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Chapter 13
Vectors and Vector-Valued Functions
Section 13.1 Vectors in the Plane
Page 1
Chapter Preview We now make a significant departure from previous chapters by stepping out of the x y-plane R2 into
three-dimensional space R3 . The fundamental concept of a vector—a quantity with magnitude and direction—is introduced in two
and three dimensions. We then put vectors in motion by introducing vector-valued functions, or simply vector functions. The calculus
of vector functions is a direct extension of everything you already know about limits, derivatives, and integrals. Also, with the
calculus of vector functions, we can solve a wealth of practical problems involving the motion of objects in space. The chapter closes
with an exploration of arc length, curvature, and tangent and normal vectors, all important features of space curves.
13.1 Vectors in the Plane
Imagine a raft drifting down a river, carried by the current. The speed and direction of the raft at a point may be represented by an
arrow (Figure 13.1). The length of the arrow represents the speed of the raft at that point; longer arrows correspond to greater speeds.
The orientation of the arrow gives the direction in which the raft is headed at that point. The arrows at points A and C in Figure 13.1
have the same length and direction indicating that the raft has the same speed and heading at these locations. The arrow at B is
shorter and points to the left, indicating that the raft slows down as it nears the rock.
FIGURE 13.1
Basic Vector Operations
The arrows that describe the raft's motion are examples of vectors—quantities that have both length (or magnitude) and direction.
Vectors arise naturally in many situations. For example, electric and magnetic fields, the flow of air over an airplane wing, and the
velocity and acceleration of elementary particles are described by vectors (Figure 13.2). In this section we examine vectors in the x yplane and then extend the concept to three dimensions in Section 13.2.
Electric field vectors due to two charges
Velocity vectors of air flowing over an
airplane wing
Tracks of elementary particles in a
cloud chamber are aligned with the
velocity vectors of the particles.
FIGURE 13.2
The vector whose tail is at the point P and whose head is at the point Q is denoted PQ (Figure 13.3). The vector QP has its tail
at Q and its head at P. We also label vectors with single, boldfaced characters such as u and v.
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Section 13.1 Vectors in the Plane
FIGURE 13.3
Two vectors u and v are equal, written u = v, if they have equal length and point in the same direction (Figure 13.4). An
important fact is that equal vectors do not necessarily have the same location. Any two vectors with the same length and direction are
equal.
FIGURE 13.4
Note
Ó
The vector v is commonly handwritten as v.
Not all quantities are represented by vectors. For example, mass, temperature, and price have magnitude, but no direction. Such
quantities are described by real numbers and are called scalars.
Note
In this book, scalar is another word for real number.
Vectors, Equal Vectors, Scalars, Zero Vector
Vectors are quantities that have both length (or magnitude) and direction. Two vectors are equal if they have the
same magnitude and direction. Quantities having magnitude but no direction are called scalars. One exception is the
zero vector, denoted 0: it has length 0 and no direction.
Note
The zero vector is handwritten 0.
Scalar Multiplication
A scalar c and a vector v can be combined using scalar-vector multiplication, or simply scalar multiplication. The resulting vector,
denoted c v, is called a scalar multiple of v. The magnitude of c v is c¤ multiplied by the magnitude of v. The vector c v has the same
direction as v if c > 0. If c < 0, then c v and v point in opposite directions. If c = 0, then 0 × v = 0 (the zero vector).
For example, the vector 3 v is three times as long as v and has the same direction as v. The vector -2 v is twice as long as v,
1
but it points in the opposite direction. The vector v points in the same direction as v and has half the length of v (Figure 13.5). The
2
1
vectors v, 3 v, -2 v, and v  2 (that is, v) are examples of parallel vectors: each one is a scalar multiple of the others.
2
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Chapter 13
Section 13.1 Vectors in the Plane
For and
example,
the vector
3 v is three times as long as v and has the same direction as v. The vector -2 v is twice as long as v,
Vectors
Vector-Valued
Functions
1
v points in the same direction as v and has half the length of v (Figure 13.5). The
but it points in the opposite direction. The vector
2
vectors v, 3 v, -2 v, and v  2 (that is,
1
v) are examples of parallel vectors: each one is a scalar multiple of the others.
2
c = -2
c = 12
2v
c=2
v
c=3
c
c= 2
Drag the symbol Å to
change v.
FIGURE 13.5
Note
For convenience, we write -u for H-1L u, -c u for H-cL u, and u  c for H1  cL u.
DEFINITION
Scalar Multiples and Parallel Vectors
Given a scalar c and a vector v, the scalar multiple c v is a vector whose magnitude is c¤ multiplied by the magnitude
of v. If c > 0, then c v has the same direction as v. If c < 0, then c v and v point in opposite directions. Two vectors are
parallel if they are scalar multiples of each other.
Notice that two vectors are parallel if they point in the same direction (for example, v and 12 v) or if they point in opposite
directions (for example, v and -2 v). Also, because 0 v = 0 for all vectors v, it follows that the zero vector is parallel to all vectors.
While it may seem counterintuitive, this result turns out to be a useful convention.
QUICK CHECK 1
Describe the magnitude and direction of the vector -5 v relative to v.
ª
EXAMPLE 1
Parallel vectors
Using Figure 13.6, write the following vectors in terms of u or v.
a.
PQ
b.
QP
c.
QR
d.
