Slide 7.3 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.3 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Vectors

Learn to represent vectors geometrically.Learn to represent vectors algebraically.Learn the definition of a unit vector. Learn to write a vector in terms of its magnitude and direction.Learn the definition of the dot product.Learn to find the angle between two vectors.Learn the definition of work.

SECTION 7.3

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VECTORSMany physical quantities, such as length, area, volume, mass, and temperature, are completely described by their magnitudes in appropriate units. Such quantities are called scalar quantities. Other physical quantities, such as velocity, acceleration, and force, are not completely determined until both a magnitude (size) and a direction are specified. For example, the movement of wind is usually described by its speed (magnitude) and the direction. The wind speed and wind direction together form a vector quantity called the wind velocity.


GEOMETRIC VECTORS

A vector can be represented geometrically by a line segment with an arrowhead. The direction of the arrow specifies the direction of the vector. The length of the arrow describes its magnitude. The tail of the arrow is called the initial point of the vector, and the tip of the arrow the terminal point. We shall denote vectors by lowercase boldface type, such as a, b, i, j, u, v, and w. When discussing vectors, we refer to real numbers as scalars. Scalars will be denoted by lower case italic type, such as a, b, x, y, and z.


GEOMETRIC VECTORS

If the initial point of a vector v is P and the terminal point is Q, we write

v PQu ruu

.

The magnitude (or norm)of a vector v PQ

u ruu,

denoted byv , or PQu ruu

,is the length of the vector v and is a scalar quantity.


EQUIVALENT VECTORS

Two vectors having the same length and same direction are called equivalent vectors.


EQUIVALENT VECTORS

Equivalent vectors are regarded as equal even though they may be located in different positions.


ZERO VECTOR

The vector of length zero is called the zero vector and is denoted by 0. The zero vector has zero magnitude and arbitrary direction. If vectors v and a, have the same length and opposite direction, then a is the opposite vector of v and we write a = –v.


VECTOR ADDITION (TRIANGLE RULE)

Let v and w be any two vectors. Place the vector w so that its initial point coincides with the terminal point of v. The vector v + w, called the resultant vector, is represented by the arrow from the initial point of v to the terminal point of w.

v

wv + w


VECTOR ADDITION (PARALLELOGRAM RULE)

Place vectors v and w so that their initial points coincide. Form a parallelogram with v and w as the adjacent sides. The vector with the same initial point as the initial points of v and w that coincides withthe diagonal of the parallelogram represents the resultant vector, v + w.

Note that:v + w = w + v


VECTOR SUBTRACTION

For any two vectors v and w, v – w = v + (–w), where –w is the opposite of w.


SCALAR MULTIPLES OF VECTORS

Let v be a vector and c a scalar (a real number). The vector cv is called the scalar multiple of v.

If c > 0, cv has the same direction as v and magnitude c||v||.

If c < 0, cv has the opposite direction as v and magnitude |c| ||v||.

If c = 0, cv = 0v = 0.


EXAMPLE 1 Geometric Vectors

Use the vectors u, v, and w to graph each vector.

a. u – 2w b. 2v – u + w

Solution

a. u – 2w


EXAMPLE 1 Geometric Vectors

Solution continued

b. 2v – u + w


ALGEBRAIC VECTORS

Specifying the terminal point of the vector will completely determine the vector. For the position vector v with initial point at the origin O and terminal point at P(v1, v2), we denote the vector by

A vector drawn with its initial point at the origin is called a position vector.

v OPu ruu

v1,v2 .


COMPONENTS OF VECTORS

A point is denoted (v1, v2),

We call v1 and v2 the components of the vector v; v1 is the first component, and v2 is the second component.

v1,v2 .we denote a vector

v v1,v2The magnitude offollows directly from the Pythagorean Theorem,

v v12 v2

2 .


EQUIVALENT VECTORS

points must coincide.

If equivalent vectors, v and w, are located so that their initial are at the origin, then their terminal

w w1,w2 .v v1,v2

v w if and only ifv1 w1 and v2 w2

v v1,v2


REPRESENTING A VECTORAS A POSITION VECTOR

The vector with initial

point P(x1, y1) and terminal

point Q(x2, y2) is equal to

the position vectorw x2 x1, y2 y1 .

PQu ruu


EXAMPLE 2 Representing a Vector in the Cartesian Plane

Let v be the vector with initial point P(4, –2) and terminal point Q(–1, 3).Write v as a position vector.Solution

v has:initial point P(4, –2), so x1 = 4 and y1 = –2terminal point Q(–1, 3), so x2 = –1 and y2 = 3

v 1 4, 3 2 v 5, 5

v x2 x1, y2 y1Thus,


ARITHMETIC OPERATIONS ON VECTORS

are vectors and c is any scalar, then

v w v1 w1,v2 w2

If v v1,v2 and w w1,w2

v w v1 w1,v2 w2

cv cv1,cv2


EXAMPLE 3 Operations on Vectors

Find each expression.a. v + w b. –2v c. 2v – w d. ||2v – w||

b. 2v 2 2, 3

Solution a. v w 2, 3 4,1

Let v 2, 3 and w 4,1 .

