Vectors and ScalarsWeek 2

PHYSICS 10 - General Physics 1

Objectives•Differentiate vectors and scalar quantities•Add vectors using graphical methods•Determine the components of a vector•Determine the magnitude and direction of

vectors using its components•Define unit vectors

Vector and Scalar

•Scalar quantity – described by a single number

•Vector quantity – has both magnitude and direction in space

•Examples:▫Temperature = 20°C (Scalar)▫Displacement = 20 m, South (Vector)

Vectors•Represented by a letter or symbol in boldface

italic type with an arrow above them

•Arrows are used to represent vectors geometrically, plotted in a Cartesian plane

•Two vectors are parallel if they same direction (otherwise they are antiparallel)

•The magnitude of a vector, is a scalar quantity

Vector addition and subtraction•Graphical methods

▫Parallelogram (or triangle law)

▫Polygon method

Vector addition and subtraction

•Resultant vector – vector sum

•Properties▫Commutative: ▫Associative:

Vector addition and subtraction• Subtracting vectors

(negating a vector is equivalent to changing its direction to the opposite)

Thus:

Review Questions1. A circular racetrack has a radius of 500 m. What is

the displacement of a cyclist when she travels around the track from the north side to the south side? When she makes one complete circle around the track? Explain your reasoning.

2. Can you find two vectors with different lengths that have a vector sum of zero? What length restrictions are required for three vectors to have a vector sum of zero? Explain your reasoning.

3. Hearing rattles from a snake, you make two rapid displacements of magnitude 1.8 m and 2.4 m. In sketches (roughly to scale), show how your two displacements might add up to give a resultant of magnitude (a) 4.2 m; (b) 0.6 m; (c) 3.0 m.

Components of Vectors

• Finding a vector’s magnitude and direction from its components.

and

• Using components to calculate the vector sum (resultant) of two or more vectors.

Components of Vectors

Exercise1. Find the magnitude and direction of the vector

represented by the following pairs of components: (a) , (b) ,

2. Vector is 2.80 cm long and is above the x-axis in the first quadrant. Vector is 1.90 cm long and is below the x-axis in the fourth quadrant. Use components to find the magnitude and direction of (a) (b) (c) . In each case, sketch the vector addition or subtraction and show that your numerical answers are in qualitative agreement with your sketch.

Unit Vectors

Unit Vectors

Position and position vector• We define position as the location of an object

in space and specified in a coordinate axis

• In determining the position of an object in space, we use the position vector,

• The position vector determines the location of the object in space

r

jrirr yxˆˆ

PositionExample:The location of the hen is

Where 8 m is the x component of the hen’s position and 4 is the vertical component of the hen’s position

5r

m j4i8 r

Displacement • Displacement is the shortest distance between two points

• Displacement is defined as the change in the position between two points

• Displacement is a vector quantity

if rrr

DisplacementExample:Determine the displacement of an object that is initially located at and final position at Solution:

mˆ0.8ˆ0.2 jiri mˆ0.3ˆ0.4 jirf

m ˆ0.8ˆ0.2m ˆ0.3ˆ0.4 jijir if rrr

mjir ˆ0.11ˆ0.2

Exercise1. (a) Write each vector in the figure in

terms of the unit vectors and and find 2. (b) Find the magnitude and direction of .

Problems1. Three horizontal ropes pull on a large stone stuck in the ground, producing the vector forces , and shown in the fig. Find the magnitude and direction of a fourth force on the stone that will make the vector sum of the four forces zero.

Problems2. A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point (Fig. P1.72). Find the magnitude and direction of the third leg of the journey. Draw the vector addition diagram and show that it is in qualitative agreement with your numerical solution.

Review

• When two vectors and represented in terms of their components, we express vector sum using unit vectors:

A

RB

)15.1(ˆˆ

ˆˆ

ˆˆˆˆ

ˆˆ

ˆˆ

jRiR

jBAiBA

jBiBjAiA

BAR

jBiBB

jAiAA

yx

yyxx

yxyx

yx

yx

• If vectors do not all lie in the xy-plane, we need a third component. We introduce a third unit vector that points in the direction of the positive z-axis.

k

)17.1(ˆˆˆ

ˆˆˆ

)16.1(ˆˆˆ

ˆˆˆ

kRjRiRR

kBAjBAiBAR

kBjBiBB

kAjAiAA

zyx

zxyyxx

zyx

zyx

Example 1.9 Using Unit Vectors

Given the two displacements

Find the magnitude of the displacement .