RS
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Section 13.1 Vectors in the Plane
PQ
QP
QR
S
RS
R
Drag the symbols Å
to change u and v.
v
O
u
P
Q
FIGURE 13.6
SOLUTION
a.
The vector PQ has the same direction and length as u; therefore, PQ = u. These two vectors are equal even though they have
different locations (Figure 13.6).
c.
Because QP and u have equal length, but opposite directions, QP = H-1L u = -u.
d.
RS points in the direction opposite to that of u with three times the length of u. Consequently, RS = -3 u.
b.
QR points in the same direction as v and is twice as long as v, so QR = 2 v.
Related Exercises 17—20
Vector Addition and Subtraction
To illustrate the idea of vector addition, consider a plane flying horizontally at a constant speed in a crosswind (Figure 13.7). The
length of vector va represents the plane's airspeed, which is the speed the plane would have in still air; va points in the direction of
the nose of the plane. The wind vector w points in the direction of the crosswind and has a length equal to the speed of the crosswind. The combined effect of the motion of the plane and the wind is the vector sum vg = va + w, which is the velocity of the plane
relative to the ground.
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Section 13.1 Vectors in the Plane
Page 5
FIGURE 13.7
QUICK CHECK 2
Sketch the sum va + w in Figure 13.7 if the direction of w is reversed.
ª
Figure 13.8 illustrates two ways to form the vector sum of two nonzero vectors u and v geometrically. The first method, called
the Triangle Rule, places the tail of v at the head of u. The sum u + v is the vector that extends from the tail of u to the head of v.
When u and v are not parallel, another way to form u + v is to use the Parallelogram Rule. The tails of u and v are connected
to form adjacent sides of a parallelogram; then, the remaining two sides of the parallelogram are sketched. The sum u + v is the
vector that coincides with the diagonal of the parallelogram, beginning at the tails of u and v. The Triangle Rule and Parallelogram
Rule each produce the same vector sum u + v.
Vector addition
Vectors u and v
Triangle Rule
Parallelogram Rule
Drag the symbol Å to
change u.
v
u
FIGURE 13.8
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Chapter 13
Vectors and Vector-Valued Functions
QUICK CHECK 3
Page 6
Section 13.1 Vectors in the Plane
Use the Triangle Rule to show that the vectors in Figure 13.8 satisfy u + v = v + u.
ª
The difference u - v is defined to be the sum u + H-vL. By the Triangle Rule, the tail of -v is placed at the head of u; then,
u - v extends from the tail of u to the head of -v (Figure 13.9). Equivalently, when the tails of u and v coincide, u - v has its tail at
the head of v and its head at the head of u.
Vector Subtraction
Vectors u and v
Triangle Rule
Direct
Drag the symbol Å to
v
change u.
u
FIGURE 13.9
EXAMPLE 2
Vector operations
Use Figure 13.10 to write the following vectors as sums of scalar multiples of v and w.
a.
OP
b.
OQ
c.
QR
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Section 13.1 Vectors in the Plane
Drag the symbols Å
to change v and w.
P
R
w
O
v
Q
FIGURE 13.10
SOLUTION
a.
Using the Triangle Rule, we start at O, move three lengths of v in the direction of v and then two lengths of w in the direction
of w to reach P. Therefore, OP = 3 v + 2 w (Figure 13.11).
b.
The vector OQ coincides with the diagonal of a parallelogram having adjacent sides equal to 3 v and -w. By the Parallelogram
Rule, OQ = 3 v - w (Figure 13.11).
b.
The vector QR lies on the diagonal of a parallelogram having adjacent sides equal to v and 2 w. Therefore, QR = v + 2 w
(Figure 13.11).
OP
OQ
QR
P
Drag the symbols Å
R
to change v and w.
w
O
v
Q
FIGURE 13.11
Related Exercises 21—22
Vector Components
So far, vectors have been examined from a geometric point of view. To do calculations with vectors, it is necessary to introduce a
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Calculus
Scientists
coordinateforsystem.
We and
beginEngineers
by considering a vector v whose
tail is
at the origin in the Cartesian plane
and whose head is at the
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point Hv1 , v2 L (Figure 13.12).
Chapter 13
Vectors and Vector-Valued Functions
Page 8
Section 13.1 Vectors in the Plane
So far, vectors have been examined from a geometric point of view. To do calculations with vectors, it is necessary to introduce a
coordinate system. We begin by considering a vector v whose tail is at the origin in the Cartesian plane and whose head is at the
point Hv1 , v2 L (Figure 13.12).
Drag the symbols Å
to change v and its
copies.
y
Position vector v = Xv1 , v2 \.
v
Copies of v at different
locations are equal.
v2
v
v
Hposition vectorL
v
x
v1
v
v
FIGURE 13.12
DEFINITION
Position Vectors and Vector Components
A vector v with its tail at the origin and head at Hv1 , v2 L is called a position vector (or is said to be in standard
position) and is written Xv1 , v2 \. The real numbers v1 and v2 are the x- and y-components of v, respectively. The
position vectors u = Xu1 , u2 \ and v = Xv1 , v2 \ are equal if and only if u1 = v1 and u2 = v2 .
Note
Round brackets Ha, bL enclose the coordinates of a point, while angle brackets Xa, b\ enclose the
components of a vector. Note that in component form, the zero vector is 0 = X0, 0\.