2 4, 31

2, 4

22, 23 4, 6


EXAMPLE 3 Operations on Vectors

Solution continuedc. 2v w 2 2, 3 4,1

d. 2v w 8,5

4,6 4,1

4 4 ,6 1

8,5

2v w 82 52

2v w 64 25 89


UNIT VECTORS

A unit vector has length 1.

1

vv.

The unit vector in the same direction as v is given by


EXAMPLE 4 Finding a Unit Vector

Find a unit vector u in the direction of v 3, 4 .

Solution

v 3 2 4 2 25 5

First, find the magnitude of v 3, 4 .

u 1

vv

1

53, 4

3

5,

4

5Now,

u 3

5

2

4

5

2

9

25

16

25

25

251

Check:


UNIT VECTORS i AND jIn a Cartesian coordinate plane, two unit vectors that lie along the coordinate axes are particularly important. These are the vectors

i 1,0 and j 0,1 .

A vector v from (0, 0) to (v1, v2) can be represented in the form

v v1i v2 j

v v12 v2

2 .with


EXAMPLE 5 Vectors Involving i and j

Find each expression for u = 4i + 7j and v = 2i + 5j.

Solutiona. u – 3v = (4i + 7j) – 3(2i + 5j)

2 2 8 2

a. u – 3v b. ||u – 3v||

= 4i + 7j – 6i – 15j = (4 – 6)i + (7 – 15)j= –2i – 8j

b. u 3v 2i 8 j

68 2 17


VECTORS IN TERMS OFMAGNITUDE AND DIRECTION

Let v v1,v2

that is the smallest positive angle that v makes with the positive x-axis. The angle is called the direction angle of v.

v v cosi sin j The formula

be a position vector, and suppose

expresses a vector v in terms of its magnitude ||v|| and its direction angle .


EXAMPLE 6Writing a Vector with Given Length and Direction Angle

Solutionv v cosi sin j

v 3 cos3

i sin3

j

v 31

2i

3

2j

v 3

2i

3 3

2j

Write the vector with magnitude 3 that makes an angle of with the positive x-axis.

3


THE DOT PRODUCT

For two vectors v v1,v2 and w w1,w2 ,

the dot product of v and w, denoted v • w, is

defined as:

v w v1,v2 w1,w2 v1w1 v2w2


EXAMPLE 7 Finding the Dot Product

Find the dot product v • w.a. v 2, 3 and w 3, 4

b. v 3i 5 j and w 2i 3j

Solution a. v w 2, 3 3, 4

2 3 3 4 6 12 6

3,5 2, 3

3 2 5 3 6 15 9

b. v w 3i 5 j 2i 3j


PROPERTIES OF THE DOT PRODUCT

1. uv v u

2. u v w uv uw3. 0v 0

4. v v v2

5. cu v c uv u cv

If u, v, and w are vectors and c is a scalar, then


THE DOT PRODUCT ANDTHE ANGLE BETWEEN TWO VECTORS

v w v w cos

Let (0 ≤ ≤ π) be the angle between two nonzero vectors v and w. Then:

cos v wv w

or


EXAMPLE 11 Using Vectors in Air Navigation

An F-15 fighter jet is flying over Mount Rushmore at an airspeed (speed in still air) of 800 miles per hour on a bearing of N 30º E. The velocity of wind is 40 miles per hour in the direction of S 45º E. Find the actual speed and direction (relative to the ground) of the plane. Round each answer to the nearest tenth.

Set up a coordinate system with north along the positive y-axis.

Solution



Letv be the air velocity (plane),w be the wind velocity, andr be the resultant ground

velocity of the plane.Then,

Solution continued

v 800 cos60º i sin 60º j w 40 cos 45º i sin 45º j

r 800 cos60º i sin 60º j 40 cos 45º i sin 45º j r v w


Solution continued

r 800 cos60º 40 cos 45º i 800sin 60º 40sin 45º j


r 800 cos60º40 cos 45º 2 800sin 60º 40sin 45º 2

r 790.6

The ground speed of the F-15 is approximately 790.6 miles per hour.

To find the actual direction (bearing) of the plane, find the angle between r and j (unit vector in the direction north).


Solution continued

cos jrj r


cos 0.8405

The direction of the F-15 relative to the ground is approximately N 32.8º E.

cos 800sin 60º 40sin 45º

790.6

cos 1 0.8405 32.8º


DEFINITION OF WORK

W FPQu ruu

The work W done by a constant force F in moving an object from a point P to a point Q is defined by


EXAMPLE 12 Computing Work

A child pulls a wagon along a level ground, with a force of 40 pounds along the handle on the wagon that makes an angle of 42º with the horizontal. How much work has she done by pulling the wagon 150 feet?

W 40 150 cos 42º

W 4458.87 foot-pounds

Solution

W FPQu ruu

W F PQu ruuu

cos

Slide 7.3 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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