Solution:Identify, Set Up and Execute:Letting , we have

mandm kjiEkjiD ˆ8ˆ5ˆ4ˆˆ3ˆ6

ED

2

m

m

m

kjiF

kji

kjikjiF

ˆ10ˆ11ˆ8

ˆ82ˆ56ˆ412

ˆ8ˆ5ˆ4ˆˆ3ˆ62

EDF

2

Example 1.9 (SOLN)

The units of the vectors , , and are meters, so the components of these vectors are also in meters. From Eqn 1.12,

Evaluate:Working with unit vectors makes vector addition and subtraction no more complicated than adding or subtraction ordinary numbers. Be sure to check for simple arithmetic errors.

D

mmmm 1710118 222

222

F

FFFF zyx

E

F

Products of VectorsKinds:1. Scalar Product (dot product) ●results to a scalar quantity

2. Vector Product (cross-product)

● yields to another vector

 

Right-hand Rule

Vector product

B

BA

Review of Products of Vectors

• We can express many physical relationships concisely by using products of vectors.

• There are two different kinds of products of vectors. The first, called the scalar product (dot product), which yields a result that is a scalar quantity. The second is the vector product (cross product) which yields another vector.

Scalar product• The scalar product of two vectors and is

denoted by . • The scalar product is also called the dot

product.

A

B

BA

Scalar product• We define to be the magnitude of

multiplied by the component of parallel to ,

BA

A

B

A

)18.1(coscos BAABBA

Scalar product• We can also express .• The scalar product of two perpendicular

vectors is always zero.

Scalar product0° to 90° Positive

90° Zero90° to 180° Negative

coscos ABABBA

Scalar product• Scalar product obeys the commutative law of

multiplication; order of the two vectors does not matter.

• We can calculate the scalar product directly if we know the x-, y-, and z-components of and . Using Eqn 1.18, we find

BA

A

B

)19.1(90cos11ˆˆˆˆˆˆ

10cos11ˆˆˆˆˆˆ

kjkiji

kkjjii

1.10 Products of Vectors

Scalar product• Now, we express and in terms of their

components, expand the product, and use the products of unit vectors

A

B

)20.1(ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

kkBAjkBAikBA

kjBAjjBAijBA

kiBAjiBAiiBA

kBjBiBkAjAiABA

zzyzxz

zyyyxy

zxyxxx

zyxzyx

1.10 Products of Vectors

Scalar product• From Eqns 1.19, six of the nine terms are

zero, thus

• Scalar product of two vectors is the sum of the products of their respective components.

• Eqn 1.21 can also be used to find the scalar product of and .

21.1zzyyxx BABABABA

A

B

Sample problem : scalar product

Find the scalar product of the two vectors shown in figure. The magnitudes of the vectors are A = 4.00 and B = 5.00.

BA

(SOLUTION)

Identify and Set Up:2 ways to calculate the scalar product. First way uses the magnitudes of the vectors and the angle between them (Eqn 1.18), and the second uses the components of the two vectors (Eqn 1.21).Execute:Using first approach, angle between the 2 vectors is =130.0 53.0 = 77.0 , so

This is positive as angle between and is between 0 and 90 .

50.40.77cos00.500.4cos ABBA

A

B

Execute:For second approach, find the components of the 2 vectors. Since angles of and are given with respect to the +x-axis, and these angles are measured in the sense from the +x-axis to the +y-axis, we use Eqns 1.17:

0

830.30.130sin00.5

214.30.130cos00.5

0

195.30.53sin00.4

407.20.53cos00.4

z

y

x

x

y

x

B

B

B

A

A

A

A

B

Example 1.10 (SOLN)

Execute:The z-components are zero because both vectors lie in the xy-plane. As in Example 1.7, we are keeping one too many significant figures in the components and will round them at the end. From Eqn 1.21,

Evaluate:We get the same result for the scalar product with both methods, as we should.

