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Copyright © 2011 Pearson Education, Inc.
Slide 10.3-1
Chapter 10: Applications of Trigonometry
and Vectors
10.1 The Law of Sines
10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex
Numbers
10.5 Powers and Roots of Complex Numbers
10.6 Polar Equations and Graphs
10.7 More Parametric Equations
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-2
10.3 Vectors and Their Applications
• Basic Terminology
– A scalar is a magnitude
• E.g. 45 pounds
– A vector quantity is a
magnitude with direction
• E.g. 50 mph east
• Vectors
– Represented in boldface type or with an arrow
over the letters
• E.g. OP, and OP represent the vector OP
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-3
10.3 Basic Terminology
• Vector OP
– First letter represents the
initial point
– Second letter represents
the terminal point
– Vector OP and vector PO
are not the same vectors.
They have the same magnitude,
but in opposite directions.
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Slide 10.3-4
10.3 Basic Terminology
•
•
Two vectors are equal if and only if they both have
the same magnitude and direction.
The sum of two vectors A and B
1. Place the initial point of B at the terminal point of A.
The vector with the same initial point as A and the same
terminal point as B is the sum A + B.
2. The parallelogram rule: place the vectors so that their
initial points coincide. Then complete the parallelogram
as shown in the figure.
Figure 22 pg 10-47
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Slide 10.3-5
10.3 Basic Terminology
• Vectors are commutative: A + B = B + A, and the
sum A + B is called the resultant of A and B.
• For every vector v there is a vector –v such that
v + (–v) = 0, the zero vector.
• The scalar product of a real number (scalar) k and a
vector u is the vector k·u, which has magnitude |k|
times the magnitude of u. If k < 0, then k·u is in the
opposite direction as u.
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Slide 10.3-6
10.3 Algebraic Interpretation of Vectors
• A vector with its initial point at the origin is called
a position vector.
• A position vector u with endpoint (a,b) is written
as u = a, b, where a is called the horizontal
component and b is called the vertical component
of u.
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Slide 10.3-7
10.3 Algebraic Interpretation of Vectors
Magnitude and Direction Angle of a Vector a, b
The magnitude (length) of vector u = a, b is given by
u  a 2 b 2 .
The direction angle  satisfies tan  = ba , where a  0.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-8
10.3 Finding the Magnitude and
Direction Angle
Example Find the magnitude and direction angle for
u = 3, –2.
Analytic Solution
u  32  (2)2  13
y 2
2
tan   

