〈1, 3〉 but a different direction. Explain how you found the vector.
Æ„
Sample answer: 具º3, 1典; methods may vary.
LOGICAL REASONING Let vector „
u = 〈r, s〉. Suppose the horizontal and
the vertical components of „
u are multiplied by a constant k. The resulting
vector is „
v = 〈kr, ks〉. How are the magnitudes and the directions of „
u and „
v
related when k is positive? when k is negative? Justify your answers. See margin.
Test
Preparation
47. When k > 0, the
magnitude of „
v is k times
the magnitude of „
u and
the directions are the
same. When k < 0, the
magnitude of „
v is |k|
times the magnitude of „
u
and the direction of „
v is
opposite the direction of
u„. Justifications may vary.
48. b. about 10.2 mi/h at a
direction of about 11.3°
north of east
48. MULTI-STEP PROBLEM A motorboat heads due east across a river at a
speed of 10 miles per hour. Vector „
u = 〈10, 0〉 represents the velocity of the
motorboat. The current of the river is flowing due north at a speed of 2 miles
per hour. Vector „
v = 〈0, 2〉 represents the velocity of the current.
ENGLISH LEARNERS
EXERCISE 47 The structure of
the language in Exercise 47 may
cause difficulties for some English
learners. You may want to read the
three informational sentences
aloud, offering explanations as
necessary, and then have students
carry out the reasoning task
independently.
48a.
a. Let „
s =„
u +„
v . Draw the vectors „
u, „
v,
N
N
„
and s in a coordinate plane.
See margin for graph.
b. Find the speed and the direction of the
motorboat as it is affected by the current.
„
c. Suppose the speed of the motorboat is
„
s
2
u
10 mi/h
W
greater than 10 miles per hour, and the
speed of the current is less than 2 miles
„
v
per hour. Describe one possible set of
current
„
„
vectors u and v that could represent the
2 mi/h
velocity of the motorboat and the
velocity of the current. Write and solve a
Sample answer: „
u = 具12, 0典
word problem that can be solved by
and „
v = 具0, 1.75典; See margin
finding the sum of the two vectors.
for word problem.
2
„
v
„
u
E
S
★
Challenge
BUMPER CARS In Exercises 49–52, use the information below.
48. c. A motorboat heads due As shown in the diagram below, a bumper car moves from point A to point B to
east across a river at a point C and back to point A. The car follows the path shown by the vectors. The
speed of 12 mi/h. The magnitude of each vector represents the distance traveled by the car from the
current of the river is initial point to the terminal point.
flowing due north at a
speed of 1.75 mi/h. Find
C
the speed and the velocity
of the motorboat as it is
affected by the current of
the river. Answer: The
speed is about 12.1 mi/h,
and the velocity is given
60 ft
v =
by the vector u„ + „
B
具12, 1.75典.
52. The answer to Ex. 50 is a
vector which gives the final
position of the bumper car,
54 ft
while the answer to Ex. 51 A
is a number which gives the
total distance traveled by
Æ„
Æ„
49. Find the sum of AB and BC . Write the sum vector in component form.
the bumper car.
Æ„
50. Add vector CA to the sum vector from Exercise 49.
EXTRA CHALLENGE
www.mcdougallittell.com
ADDITIONAL PRACTICE
AND RETEACHING
具18, 60典
具0, 0典
51. Find the total distance traveled by the car. about 173 ft
52. Compare your answers to Exercises 50 and 51. Why are they different?
See margin.
9.7 Vectors
579
For Lesson 9.7:
• Practice Levels A, B, and C
(Chapter 9 Resource Book, p. 103)
• Reteaching with Practice
(Chapter 9 Resource Book, p. 106)
•
See Lesson 9.7 of the
Personal Student Tutor
For more Mixed Review:
•
Search the Test and Practice
Generator for key words or
specific lessons.
579
4 ASSESS
DAILY HOMEWORK QUIZ
Transparency Available
„
Draw vector PQ in a coordinate
plane. Write the component form
of the vector and find its magnitude. Round your answer to the
nearest tenth. Check students'
drawings.
1. P(–2, –3), Q(1, 4) 具3, 7典; 7.6
2. P(–4, 5), Q(3, –1) 具7, –6典; 9.2
„
Let „
a = 具4, –2典, b = 具–3, 1典, and
„
c = 具7, 2典. Find the given sum.