There are infinitely many vectors equal to the position vector v, all with the same length and direction (Figure 13.12). It is
important to abide by the convention that v = Xv1 , v2 \ refers to the position vector v or to any other vector equal to v.
Now consider the vector PQ, equal to v but not in standard position, with its tail at the point PHx1 , y1 L and its head at the point
QHx2 , y2 L. The x-component of PQ is the difference in the x-coordinates of Q and P, or x2 - x1 . The y-component of PQ is the
difference in the y-coordinates, y2 - y1 (Figure 13.13). Therefore, PQ has the same length and direction as the position vector
Xv1 , v2 \ = Xx2 - x1 , y2 - y1 \ and we write PQ = Xx2 - x1 , y2 - y1 \.
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Section 13.1 Vectors in the Plane
Drag the symbols Å
y
to change PQ.
y2
PQ = Xx2 - x1 , y2 - y1 \
PQ¤ =
QHx2 , y2 L
Hx2 - x1 L2 + Hy2 - y1 L2
v = Xv1 , v2 \
v2
y2 - y1
PHx1 , y1 L
y1
x2 - x1
x
v1
x1
x2
FIGURE 13.13
QUICK CHECK 4
Given the points PH2, 3L and QH-4, 1L, find the components of PQ.
ª
As already noted, there are infinitely many vectors equal to a given position vector. All these vectors have the same length and
direction; therefore, they are all equal. In other words, two arbitrary vectors are equal if they are equal to the same position vector.
For example, the vector PQ from PH2, 5L to QH6, 3L and the vector AB from AH7, 12L to BH11, 10L are equal because they are both
equal to the position vector X4, -2\.
Magnitude
The magnitude of a vector is simply its length. By the Pythagorean Theorem and Figure 13.13, we have the following definition.
DEFINITION
Magnitude of a Vector
Given the points PHx1 , y1 L and QHx2 , y2 L, the magnitude, or length, of PQ = Xx2 - x1 , y2 - y1 \, denoted by PQ¤, is
the distance between P and Q:
PQ¤ =
Hx2 - x1 L2 + Hy2 - y1 L2 .
The magnitude of the position vector v = Xv1 , v2 \ is v¤ =
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v21 + v22 .
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Chapter 13
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Section 13.1 Vectors in the Plane
Note
Just as the absolute value p - q¤ gives the distance between two points on the number line, the
magnitude PQ¤ denotes the distance between the points P and Q. The magnitude of a vector is
also called its norm.
EXAMPLE 3
Calculating components and magnitude
Given the points OH0, 0L, PH-3, 4L, and QH6, 5L, find the components and magnitude of the following vectors.
a.
OP
b.
PQ
SOLUTION
a.
The vector OP is the position vector whose head is located at PH-3, 4L. Therefore, OP = X-3, 4\ and the magnitude is
OP¤ =
b.
H-3L2 + 42 = 5.
PQ = X6 - H-3L, 5 - 4\ = X9, 1\ and PQ¤ =
92 + 12 =
82 .
Related Exercises 23—27
Vector Operations in Terms of Components
We now show how vector addition, vector subtraction, and scalar multiplication are performed using components. Suppose
u = Xu1 , u2 \ and v = Xv1 , v2 \. The vector sum of u and v is u + v = Xu1 + v1 , u2 + v2 \. This definition of a vector sum is consistent
with the Parallelogram Rule given earlier (Figure 13.14).
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Section 13.1 Vectors in the Plane
Drag the symbols Å
y
to change u and v.
u + v = Xu1 + v1 , u2 + v2 \ by the
Parallelogram Rule
Hu1 + v1 , u2 + v2 L
u2 + v2
v2
u+v
v
u2
u
x
v1
u1
u1 + v1
FIGURE 13.14
For a scalar c and a vector u, the scalar multiple c u is c u = Xc u1 , c u2 \; that is, the scalar c multiplies each component of u. If
c > 0, u and c u have the same direction (Figure 13.15). If c < 0, u and c u have opposite directions. In either case, c u¤ = c¤ u¤
(Exercise 87).
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Chapter 13
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Section 13.1 Vectors in the Plane
c
y
c = -1
Drag the symbol Å to
change u.
u2
u = Xu1 , u2 \
c u1
x
u1
cu = Xc u1 , c u2 \
c u2
FIGURE 13.15
Notice that u - v = u + H-vL, where -v = X-v1 , -v2 \. Therefore, the vector difference of u and v is u - v = Xu1 - v1 , u2 - v2 \.
DEFINITION
Vector Operations in R2
Suppose c is a scalar, u = Xu1 , u2 \ and v = Xv1 , v2 \.
u + v = Xu1 + v1 , u2 + v2 \ Vector addition
u - v = Xu1 - v1 , u2 - v2 \ Vector subtraction
c u = Xc u1 , c u2 \
Scalar multiplication
Note
Recall that R2 (pronounced R-two) stands for the x y-plane or the set of all ordered pairs of real
numbers.
EXAMPLE 4
Vector operations
Let u = X-1, 2\ and v = X2, 3\.
b.
Evaluate u + v¤.
c.
Find two vectors half as long as u and parallel to u.
a.
Simplify 2 u - 3 v.
SOLUTION
a.
b.
Because u + v = X-1, 2\ + X2, 3\ = X1, 5\, we have u + v¤ =
2 u - 3 v = 2 X-1, 2\ - 3 X2, 3\ = X-2, 4\ - X6, 9\ = X-8, -5\.
