CHAPTER 1.4: SCALAR AND VECTORCHAPTER 1.4: SCALAR AND VECTOR

SCALAR AND VECTOR

Scalars are quantities which have magnitude without direction

Examples of scalars

• temperature• mass• kinetic energy

• time• amount• density• charge

Scalars

Vector

A vector is a quantity that has both magnitude (size) and direction

it is represented by an arrow whereby– the length of the arrow is the magnitude, and– the arrow itself indicates the direction

The symbol for a vector is a letter with an arrow over it

A

Example

Two ways to specify a vector

It is either given by• a magnitude A, and• a direction

Or it is given in the x and y components as

• Ax

• Ay

y

x

A

A

Ay

x

Ax

Ay

y

x

AAx

AyA

Ax = A cos

Ay = A sin

│A │ =√ ( Ax2

+ Ay2

)

The magnitude (length) of A is found by using the Pythagorean Theorem

The length of a vector clearly does not depend on its direction.

y

x

AAx

AyA

The direction of A can be stated as

tan = Ay / Ax

=tan-1(Ay / Ax)

Some Properties of Vectors

Equality of Two Vectors

Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same directions. i.e. A = B

A BA

A

B

B

Negative of a Vector

The negative of vector A is defined as giving the vector sum of zero value when added to A . That is, A + (- A) = 0. The vector A and –A have the same magnitude but are in opposite directions.

A

-A

Scalar Multiplication

The multiplication of a vector Aby a scalar

- will result in a vector B

B = A- whereby the magnitude is changed but not the direction

• Do flip the direction if is negative

B = A

If = 0, therefore B = A = 0, which is also known as a zero vector

(A) = A = (A)

(+)A = A + A

Example

The addition of two vectors A and B

- will result in a third vector C called the resultant

C = A + B

A

BC

Geometrically (triangle method of addition)

• put the tail-end of B at the top-end of A• C connects the tail-end of A to the top-end of B

We can arrange the vectors as we like, as long as we maintain their length and direction

Vector Addition

Example

More than two vectors?

x1

x5

x4

x3

x2

xi

xi = x1 + x2 + x3 + x4 + x5

Example

Vector Subtraction

Equivalent to adding the negative vector

A

-BA - B

B

A BC =

A + (-B)C =

Example

Rules of Vector Addition

commutative

A + B = B + A

A

B

A + BB

A A + B

associative

(A + B) + C = A + (B + C)

B

CA

B

CA A + B

(A + B) + CA + (B + C)

B + C

distributive

m(A + B) = mA + mB

A

B

A + B mA

mB

m(A + B)

Parallelogram method of addition (tailtotail)

A

B

A + B

The magnitude of the resultant depends on the relative directions of the vectors

a vector whose magnitude is 1 and dimensionless

the magnitude of each unit vector equals a unity; that is, │ │= │ │= │ │= 1

i a unit vector pointing in the x direction

j a unit vector pointing in the y direction

k a unit vector pointing in the z direction

and defined as

Unit Vectors

k

j

i

Useful examples for the Cartesian unit vectors [ i, j, ki, j, k ] - they point in the direction of the x, y and z axes respectively

x

y

z

ii

jj

kk

Component of a Vector in 2-D

vector A can be resolved into two components

Ax and Ay

x- axis

y- axis

Ay

Ax

A

θ

A = Ax + Ay

The component of A are

│Ax│ = Ax = A cos θ

│Ay│ = Ay = A sin θ

The magnitude of A

A = √Ax2 + Ay

2

tan = Ay / Ax

=tan-1(Ay / Ax)

The direction of A

x- axis

y- axis

Ay

Ax

A

θ

ExampleExample

The unit vector notation for the vector A is written

A = Axi + Ayj

x- axis

y- axis

Ax

Ay

θ

A

i

j

ExampleExample

Component of a Vector in 3-D

vector A can be resolved into three components

Ax , Ay and Az

A

Ax

Ay

Az

z- axis

y- axis

x- axis

ij

k

A = Axi + Ayj + Azk

if

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A + B = C sum of the vectors A and B can then be obtained as vector C

C = (Axi + Ayj + Azk) + (Bxi + Byj + Bzk)

C = (Ax + Bx)i+ (Ay + By)j + (Az + Bz)kC = Cxi + Cyj + Czk Example

Dot product (scalar) of two vectors

The definition:

θ

B

AA · B = │A││B │cos θ

if θ = 900 (normal vectors) then the dot product is zero

Dot product (scalar product) properties:

if θ = 00 (parallel vectors) it gets its maximum

value of 1

and i · j = j · k = i · k = 0|A · B| = AB cos 90 = 0

|A · B| = AB cos 0 = 1 and i · j = j · k = i · k = 1

A + B = B + A

the dot product is commutative

Use the distributive law to evaluate the dot product if the components are known

A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk)

A. B = (AxBx) i.i + (AyBy) j.j + (AzBz) k.k

A . B = AxBx + AyBy + AzBz

Example

Cross product (vector) of two vectorsThe magnitude of the cross product given by

the vector product creates a new vector

this vector is normal to the plane defined by the

original vectors and its direction is found by using the

right hand rule

│C │= │A x B│ = │A││B │sin θ

θ

A

BC

if θ = 00 (parallel vectors) then the cross

product is zero

Cross product (vector product) properties:

if θ = 900 (normal vectors) it gets its maximum

value

and i x i = j x j = k x k = 0|A x B| = AB sin 0 = 0

|A x B| = AB sin 90 = 1 and i x i = j x j = k x k = 1

the relationship between vectors i , j and k can

be described as

i x j = - j x i = k

j x k = - k x j = i

k x i = - i x k = j

Example

Measurement and Error

THE END

Vectors are represented by an arrow

A- B

BA

A

θ

Conceptual Example

If B is added to A, under what condition does the resultant vector A + B have the magnitude equal to A + B ?

