Vectors and Scalars
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Vectors and Scalars. Vectors & Scalars. Vectors are the physical quantities having ; 1- Magnitude 2- Direction 3- Should obey the parallelogram law of addition

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Vectors scalars

Vectors & Scalars

Vectors are the physical quantities having ;

1- Magnitude

2- Direction

3- Should obey the parallelogram law of addition

For Ex- Force ,Acceleration , weight are vectors because to define them direction and magnitude both are needed . They also follow the parallelogram law of addition.

Scalar quantity :- Those physical quantities which requires only magnitude to explain it .

Mass, distance , work , energy are scalar quantities ,because to define them magnitude is sufficient.


Distance is scalar quantity whereas displacement is vector quantity

Displacement is vector quantity :- Aero plane and ship required a fixed direction to go ahead as there are no roads in the sky or in ocean. So distance cover by then is known as displacement , which is vector quantity.

Distance is scalar quantity :- In roads Bus generally moves according to zig-zag path ,because the Bus does not follow a fixed direction so distance cover by it is treated as scalar.


C quantity

C

C

C

Click to start

Mass and Weight

Mass- It is represented by m. It is the amount of matter which a body possess. The more mass an object has, the more force is needed to get it moving or slow it. It is scalar quantity.

Weight -: It is the force by which earth attracts body towards its centre. It is represented by mg, where g ( for earth g is about 9.8 m/sec2) is the gravitational acceleration. It is vector quantity.

Direction of Earth Pull

Earth


Types of vectors quantity

Unit Vector – It is the vector having magnitude one ‘1’. If displacement is 5m east . It means that its unit vector is 1m east.

If a = a1î+a2Ĵ+a3kˆ then unit vector of it is represented as â (To distinguish this vector from others it is represented as â.), where

â = (a1î+a2Ĵ+a3k ˆ )/(a12+a22+a32)

If b = 2î+3Ĵ+4kˆ then

bˆ = (2î+3Ĵ+4kˆ)/  (22+32+42)

= (2î+3Ĵ+4kˆ)/  39

Zero Vector – It is the vector having magnitude zero. It may have any direction. It may be represented as AA, BB or CC etc.


Parallel vectors:- quantity If the vectors are parallel to each other then they are called parallel vectors. Here vector l , m ,n are parallel to each other. Vector product of two parallel vectors is zero, i.e. is l x m =0

m = l and n = l

l

m

n

Perpendicular vectors :- If the vectors are perpendicular to each other than they are called perpendicular vectors. Scalar product of two perpendicular vectors ( a & b) is zero i.e. a.b =0 Here vectors a & b are perpendicular to each other.

l

m


Parallelogram law of addition quantity

R

B

C

Q

α

α

O

D

P

A

Click to start

QSinα

Q

QCosα

In OBD R2 = (P+QCosα)2 + Q2Sin2α

R2 = (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α

= (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α

= P2+Q2 ( Cos2α +Sin2α)+2PQCosα

= P2+Q2 +2PQCosα


R= 5 quantity

i2= 3Amp

I1 = 3Amp

  • On the other part current despite of having direction and magnitude does not obey the parallelogram law of addition

In the above figure the resultant current should be 5 Amp, but it is actually 7 amp(3+4). It shows that current is a scalar quantity.


Addition of many vectors quantity

α3

c

c

c

c

α2

d

d

d

d

R

R

b

b

b

b

α1

a

a

a

a

See adding of vectors

See angle between different vectors

b

c

α2

α1

α3

Here R = a + b+ c+ d

Click to start

Given vectors

Here we are interested to get a+b+c+d

Step wise way of addition


RSin quantityα

R

α

RCosα

P

Mg

Click to start

How a Person moves?

When a person push the earth backward, the earth provides the force of reaction “R” to person. R can be divided into two components, RSinα and RCosα . RSinα helps to make the weight of person light and RCosα helps to push a person forward.

R is the reaction force given by earth.

Here apparent weight of the body = Mg - Sinα

Pushing the earth backward


F quantity2

F3

4

3

2

1

F1

F5

F4

R

If many forces are working at a point, then finding of resultant

Here these forces can be break in their perpendicular components X & Y.

