Vector and scalar

Chapter 1(b)

Scalar and vectors Definitions Addition n subtraction rules Scalar and dot product

Learning Outcomes

Define scalar and vector quantities, unit vector in Cartesian coordinates.

Explain vector addition and subtraction n their rules.

Define and use dot and cross product (multiplying vector)

Trigonometry

Trigonometry

h

hosin

h

hacos

a

o

h

htan

To find the length

Trigonometry

h

ho1sin

h

ha1cos

a

o

h

h1tan

To find the angle

Characteristics of a Scalar

Quantity Only has magnitude

Requires 2 things:

1. A value

2. Appropriate units

Ex. Mass: 5kg

Temp: 21 C

Speed: 65 m/s

Characteristics of a Vector

Quantity Has magnitude & direction

Requires 3 things:

1. A value

2. Appropriate units

3. A direction!

Ex. Acceleration: 9.8 m/s2 down

Velocity: 25 m/s West

Scalars and Vectors

By convention, the length of a vector

arrow is proportional to the magnitude

of the vector.

8 N 4 N

Arrows are used to represent vectors. The

direction of the arrow gives the direction of

the vector.

Scalar and Vector Quantities

The car moved a distance of 2 km in a

direction 30o north of east

3-2 The Components of a Vector

. ofcomponent vector theand

component vector thecalled are and

r

yx

y

x

The Components of a Vector

.AAA

AA

A

yx

that soy vectoriall together add and

axes, and the toparallel are that and vectors

larperpendicu twoare of components vector The

yxyx

The Components of a Vector

It is often easier to work with the scalar components

rather than the vector components.

. of

componentsscalar theare and

A

yx AA

1. magnitude with rsunit vecto are and yx

yxA yx AA

The Components of a Vector

Example

A displacement vector has a magnitude of 175 m and points at

an angle of 50.0 degrees relative to the x axis. Find the x and y

components of this vector.

r

ysin

m 1340.50sinm 175sin ry

r

xcos

m 1120.50cosm 175cos rx

yxr m 134m 112

Signs of vector components:

The components of a vector

EXERCISE

1)The vector A has a magnitude of 7.25 m.

Find its components for direction of

angles of

(a)=5.0o

(b)=125o

(c)= 245o

(d) = 335o

Answer

(a)Ax=7.22m, Ay=0.632m

(b) Ax=-4.16m, Ay=5.94m

(c) Ax=-3.06m, Ay=-6.57m

(d) Ax=6.57m, Ay=-3.06m

IMPORTANT FOR VECTOR

COMPONENTS

Given the components of a vector, find its

magnitude and direction:

x

y

yx

A

A

AAA

1

22

tan

cosAAx

sinAAy

Ay= 2.00 m

Ax = 6.00 m

EXAMPLE:

Length, angle, and components can

be calculated from each other using

trigonometry:

Given vector component x is 6.00m and vector

component y is 2.00m, find the magnitude and

direction of vector A.

A

2.00 m

6.00 m

222 m 00.6m 00.2 R

R

m32.6m 00.6m 00.2 22 R

2.00 m

6.00 m

00.600.2tan

4.1800.600.2tan 1

If each component of a

vector is doubled, what

happens to the angle of

that vector?

1) it doubles

2) it increases, but by less than double

3) it does not change

4) it is reduced by half

5) it decreases, but not as much as half

Question 3.4 Vector Components I

If each component of a

vector is doubled, what

happens to the angle of

that vector?

1) it doubles

2) it increases, but by less than double

3) it does not change

4) it is reduced by half

5) it decreases, but not as much as half

The magnitude of the vector clearly doubles if each of its

components is doubled. But the angle of the vector is given by tan

= 2y/2x, which is the same as tan = y/x (the original angle).

Question 3.4 Vector Components I

Vector Addition and Subtraction

Often it is necessary to add one vector to another.

Vector Addition and Subtraction

5 m 3 m

8 m

Vector Addition and Subtraction

A

B

BA

A

B

BA

Component Method of Vector Addition

Treat each vector separately:

1. To find the X component, you must:

Ax = Acos

2. To find the Y component, you must:

Ay = Asin

3. Repeat steps 1 & 2 for all vectors

Component Method (cont.)

4. Add all the X components (Rx)

5. Add all the Y components (Ry)

6. The magnitude of the Resultant Vector is

found by using Rx, Ry & the Pythagorean

Theorem:

R2 = Rx2 + Ry2

7. To find direction: Tan = Ry / Rx

Vector Addition and Subtraction

Adding vectors graphically: Place the tail of the second at the head of

the first. The sum points from the tail of the first to the head of the last.

yxA yx AA

Addition Rule for Two Vectors

BAC

1. Find the components of each vector to be added.

yxB yx BB

yxyxyxC

yyxx

yxyx

BABA

BBAA

xxx BAC

yyy BAC

Addition Rule for Two Vectors

2. Add the x- and y-components separately.

3. Find the resultant vector.

Subtracting Vectors

Subtracting Vectors: The negative of a vector is a vector of the same

magnitude pointing in the opposite direction. Here,

D= A B

Lets try

)5(

)22332(

)223()32())(

)35(

)22332(

)223()32())(

zyx

zyxzyx

zyxzyxBAii

zyx

zyxzyx

zyxzyxBAi

Component Method (cont.)

Lets try!

A = 2 m/s 30 N of E

B = 3 m/s 40 N of W

(this is easy!)

Find: Magnitude & Direction

Magnitude = 2.98 m/s

Direction = 79.02 N of W@

180-79.02 =100.98

Component Method (cont.)

Lets try!

F1 = 37N 54 N of E

F2 = 50N 18 N of W

F3 = 67 N 86 S of W

(this is not so easy!)

Find: Magnitude & Direction

Magnitude =37.3 N

Direction = 35.1 S of W @

180+35.1=215.1

3-4 Unit Vectors

Unit vector is dimensionless vectors of unit length (magnitude of 1) with a function to indicate direction.

Unit vector - indicates the x-direction

Unit vector - indicates the y-direction

Unit vector - indicates the z-direction

x

zy

EXERCISE

To find a magnitude of a vector

1) Find magnitude vector A and vector B respectively

2) Find magnitude of vector A +B

zyxB

zyxA

223

32

To find a magnitude of a vector

1) Find magnitude vector A and vector B respectively

units

units

2) Find magnitude of vector A +B

zyx

zyxzyxBA

35

22332)(

units

BA

9.535

135222

zyxB

zyxA

223

32

There are two distinct ways to multiply vectors, referred to as the dot product and the cross product.

The dot product yields a scalar (a number) as the result.

The cross product yields a vector as the result.

Vector Multiplication

Definition of the scalar, or dot, product:

Application example:

Work is the dot product of force and displacement

zByBxBzAyAxABA zyxzyx

zzyyxx BABABA

Dot Product

Exercise

yxb

yxa

52

86

1) Find a.b

2) Find angle between a and b

The vector cross product is defined as:

The direction of the cross product is defined by a

right-hand rule:

Cross Product

The cross product can also be written in determinant

form:

Cross Product

Application example: The relation of the magnetic force

on a charge q with a velocity in a magnetic field is v

B

sinqvBBvqF

x zy

zBABAyBABAxBABA xyyxxzzxyzzy

Exercise

zyxb

zyxa

843

562

1) Find axb

2) Find the angle between a and b