„
3. b + c„ 具4, 3典
4. „
a + c„ 具11, 0典
MIXED REVIEW
53. Since ™D and ™E are rt.
√ and all rt. √ are £,
™D £ ™E. Since
¤ABC
Æ
Æ
is
equilateral, AB £ BC .
Æ Æ
DE ∞ AC , so ™DBA £
™BAC and ™EBC £
™BCA by the Alternate
Interior Angles Thm. An
equilateral triangle is
also equiangular, so
m™BAC = m™BCA =
60°. By the def. of £ √
and the substitution prop.
of equality, ™DBA £
™EBC. ¤ADB £ ¤CEB
by the AAS Congruence
Thm. Corresponding parts
of
£ ◊Æ
are £ , so
Æ
DB £ EB . By the def. of
midpoint, B is
the
Æ
midpoint of DE .
53.
PROOF Use the information and the
diagram to write a proof. (Review 4.5)
Challenge problems for
Lesson 9.7 are available in
blackline format in the Chapter 9
Resource Book, p. 110 and at
www.mcdougallittell.com.
ADDITIONAL TEST
PREPARATION
1. WRITING Summarize how to
identify equal vectors and parallel vectors. Sample answer:
Two vectors are equal if they have
the same magnitude and direction. Two vectors are parallel if
they have the same or opposite
directions.
11.
1. a = 41.7, b = 19.4
m™A = 65°
2. y = 12, z = 17.0
m™Y = 45°
3. m = 13.4, q = 20.9
m™N = 50°
4. p = 7.7, q = 2.1
m™Q = 15°
5. f = 4.7, m™F = 37.9°
m™G = 52.1°
6. l = 12.0, m™K = 14.0°
m™L = 76.0°
Æ
Æ
¤ABC is equilateral; DE ∞ AC
Æ
A
PROVE 䉴 B is the midpoint of DE .
xy USING ALGEBRA Find the values of x and y. (Review 4.6)
54.
55.
yⴗ
56.
x = 45, y = 90
xⴗ
yⴗ
xⴗ
xⴗ
yⴗ
x = 120, y = 30
x = 30, y = 60
xy USING ALGEBRA Find the product. (Skills Review, p. 798, for 10.1)
57. (x + 1)2
x 2 + 2x + 1
58. (x + 7)2
x 2 + 14x + 49
59. (x + 11)2
60. (7 + x)2
49 + 14x + x2
x 2 + 22x + 121
Self-Test for Lessons 9.6 and 9.7
Solve the right triangle. Round decimals to the nearest tenth. (Lesson 9.6)
1.
2.
A
46
b
a
B
12
4. P
75ⴗ
q
5.
E
8
p
q
N
6. J
L
K
3
L
f
q
F
M
Z
y
6
R
40ⴗ
45ⴗ
X
m
16
z
25ⴗ
C
q
3.
Y
7. P(3, 4), Q(º2, 3) 具º5, º1典; 5.1
9. P(0, º1), Q(3, 4) 具3, 5典; 5.8
Æ„
7.6
12.4
G
8. P(º2, 2), Q(4, º3) 具6, º5典; 7.8
10. P(2, 6), Q(º5, º5) 具º7, º11典; 13.0
Æ„
11. Vector ST = 〈3, 8〉. Draw ST in a coordinate plane and find its direction
relative to east. (Lesson 9.7) See margin. about 69° north of east
2
x
„
„
„
Let u
= 〈0, º5〉, v
= 〈4, 7〉, w
= 〈º2, º3〉, and z„= 〈2, 6〉. Find the given sum.
(Lesson 9.7)
580
580
C
Æ„
T
2
E
Draw vector PQ in a coordinate plane. Write the component form of the
vector and find its magnitude. Round your answer to the nearest tenth.
(Lesson 9.7)
y
S
B
GIVEN 䉴 ™D and ™E are right angles;
QUIZ 3
EXTRA CHALLENGE NOTE
D
12. „
u +„
v 具4, 2典
13. „
v +„
w 具2, 4典
14. „
u+„
w 具º2, º8典
15. „
u+„
z 具2, 1典
16. „
v +„
z
17. „
w+„
z 具0, 3典
Chapter 9 Right Triangles and Trigonometry
具6, 13典