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c.
Thefor
vectors
u =andX-1,
2\ = [- , 1_ and -
1
1
Briggs, Cochran, Gillett,
2 Schulz
2
1
2
12 + 52 =
26 .
1
1
© 2011,
Pearson Education,
u = -Printed:
X-1,12/27/13
2\ = [ , -1_ have half the lengthCopyright
of u and
are parallel
to u. Inc.
2
2
2
1
Chapter 13
Vectors and Vector-Valued Functions
1
c.
Page 13
1
1
1
1
X-1, 2\ = [- , 1_ and - u = - X-1, 2\ = [ , -1_ have half the length of u and are parallel to u.
2
2
2
2
2
1
u=
The vectors
Section 13.1 Vectors in the Plane
2
Related Exercises 28—41
Unit Vectors
A unit vector is any vector with length 1. Two useful unit vectors are the coordinate unit vectors i = X1, 0\ and j = X0, 1\ (Figure
13.16).
FIGURE 13.16
These vectors are directed along the coordinate axes and allow us to express all vectors in an alternate form. For example, by the
Triangle Rule (Figure 13.17).
X3, 4\ = 3 X1, 0\ + 4 X0, 1\ = 3 i + 4 j.
In general, the vector v = Xv1 , v2 \ is also written
v = v1 X1, 0\ + v2 X0, 1\ = v1 i + v2 j.
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Vectors and Vector-Valued Functions
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Section 13.1 Vectors in the Plane
Drag the symbols Å
y
to change v.
Xv1 , v2 \ = v1 i + v2 j
X3, 4\ = 3i + 4j
5
v2 4
v1 i +v2 j
3
v2 j
2
1
j
v1 i
i
1
2
3
v1
4
5
x
FIGURE 13.17
Note
Coordinate unit vectors are also called standard basis vectors.
Given a nonzero vector v, we sometimes need to construct a new vector parallel to v of a specified length. Dividing v by its
v
length, we obtain the vector u = . Because u is a positive scalar multiple of v, it follows that u has the same direction as v.
v¤
v¤
v
v
Furthermore, u is a unit vector because u¤ =
= 1. The vector -u = is also a unit vector (Figure 13.18). Therefore, ±
are
v¤
v¤
v¤
unit vectors parallel to v that point in opposite directions.
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Section 13.1 Vectors in the Plane
FIGURE 13.18
cv
To construct a vector that points in the direction of v and has a specified length c > 0, we form the vector
= c¤
cv
scalar multiple of v, so it points in the direction of v, and its length is
direction and also has length c.
QUICK CHECK 5
ª
EXAMPLE 5
v¤
v¤
cv
v¤
= c. The vector -
3 4
Find vectors of length 10 parallel to the unit vector u = [ , _.
5 5
v¤
v¤
: It is a positive
points in the opposite
Magnitude and unit vectors
Consider the points PH1, -2L and QH6, 10L.
a.
Find PQ and two unit vectors parallel to PQ.
b.
Find two vectors of length 2 parallel to PQ.
SOLUTION
a.
PQ = X6 - 1, 10 - H-2L\ = X5, 12\, or 5 i + 12 j. Because PQ¤ =
PQ
PQ¤
=
X5, 12\
13
=[
5
12
,
13 13
52 + 122 =
_=
Another unit vector parallel to PQ but having the opposite direction is [b.
5
169 = 13, a unit vector parallel to PQ is
12
i+
j.
13
13
5
12
,-
13
13
_.
5
12
i+
To obtain two vectors of length 2 that are parallel to PQ, we multiply the unit vector
13
5
2
12
i+
13
13
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10
j =
24
i+
13
5
j and -2
12
i+
13 Printed: 12/27/13
13
10
13
24
i-
j =13
j by ± 2:
13
j
13
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Chapter 13
Vectors and Vector-Valued Functions
5
12
i+
2
13
Page 16
Section 13.1 Vectors in the Plane
10
j =
13
24
i+
13
5
j and -2
13
12
i+
13
10
13
24
i-
j =13
j
13
Related Exercises 42—47
QUICK CHECK 6
ª
Verify that the vector [
5
12
,
13 13
_ has length 1.
Properties of Vector Operations
When we stand back and look at vector operations, ten general properties emerge. For example, the first property says that vector
addition is commutative, which means u + v = v + u. This property is proved by letting u = Xu1 , u2 \ and v = Xv1 , v2 \. By the commutative property of addition for real numbers,
u + v = Xu1 + v1 , u2 + v2 \ = Xv1 + u1 , v2 + u2 \ = v + u.
Note
The Parallelogram Rule illustrates the commutative property u + v = v + u.
The proofs of other properties are outlined in Exercises 82-85.
SUMMARY
Properties of Vector Operations
Suppose u, v, and w are vectors and a and c are scalars. Then the following properties hold (for vectors in any number
of dimensions).
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
u+v=v+u
Hu + vL + w = u + Hv + wL
v+0=v
v + H-vL = 0
cHu + vL = c u + c v
Ha + cL v = a v + c v
0v=0
c0=0
1v=v
aHc vL = Ha cL v
Commutative property of addition
Associative property of addition
Additive identity
Additive inverse
Distributive property 1
Distributive property 2
Multiplication by zero scalar
Multiplication by zero vector
Multiplicative identity
Associative property of scalar multiplication
These properties allow us to solve vector equations. For example, to solve the equation u + v = w for u, we proceed as follows:
Hu + vL + H-vL = w + H-vL Add - v to both sides.
u + @v + H-vLD = w + H-vL Property 2
0
u+0=w-v
u=w-v
QUICK CHECK 7
Property 4
Property 3
Solve 3 u + 4 v = 12 w for u.