50.4

00830.3195.3214.3407.2

zzyyxx BABABABA

Key Equations

)10.1(yyy

xxx

BAR

BAR

)16.1(ˆˆˆ

ˆˆˆ

kBjBiBB

kAjAiAA

zyx

zyx

)22.1(sinABC

27.1xyyxz

zxxzy

yzzyx

BABAC

BABAC

BABAC

)18.1(coscos BAABBA

21.1zzyyxx BABABABA

Finding Angles with Scalar Product

Find the angle between the two vectors

Solution:Identify:The scalar product of two vectors and is related to the angle between them and to the magnitudes A and B. The scalar product is also related to the components of the two vectors. If we are given the components of the vectors, we first determine the scalar product and the values of A and B, then determine the target variable .

kjiBkjiA ˆˆ2ˆ4ˆˆ3ˆ2

and

BA

A

B

Finding Angles with Scalar Product

Set Up and Execute:Use either Eqns 1.18 or 1.21. Equating these two and rearranging,

This formula can be used to find the angle between any two vectors and . The components of are Ax = 2, Ay = 3 and Az = 1, and components of are Bx = 4, By = 2 and Bz = 1.

AB

BABABA zzyyxx cos A

B

A

B

A

B

Finding Angles with Scalar Product

Set Up and Execute:Thus,

100

175.02114

3cos

21124

14132

3112342

222222

222222

AB

BABABA

BBBB

AAAA

BABABABA

zzyyxx

zyx

zyx

zzyyxx

Example 1.11 Finding Angles with Scalar Product

Evaluate:As a check on this result, note that the scalar product is negative. This means that is between 90 and 180 , in agreement with our answer.

BA

1.10 Products of Vectors

Vector product• Vector product of two vectors and , also

called the cross product, denoted by .• To define the vector product, we draw the

vectors as shown.

A

B

BA

1.10 Products of Vectors

Vector product• The vector product is defined as a vector

quantity with a direction perpendicular to the plane (both and ) and a magnitude equal to AB sin .

• If

• The vector product of any two parallel or antiparallel vectors is always zero.

• The vector product of any vector with itself is zero.

A

B

BAC

)22.1(sinABC

1.10 Products of Vectors

Vector product• There are always two directions

perpendicular to a given plane, one on each side of the plane. We choose which of these is the direction of .

• This right-hand rule is what we use to determine the direction of the vector product.

• Note that vector product is not commutative. In fact, for any two vectors and

BA

)23.1(ABBA

A

B

1.10 Products of Vectors

Vector product• If we know the components of and , we

can calculate the components of the vector product, using a procedure similar to scalar product.

• First, we work out the multiplication table for the unit vectors The vector product of any vector with itself is zero, so

• The boldface zero illustrates that each product is a zero vector, that is, all components equal to zero with an undefined direction.

.ˆ,ˆ,ˆ kji

A

B

0 kkjjii ˆˆˆˆˆˆ

1.10 Products of Vectors

Vector product• Using Eqns 1.22 and 1.23 and the right-

hand rule, we find

24.1ˆˆˆˆˆ

ˆˆˆˆˆ

ˆˆˆˆˆ

jkiik

ijkkj

kijji

1.10 Products of Vectors

Vector product• Express and in terms of their

components and corresponding unit vectors,

A

B

25.1ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

)ˆˆˆ()ˆˆˆ(

kBkAjBkAiBkA

kBjAjBjAiBjA

kBiAjBiAiBiA

kBjBiBkAjAiABA

zzyzxz

zyyyxy

zxyxxx

zyxzyx

1.10 Products of Vectors

Vector product• Rewrite individual terms as

and so on. Evaluating with the multiplication table for the unit vectors and grouping the terms, we find

• Thus the components of are given by

27.1xyyxz

zxxzyyzzyx

BABAC

BABACBABAC

)26.1(ˆ

ˆˆ

kBABA

jBABAiBABABA

zyyx

zxxzyzzy

jiBAjBiA yxyx ˆˆˆˆ

BAC

1.10 Products of Vectors

Vector product• Vector product can also be expressed in

determinant form as:

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

1.10 Products of Vectors

Vector product• All vector products of the unit vectors

would have signs opposite to those in Eqn 1.24.