x
3
3
2
1 
tan     33.7
 3
Vector u has a positive x-component and a negative ycomponent, placing u in quadrant IV. Adding 360°
yields the direction of   33.7  360  326.3.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-9
10.3 Finding Horizontal and Vertical
Components
Horizontal and Vertical Components
The horizontal and vertical components, respectively,
of a vector u having magnitude |u| and direction angle
 are given by
a = |u| cos 
and b = |u| sin .
That is, u = a, b =  |u| cos  , |u| sin  .
.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-10
10.3 Finding Horizontal and Vertical
Components
Example From the figure,
the horizontal component is
a = 25.0 cos 41.7°  18.7.
The vertical component is
b = 25.0 sin 41.7°  16.6.
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Slide 10.3-11
10.3 Finding the Magnitude of a Resultant
Example Two forces of 15 and 22 newtons act on a point in
the plane. If the angle between the forces is 100°, find the
magnitude of the resultant force.
Solution
From the figure, the angles
of the parallelogram adjacent to angle P
each measure 80º, since they are
supplementary to angle P. The resultant
force divides the parallelogram into two
triangles. Use the law of cosines on
either triangle.
|v|2 = 152 + 222 –2(15)(22) cos 80º
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
|v|  24 newtons
Slide 10.3-12
10.3 Some Properties of a Parallelogram
•
Properties of Parallelograms
1. A parallelogram is a quadrilateral whose
opposite sides are parallel.
2. The opposite sides and opposite angles of a
parallelogram are equal, and adjacent angles
are supplementary.
3. The diagonals of a parallelogram bisect each
other but do not necessarily bisect the angles.
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Slide 10.3-13
10.3 Operations with Vectors
m  a,b , n  c,d , and p  a c,b  d .
a  c,b  d  a,b  c,d
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Slide 10.3-14
10.3 Operations with Vectors
Vector Operations
For any real numbers a, b, c, d, and k,
a,b  c,d  a  c,b  d
k  a,b  ka,kb .
If a  a1 ,a2 , then  a   a1 , a2 .
a,b  c,d  a,b   c,d  a c,b  d
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-15
10.3 Performing Vector Operations
Example Let u = –2, 1 and v = 4, 3. Find each of
the following: (a) u + v, (b) –2u, (c) 4u – 3v.
Solution
(a) u + v = –2, 1 + 4, 3 = –2 + 4, 1 + 3 = 2, 4
(b) –2u = –2 · –2, 1 = –2(–2), –2(1) = 4, –2
(c) 4u – 3v = 4 · –2, 1 – 3 · 4, 3
= 4(–2) – 3(4), 4(1) – 3(3)
= –8 – 12, 4 –9
= –20, –5
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Slide 10.3-16
10.3 The Unit Vector
• A unit vector is a vector that has magnitude 1.
• Two very useful unit vectors
are defined as
i = 1, 0 and j = 0, 1.
i, j Forms for Unit Vectors
If v = a, b, then v = ai + bj.
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Slide 10.3-17
10.3 Dot Product
Dot Product
The dot product of two vectors u = a, b and
v = c, d  is denoted u · v, read “u dot v,” and given
by
u · v = ac + bd.
Example Find the dot product 2, 3 · 4, –1.
Solution
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2, 3 · 4, –1 = 2(4) + 3(–1)
=8–3
=5
Slide 10.3-18
10.3 Dot Product
Properties of the Dot Product
For all vectors u, v, and w and real numbers k,
a)
b)
c)
d)
e)
f)
u·v= v·u
u · (v + w) = u · v + u · w
(u + v) · w = u · w + v · w
(ku) · v = k(u · v) = u · (kv)
0·u=0
u · u = |u|2.
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Slide 10.3-19
10.3 Dot Product
Geometric Interpretation of the Dot Product
If  is the angle between the two nonzero vectors
u and v, where 0º    180º, then
u  v  u v cos 
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u v
or cos  
.
uv
Slide 10.3-20
10.3 Finding the Angle Between Two
Vectors
Example
Solution
Find the angle between the two vectors
u = 3, 4 and v = 2, 1.
3, 4  2,1
u v
cos  

uv
3, 4 2,1
3( 2)  4( 2)

9 16 4 1
10

 .894427191
5 5
Therefore,   26.57º.
NOTE
If a · b = 0, then cos  = 0 and  = 90º. Thus a and b
are perpendicular or orthogonal vectors.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-21
10.3 Applying Vectors to a
Navigation Problem
Example
A plane with an airspeed of 192 mph is headed on a
bearing of 121º. A north wind is blowing (from north
to south) at 15.9 mph. Find the groundspeed and the
actual bearing of the plane.
Solution
Let |x| be groundspeed. We
must find angle . Angle AOC = 121º.
Find |x| using the law of cosines .
x  192  15.9  2(192)(15.9) cos 121
2
2

2
 40,261 
x  200.7 mph
sin  sin 121

15.9
200.7
sin   .06792320    3.89
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Slide 10.3-22
10.3 Finding a Required Force
Example
Find the force required to pull a wagon weighing
50 lbs up a ramp inclined at 20º to the horizontal.
(Assume no friction.)
Solution
The vertical 50 lb force BA
represents the force of gravity. BA is the
sum of the vectors BC and –AC. Vector
BC represents the force with which the
weight pushes against the ramp. Vector
BF represents the force required to pull
the weight up the ramp. Since BF and AC are equal, | AC | gives
the magnitude of the required force.
AC
sin20 
50
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
AC  50sin20  2017 lbs
Slide 10.3-23