ª
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Chapter 13
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Section 13.1 Vectors in the Plane
Page 17
Applications of Vectors
Vectors have countless practical applications, particularly in the physical sciences and engineering. These applications are explored
throughout the remainder of the book. For now, we present two common uses of vectors: to describe velocities and forces.
Velocity Vectors
Consider a motorboat crossing a river whose current is everywhere represented by the constant vector w (Figure 13.19); this means
that w¤ is the speed of the moving water and w points in the direction of the moving water. Assume that the vector vw gives the
direction and speed of the boat relative to the water. The combined effect of w and vw is the sum vg = vw + w, which gives the speed
and direction of the boat that would be observed by someone on the shore (or on the ground).
FIGURE 13.19
Note
Speed of the boat relative to the water means the speed the boat would have in still water (or
relative to someone traveling with the current).
EXAMPLE 6
Speed of a boat in a current
Assume the water in a river moves southwest (45 ° west of south) at 4 mi  hr. If a motorboat is traveling due east at 15 mi  hr relative
to the shore, determine the speed of the boat and its heading relative to the moving water (Figure 13.19).
SOLUTION
To solve this problem, the vectors are placed in a coordinate system (Figure 13.20). Because the boat is moving east at 15 mi  hr, the
velocity relative to the shore is vg = X15, 0\. To obtain the components of w = Ywx , w y ], observe that w¤ = 4 and the lengths of the
sides of the 45-45-90 triangle in Figure 13.20 are
wx ¤ = ¡w y ¥ = w¤ cos 45 ° =
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4
=2
2.
2
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Section 13.1 Vectors in the Plane
FIGURE 13.20
Note
Recall that the lengths of the legs of a 45-45-90 triangle are equal and are J1 ’
2 ) times the
length of the hypotenuse.
Given the orientation of w (southwest), w = [-2
2 _. Because vg = vw + w (Figure 13.19),
2 , -2
vw = vg - w = X15, 0\ - [-2
The magnitude of vw is
= [15 + 2
vw ¤ =
2,2
K15 + 2
2 , -2
2 _.
2 O + K2
2
2O
2_
2
Therefore, the speed of the boat relative to the water is approximately 18 mi  hr.
» 18.
The heading of the boat is given by the angle Θ between vw and the positive x-axis. The x-component of vw is 15 + 2
component is 2
2 and the y-
2 : therefore,
Θ = tan-1
2
2
» 9 °.
15 + 2
2
The heading of the boat is approximately 9 ° north of east, and its speed relative to the water is approximately 18 mi  hr.
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Related Exercises 48—53
Force Vectors
Suppose a child pulls on the handle of a wagon at an angle of Θ with the horizontal (Figure 13.21a). The vector F represents the force
exerted on the wagon; it has a magnitude F¤ and a direction given by Θ.
Note
The magnitude of F is typically measured in pounds (lb) or newtons (N), where 1 N = 1 kg × m ‘ s2 .
We denote the horizontal and vertical components of F by Fx and F y , respectively. Then, Fx = F¤ cos Θ, F y = F¤ sin Θ, and the force
vector is F = X F¤ cos Θ, F¤ sin Θ\ (Figure 13.21b).
Note
The vector Xcos Θ, sin Θ\ is a unit vector. Therefore, any position vector v may be written
v = X v¤ cos Θ, v¤ sin Θ\, where Θ is the angle that v makes with the positive x-axis.
FIGURE 13.21
EXAMPLE 7
Finding force vectors
A child pulls a wagon (Figure 13.21) with a force of F¤ = 20 lb at an angle of Θ = 30 ° to the horizontal. Find the force vector F.
SOLUTION
The force vector (Figure 13.22) is
F = X F¤ cos Θ, F¤ sin Θ\ = X20 cos 30 °, 20 sin 30 °\ = [10
3 , 10_.
FIGURE 13.22
Related Exercises 54—58
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EXAMPLE 8
Page 20
Section 13.1 Vectors in the Plane
Balancing forces
A 400–lb engine is suspended from two chains that form 60 ° angles with a horizontal ceiling (Figure 13.23). How much weight does
each chain support?
FIGURE 13.23
SOLUTION
Let F1 and F2 denote the forces exerted by the chains on the engine and let F3 be the downward force due to the weight of the engine
(Figure 13.23). Placing the vectors in a standard coordinate system (Figure 13.24), we find that F1 = X F1 ¤ cos 60 °, F1 ¤ sin 60 °\,
F2 = X- F2 ¤ cos 60 °, F2 ¤ sin 60 °\ and F3 = X0, -400\.
FIGURE 13.24
If the engine is in equilibrium (so the chains and engine are stationary), the sum of the forces must be zero; that is, F1 + F2 + F3 = 0
or F1 + F2 = -F3 . Therefore,
X F1 ¤ cos 60 ° - F2 ¤ cos 60 °, F1 ¤ sin 60 ° + F2 ¤ sin 60 °\ = X0, 400\.