• Axis system where is called a right-handed system. The usual practice is to use only right-handed systems.

kji ˆ,ˆ, and

kji ˆˆˆ

Example 1.12 Calculating a vector product

Vector has magnitude 6 units and is in the direction of the +x-axis. Vector has magnitude 4 units and lies in the xy-plane, making an angle of 30 with the +x-axis. Find the vector product .

A

B

BA

Example 1.12 (SOLN)

Identify and Set Up:Find the vector product in one of two ways. First way is to use Eqn 1.22 to determine the magnitude ofand then use the right-hand rule to find the direction of the vector product. Second way is to use the components of and to find the components of the vector product using Eqn 1.27.Execute:With first approach using Eqn 1.22,

BA

A

B

BAC

1230sin46sin AB

Example 1.12 (SOLN)

Execute:From right-hand rule, direction of is along the +z-axis, so we have .For second approach, we first write the components of

Defining , we have

030sin43230cos4

006

zyx

zyx

BBB

AAA

kBA ˆ12 BA

BA

and

BAC

1232026

006320

02000

z

y

x

C

C

C

Example 1.12 (SOLN)

Execute:The vector product has only a z-component, and it lies along the +z-axis. The magnitude agrees with the result we obtained with the first approach, as it should.Evaluate:The first approach was more direct because the magnitudes of each vector and angle between them was known, and both vectors lay in one of the planes of the coordinate system. Sometimes, we need to find the vector product of 2 vectors that are not conveniently oriented or which only components are given. We use the second approach for such cases.

C

Concepts Summary

• Fundamental physical quantities of mechanics are mass, length and time.

• Corresponding basic SI units are the kilogram, the meter, and the second.

• Other units for these quantities, related by powers of 10, are identified by adding prefixes to the basic units.

• Derived units for other physical quantities are products or quotients of the basic units.

• Equations must by dimensionally consistent; two terms can be added only when they have the same units.

Concepts Summary

• Accuracy of a measurement can be indicated by the number of significant figures or by a stated uncertainty.

• Result of a calculation usually has no more significant figures than the input data.

• When only crude estimates are available for input data, we can often make useful order-of-magnitude estimates.

• Scalar quantities are numbers, and combine with usual rules of arithmetic.

Concepts Summary

• Vector quantities have direction as well as magnitude, and combine according to the rules of vector addition.

• Graphically, two vectors and are added by placing the tail of at the head, or tip, of .

• The vector sum then extends from the tail of to the head of .

• Vector addition can be carried out using components of vectors.

• The x-component of is the sum of the x-components of and , and likewise for y- and z-components.

A

B

BA

BAR

B

A

A

B

A

B

Concepts Summary

• Unit vectors describe directions in space.• A unit vector has a magnitude of one, with no

units.• The unit vectors, aligned with the x-, y-,

and z-axes of a rectangular coordinate system, are especially useful.

• The scalar product of two vectors and is a scalar quantity .

• It can be expressed in two ways: in terms of the magnitudes of and and the angle between the two vectors, or in terms of the components of and .

BAC

kji ˆ,ˆ,ˆ

A

B

A

B

A

B

Concepts Summary

• The scalar product is commutative; for any two vectors and , .

• The scalar product of two perpendicular vectors is zero.

• The vector product of two vectors and is another vector .

• The magnitude of depends on the magnitudes of and and the angle between the two vectors.

• The direction of is perpendicular to the plane of the two vectors being multiplied, as given by the right-hand rule.

ABBA

BAC

C

BA

A

B

A

B

A

B

BA

Concepts Summary

• The components of can be expressed in terms of the components of and .

• The vector product is not commutative; for any two vectors and , .

• The vector product of two parallel or antiparallel vectors is zero.

BAC

ABBA

A

B

A

B

Key Equations

)10.1(yyy

xxx

BAR

BAR

)16.1(ˆˆˆ

ˆˆˆ

kBjBiBB

kAjAiAA

zyx

zyx

)22.1(sinABC

27.1xyyxz

zxxzy

yzzyx

BABAC

BABAC

BABAC

)18.1(coscos BAABBA

21.1zzyyxx BABABABA