Under what conditions is the resultant vector equal to zero?

*

Example (1Dimension)

x1 = 5

x1 x2

x2 = 3

x = x2 - x1 = 2

x1 + x2

x1

x2

x = x2 - x1

x1 + x2 = 8

MORE EXAMPLEMORE EXAMPLE

Example 1 (2 Dimension)

If the magnitude of vector A and B are equal to 2 cm and 3 cm respectively , determine the magnitude and direction of the resultant vector, C for

B

Aa) A + B

b) 2A + B

SOLUTIONSOLUTION

Solution

a) |A + B| = √A2 + B2

= √22 + 32

= 3.6 cm

The vector direction

tan θ = B / A

θ = 56.3

b) |2A + B| = √(2A)2 + B2

= √42 + 32

= 5.0 cm

The vector direction

tan θ = B / 2A

θ = 36.9

MORE EXAMPLEMORE EXAMPLE

Example 2 (A Vacation Trip)

A car travels 20.0 km due north and then 35.0 km in a direction 600 west of north. Find the magnitude and direction of the car’s resultant displacement.

SOLUTIONSOLUTION

Solution

The magnitude of R can be obtained using the law of cosines as in figure

Since θ =1800 – 600 = 1200 and

C2 = A2 + B2 – 2AB cos θ, we find that

C = √A2 + B2 – 2AB cos θ

C = √202 + 352 – 2(20)(35) cos 1200

C = 48.2 km

C

A

B60

θβ

ContinueContinue

The direction of C measured from the northerly direction can be obtained from the sines law

CB

sinsin

629.0120sin2.48

0.35sinsin 0

C

B

β = 38.90

Therefore, the resultant displacement of the car is 48.2 km in direction 38.90 west of north

If one component of a vector is not zero, can its magnitude be zero? Explain.

*Conceptual Example

MORE EXAMPLEMORE EXAMPLE

Conceptual Example

If A + B = 0, what can you say about the components of the two vectors?

*

Example 1

Find the sum of two vectors A and B lying in the xy plane and given by

A = 2.0i + 2.0j and B = 2.0i – 4.0j

SOLUTIONSOLUTION

Solution

Comparing the above expression for A with the general relation A = Axi + Ayj , we see that Ax= 2.0 and Ay= 2.0. Likewise, Bx= 2.0, and By= -4.0 Therefore, the resultant vector C is obtained by using EquationC = A + B + (2.0 + 2.0)i + (2.0 - 4.0)j = 4.0i -2.0j

or Cx = 4.0 Cy = -2.0

The magnitude of C given by equation

C = √Cx2 + Cy

2 = √20 = 4.5

*

Find the angle θ that C makes with the positive x axis

Exercise

Example

A particle undergoes three consecutive displacements d1 = (1.5i + 3.0j – 1.2k) cm,

d2 = (2.3i – 1.4j – 3.6k) cm d3 = (-1.3i + 1.5j) cm. Find the component and its magnitude.

Solution

R = d1 + d2 + d3

= (1.5 + 2.3 – 1.3)i + (3.0 – 1.4 + 1.5)j + (-1.2 – 3.6 + 0)k

= (2.5i + 3.1j – 4.8k) cm

That is, the resultant displacement has component

Rx = 2.5 cm Ry = 3.1 cm and Rz = -4.8 cm

Its magnitude is

R = √ Rx2 + Ry

2 + Rz2

= 6. 2 cm

Example - 2D [headtotail]

x1

x2

x1 + x2

(1, 0)

(2, 2)

x1 + x2 = (1, 0) + (2, 2)= (3, 2)

Example - 2D [tailtotail]

x1 - x2?

(x2)

x1

x1 + x2x2

(1, 0)

(2, 2)

x1 + x2 = (1, 0) + (2, 2)= (3, 2)

Example of 2D (subtraction)

(1, 0)

(2, 2)

x1

x2

x1 + x2

Example -2D for subtraction

x1

-x2x1 - x2

(1, 0)

(2, 2)

x1 - x2 = (1, 0) - (2, 2)= (-1, -2)

x1 - x2 = x1 + (-x2)

Not given the components?

1 m

22 m45o

X1 = (1, 0)X2 = (x2E, x2N) = (22cos(45o), 22sin(45o)) = (2, 2)

x1

-x2x1 - x2

22 m

1 m

45o

Cosine rule:a2=b2 + c2 - 2bccosA = 1 + 8 - 22(1/ 2)a = 5 m