Equating corresponding components, we obtain the following two equations to be solved for F1 ¤ and F2 ¤:
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F1 ¤ cos 60 ° - F2 ¤ cos 60 ° = 0
F1 ¤ sin 60 Printed:
° + F2 ¤ 12/27/13
sin 60 ° = 400
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F1 ¤ cos 60 ° - F2 ¤ cos 60 ° = 0
F1 ¤ sin 60 ° + F2 ¤ sin 60 ° = 400
Factoring the first equation, we find that H F1 ¤ - F2 ¤L cos 60 ° = 0, which implies that F1 ¤ = F2 ¤. Replacing F2 ¤ by F1 ¤ in the second
equation gives 2 F1 ¤ sin 60 ° = 400. Noting that sin 60 ° =
3 “ 2 and solving for F1 ¤, we find that F1 ¤ = 400 “
3 » 231. Each
chain must be able to withstand a weight of approximately 231 lb.
Related Exercises 54—58
Quick Quiz
1. Given two points PH3, 2L and QH- 5, 1L, the vector PQ is equal to the position vector
HaL X8, 1\.
HbL X- 8, - 1\.
HcL X- 8, 1\.
2. Assuming the x-axis points east and the y-axis points north, the direction of the vector X3, - 3\ is
HaL northwest.
HbL southwest.
HcL southeast.
3. The magnitude of PQ, where P is H- 2, 5L and Q is H4, - 1L is
HaL 6
HbL
HcL
2.
32 .
2.
4. If u = X2, 5\ and v = X- 9, 4\, then 3 u - 2 v equals
HaL X12, 23\.
HbL X24, 7\.
HcL X24, 23\.
5. If u = X1, - 4\ and v = X- 1, 3\, then u + v¤ equals
HaL 1.
HbL
HcL 7.
5.
6. The vector u = X3, 4\ can also be written
HaL 4 i + 3 j.
HbL 3 i + 4 j.
HcL 4 i - 3 j.
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7. A unit vector in the direction of X3, 4\ is
3 4
HaL [ , _.
7 7
3 4
HbL [ ,
_.
25 25
3 4
HcL [ , _.
5 5
8. A vector of length 5 in the direction of X1, 2\ is
HaL X1, 2\.
HbL
5 X1, 2\.
HcL X5, 10\.
9. A force of 20 points directed at an angle of 60 ° above the horizontal has a horizontal component of
HaL 10 pounds.
HbL 20
HcL 10
3 pounds.
3 pounds.
10. A boat travels to the east at 5 mi  hr relative to the water in a side wind from the north at 5
Its speed relative to the shore is
HaL 20 mi  hr.
3 mi  hr.
HbL 10 mi  hr.
HcL 10
3 mi  hr.
SECTION 13.1 EXERCISES
Review Questions
1.
Interpret the following statement: Points have a location but no size or direction; nonzero vectors have a size and
direction, but no location.
2.
What is a position vector?
3.
Draw x- and y-axes on a page and mark two points P and Q. Then draw PQ and QP.
4.
On the diagram of Exercise 3, draw the position vector that is equal to PQ.
5.
Given a position vector v, why are there infinitely many vectors equal to v?
6.
Explain how to add two vectors geometrically.
7.
Explain how to find a scalar multiple of a vector geometrically.
8.
Given two points P and Q, how are the components of PQ determined?
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9.
Page 23
Section 13.1 Vectors in the Plane
If u = Xu1 , u2 \ and v = Xv1 , v2 \, how do you find u + v?
10. If v = Xv1 , v2 \ and c is a scalar, how do you find c v?
11. How do you compute the magnitude of v = Xv1 , v2 \?
12. Express the vector v = Xv1 , v2 \ in terms of the unit vectors i and j.
13. How do you compute PQ¤ from the coordinates of the points P and Q?
14. Explain how to find two unit vectors parallel to a vector v.
15. How do you find a vector of length 10 in the direction of v = X3, -2\?
16. If a force has magnitude 100 and is directed 45 ° south of east, what are its components?
Basic Skills
17-20. Vector operations Refer to the figure and carry out the following vector operations.
17. Scalar Multiples Which of the following vectors equals CE? (There may be more than one correct answer.)
a. v
1
b.
HI
2
1
c.
OA
3
d. u
1
e.
IH
2
18.
Scalar Multiples Which of the following vectors equals BK? (There may be more than one correct answer.)
a. 6v
b. -6 v
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Section 13.1 Vectors in the Plane
c. 3 H I
d. 3 I H
e. 3 AO
19. Scalar multiples Write the following vectors as scalar multiples of u or v.
a. O A
b. OD
c. OH
d. AG
e. CE
20. Scalar multiples Write the following vectors as scalar multiples of u or v.
a. I H
b. H I
c. J K
d. F D
e. E A
21. Vector addition Write the following vectors as sums of scalar multiples of u and v.
a. OE
b. OB
c. OF
d. OG
e. OC
f. OI
g. OJ
h. OK
i.
OL
22. Vector addition Write the following vectors as sums of scalar multiples of u and v.
a. BF
b. DE
c. AF
d. AD
e. CD
f. J D
g. J I
h. DB
i.
IL
23. Components and magnitudes Define the points OH0, 0L, PH3, 2L, QH4, 2L, and RH-6, -1L. For each vector, do the
following.
(i) Sketch the vector in an x y-coordinate system.
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Section 13.1 Vectors in the Plane
(ii) Compute the magnitude of the vector.
a. OP
b. QP
c. RQ
24-27. Components and equality Define the points PH-3, -1L, QH-1, 2L, RH1, 2L, SH3, 5L, TH4, 2L, and U H6, 4L.
24. Sketch PU , T R, and SQ and the corresponding position vectors.
25. Sketch QU , PT, and RS and the corresponding position vectors.
26. Find the equal vectors among PQ, RS, and T U .
27. Which of the vectors QT or SU is equal to X5, 0\?
28-33. Vector operations Let u = X4, -2\, v = X-4, 6\, and w = X0, 8\ . Express the following vectors in the form Xa, b\.
28. u + v
29. w -u
30. 2 u + 3 v
31. w -3 v
32. 10 u - 3 v + w
33. 8 w + v - 6 u
34-41. Vector operations Let u = X3, -4\, v = X1, 1\, and w = X-1, 0\. Carry out the following computations.
34. Find u + v¤.
35. Find -2 v¤.
36. Find u + v + w¤.
37. Find 2 u + 3 v - 4 w¤.
38. Find two vectors parallel to u with four times the magnitude of u.
39. Find two vectors parallel to v with three times the magnitude of v.
40. Which has the greatest magnitude, 2 u or 7 v?
41. Which has the greater magnitude, u - v or w -u?
42-47. Unit vectors Define the points PH-4, 1L, QH3, -4L, and RH2, 6L. Carry out the following calculations.
42. Express PQ in the form a i + b j.
43. Express QR in the form a i + b j.
44. Find the unit vector with the same direction as QR.
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45. Find two unit vectors parallel to PR.
46. Find two vectors parallel to RP with length 4.
47. Find two vectors parallel to QP with length 4.
48. A boat in a current The water in a river moves south at 10 mi  hr. If a motorboat is traveling due east at a speed of
20 mi  hr relative to the shore, determine the speed and direction of the boat relative to the moving water.
49. Another boat in a current The water in a river moves south at 5 km hr. If a motorboat is traveling due east at a speed
of 40 km hr relative to the water, determine the speed of the boat relative to the shore.
50. Parachute in the wind In still air, a parachute with a payload would fall vertically at a terminal speed of 4 m  s. Find
the direction and magnitude of its terminal velocity relative to the ground if it falls in a steady wind blowing horizontally
from west to east at 10 m  s.
51. Airplane in a wind An airplane flies horizontally from east to west at 320 mi  hr relative to the air. If it flies in a steady
40 mi  hr wind that blows horizontally toward the southwest (45 ° south of west), find the speed and direction of the
airplane relative to the ground.
52. Canoe in a current A woman in a canoe paddles due west at 4 mi  hr relative to the water in a current that flows
northwest at 2 mi  hr. Find the speed and direction of the canoe relative to the shore.
53. Boat in a wind A sailboat floats in a current that flows due east at 1 m  s. Due to a wind, the boat's actual speed relative
to the shore is
3 m  s in a direction 30 ° north of east. Find the speed and direction of the wind.
54. Towing a boat A boat is towed with a force of 150 lb with a rope that makes an angle of 30 ° to the horizontal. Find the
horizontal and vertical components of the force.
55. Pulling a suitcase Suppose you pull a suitcase with a strap that makes a 60 ° angle with the horizontal. The magnitude
of the force you exert on the suitcase is 40 lb.
a. Find the horizontal and vertical components of the force.
b. Is the horizontal component of the force greater if the angle of the strap is 45 ° instead of 60 °?
c. Is the vertical component of the force greater if the angle of the strap is 45 ° instead of 60 °?
56. Which is greater? Which has a greater horizontal component, a 100 –N force directed at an angle of 60 ° above the
horizontal or a 60 –N force directed at an angle of 30 ° above the horizontal?
57. Suspended load If a 500-lb load is suspended by two chains (see figure), what is the magnitude of the force each chain
must be able to support?
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58. Net force Three forces are applied to an object, as shown in the figure. Find the magnitude and direction of the sum of
the forces.
Further Explorations
59. Explain why or why not Determine whether the following statements are true and give an explanation or
counterexample.
a. José travels from point A to point B in the plane by following vector u, then vector v, and then vector w. If he starts
at A and follows w, then v, and then u, he still arrives at B.
b. Maria travels from A to B in the plane by following the vector u. By following -u, she returns from B to A.
c. The magnitude of u + v is at least the magnitude of u.
d. The magnitude of u + v is at least the magnitude of u plus the magnitude of v.
e. Parallel vectors have the same length.
f. If AB = CD, then A = C and B = D.
g. If u and v are perpendicular, then u + v¤ = u¤ + v¤.
h. If u and v are parallel and have the same direction, then u + v¤ = u¤ + v¤.
60. Finding vectors from two points Given the points AH-2, 0L, BH6, 16L, CH1, 4L, DH5, 4L, EK
FK3
2 , -4
a. AB
2 O, find the position vector equal to the following vectors.
2,
2 O, and
b. AC
c. EF
d. CD
61. Unit vectors
a. Find two unit vectors parallel to v = 6 i - 8 j.
b. Find b if v = X1  3, b\ is a unit vector.
a
c. Find all values of a such that w = a i -
j is a unit vector.
3
62. Equal vectors For the points AH3, 4L, BH6, 10L, CHa + 2, b + 5L, and DHb + 4, a - 2L, find the values of a and b such that
AB = CD.
63-67. Vector equations Use the properties of vectors to solve the following equations for the unknown vector x = Xa, b\.
Let u = X2, -3\ and v = X-4, 1\.
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63. 10 x = u
64. 2 x + u = v
65. 3 x - 4 u = v
66. -4 x = u -8 v
67-69. Linear Combinations A sum of scalar multiples of two or more vectors (such as c1 u +c2 v + c3 w, where ci are
scalars) is called a linear combination of the vectors. Let i = X1, 0\, j = X0, 1\, u = X1, 1\, and v = X-1, 1\.
67. Express X4, -8\ as a linear combination of i and j (that is, find scalars c1 and c2 such that X4, -8\ = c1 i + c2 j).
68. Express X4, -8\ as a linear combination of u and v.
69. For arbitrary real numbers a and b, express Xa, b\ as a linear combination of u and v.
70-71. Solving vector equations Solve the following pairs of equations for the vectors u and v. Assume i = X1, 0\ and
j = X0, 1\.
70. 2 u = i, u - 4 v = j
71. 2 u + 3 v = i, u - v = j
72-75. Designer vectors Find the following vectors.
72. The vector that is 3 times X3, -5\ plus -9 times X6, 0\.
73. The vector in the direction of X5, -12\ with length 3.
74. The vector in the direction opposite to that of X6, -8\ with length 10.
75. The position vector for your final location if you start at the origin and walk along X4, -6\ followed by X5, 9\.
Applications
76. Ant on a page An ant is walking due east at a constant speed of 2 mi  hr on a sheet of paper that rests on a table.
Suddenly the sheet of paper starts moving southeast at
2 mi  hr. Describe the motion of the ant relative to the table.
77. Clock vectors Consider the 12 vectors that have their tails at the center of a (circular) clock and their heads at the
numbers on the edge of the clock.
a. What is the sum of these 12 vectors?
b. If the 12:00 vector is removed, what is the sum of the remaining 11 vectors?
c. By removing one or more of these 12 clock vectors, explain how to make the sum of the remaining vectors as large
as possible in magnitude.
d. If the clock vectors originate at 12:00 and point to the other 11 numbers, what is the sum of the vectors?
(Source: Calculus, by Gilbert Strang. Wellesley-Cambridge Press, 1991.)
78. Three-way tug-of-war Three people located at A, B, and C pull on ropes tied to a ring. Find the magnitude and
direction of the force with which C must pull so that no one moves (the system is at equilibrium).
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79. Net force Jack pulls east on a rope attached to a camel with a force of 40 lb. Jill pulls north on a rope attached to the
same camel with a force of 30 lb. What is the magnitude and direction of the force on the camel? Assume the vectors lie
in a horizontal plane.
80. Mass on a plane A 100 –kg object rests on an inclined plane at an angle of 30 ° to the floor. Find the components of the
force perpendicular to and parallel to the plane. (The vertical component of the force exerted by an object of mass m is
its weight, which is m g, where g = 9.8 m ‘s2 is the acceleration due to gravity.)
Additional Exercises
81-85. Vector properties Prove the following vector properties using components. Then, make a sketch to illustrate the
property geometrically. Suppose u, v, and w are vectors in the x y-plane and a and c are scalars.
81. Commutative property: u + v = v + u
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82. Associative property: Hu + vL +w = u + Hv + wL
83. Associative property: aHc vL = Ha cL v
84. Distributive property 1: aHu + vL = a u + a v
85. Distributive property 2: Ha + cL v = a v + c v
86. Midpoint of a line segment Use vectors to show that the midpoint of the line segment joining PHx1 , y1 L and QHx2 , y2 L
is the point HHx1 + x2 L  2, Hy1 + y2 L  2L. (Hint: Let O be the origin and let M be the midpoint of PQ. Draw a picture and
1
1
show that OM = OP + PQ = OP + JOQ - OPN.)
2
2
87. Magnitude of scalar multiple Prove that c v¤ = c¤ v¤, where c is a scalar and v is a vector.
88. Equality of vectors Assume PQ equals RS. Does it follow that PR is equal to QS? Prove your conclusion.
89. Linear independence A pair of nonzero vectors in the plane is linearly dependent if one vector is a scalar multiple of
the other. Otherwise, the pair is linearly independent.
a. Which pairs of the following vectors are linearly dependent and which are linearly independent: u = X2, -3\,
v = X-12, 18\, and w = X4, 6\?
b. Explain geometrically what it means for a pair of vectors in the plane to be linearly dependent and independent.
c. Prove that if a pair of vectors u and v is linearly independent, then given any vector w, there are constants c1 and c2
such that w = c1 u +c2 v.
90. Perpendicular vectors Show that two nonzero vectors u = Xu1 , u2 \ and v = Xv1 , v2 \ are perpendicular to each other if
u1 v1 + u2 v2 = 0.
91. Parallel and perpendicular vectors Let u = Xa, 5\ and v = X2, 6\.
a. Find the value of a such that u is parallel to v.
b. Find the value of a such that u is perpendicular to v.
92. The Triangle Inequality Suppose u and v are vectors in the plane.
a. Use the Triangle Rule for adding vectors to explain why u + v¤ £ u¤ + v¤. This result is known as the Triangle
Inequality.
b. Under what conditions is u + v¤ = u¤ + v¤?
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