X = F1 Cos0 + F2 Cos1 + F3Cos 2 +F4Cos 3 + F5Cos 4

Y = F1 Sin0 + F2 Sin1 + +F3Sin 2+ F4Sin 3 + F5Sin 4

So the resultant vector R =  X2 + Y2

I f resultant R makes an angle with X axis then tan = Y/X


The another way of addition may be quantity

3-2

F4

F3

2-1

4

F5

4-3

F2

F3

3

F2

R = F1+F2+F3+F4+F5

2

1

1

F1

F1

F5

F4


Click to start quantity

Click to start

Force is a vector quantity because it is either pull or push. Pull or push have direction also in addition to magnitude, so it is vector quantity.

It is pulling. Direction of pulling is shown.

It is pushing. Direction of force is shown in figure.


Pulling of Roller is easier than pushing quantity

Click to start

F Sinα

F (Pulling)

α

F Cosα

Pulling

This force can be divided into two components.

Here net weight (downward force on roller )of roller after applying force= Mg – FSinα

The net horizontal pulling force = FCosα

It shows that here the apparent weight of roller decreases, So it is easier to pull a roller instead of pushing.

Roller

Mg


Click to start quantity

F Sinα

F (Pushing)

α

F Cosα

Pushing

Here net weight (downward force on roller )of roller = Mg + FSinα

The net horizontal pulling force = FCosα

This force can be divided into two components.

It shows that here the apparent weight of roller increases, So it is easier to pull a roller instead of pushing.

Roller

Note- In pushing or pulling angle ‘α’ remains same.

Mg


Path of Projectile quantity

v

vSin1

1

vCos1

H= It is the maximum height attained by the projectile.

u

uSin

uCos

C

uCos = vCos1 ( Here vertical component becomes zero.)

B

vCos2

2

A

v

vSin2

(Range ‘R’:- It is the maximum distance cover by the projectile )

uCos = vCos1 = vCos2(Horizontal components of velocity always remains same.)

vSin1 = uSint1 -1/2gt12 Relation between u & v at point A & B


u quantity

u

Position of particle after time ‘t’

uSin

uSinα

y

x

uCos

uCosα

When a particle/body is thrown in the sky it’s path becomes parabolic.

x= uCos.t( Horizontal component of velocity (uCos) is responsible for covering horizontal distance ‘x’.)

y= uSInt – ½ gt2( Vertical component of velocity (uSIn) is responsible for covering vertical distance ‘y’.)

Or y = uSin(x/uCos ) – 1/2g(x/uCos)2

OrY = x tan - (gSec2/2u2)x2

This equation is of the form Y = Ax + BX2

Which is the equation of parabola. Hence the path of projectile is parabolic.

(x,y)


Flight Time- quantity-Let a body takes time ‘t’ to reach in the highest point C. At C the vertical velocity of the body will be zero. So

v = uSin - gt

0 = uSin - gt

or t= uSin/g

So total time taken T =2t = 2uSin/g

RangeR = horizontal velocity x total flight time

Range R = uCos x T (horizontal component of velocity is responsible for the horizontal distance covered by the body.)

=uCos x 2uSin/g

= u2 2SinCos/g

= u2 Sin2/g Max value of Sin2 =1 So 2= 900 or  =450

Therefore max R = u2/2g

Vertical Height H

At point C the vertical component of velocity =0 ,So by

v2= u2 - 2gH

0= u2Sin2 - 2gH

or H= u2Sin2/2g Max value of Sin =1 so  = 900

Therefore maximum vertical height which can be attained H= u2/2g


quantity


45 quantity0

R= longest range

When  = 450 , then body covers the longest range.

A hammer thrower should throw the hammer at the angle of 450 in order to cover maximum distance.


70 quantity0


When quantity = 900 , then body get the greatest height.

High jumper should take lift at an angle of 900 in order to cover maximum vertical distance.


Vector represents in 3 dimension quantity

Click to start

Y

2i + 3j + 4k

X

900

900

900

i

j

k

Z

If a vector is 2i + 3j +4k then it means that component of vector which is in the direction of X ,Y and Z axis are 2,3 and 4 respectively.


Vector Products quantity

^

^

^

^

^

n

n

n

n

n

b

b

a

a

Direction of aXb is perpendicular to both a & b. It is shown by .

Area covered by a & b can be calculated as aXb

b

a

If two vectors are parallel to each other

then  = 0 , so a x b = absin0 = 0 .

Click to start

Vectors are multiplied in two ways

1-Vector product

2- Scalar Product

1-Vector product :- If a and b are two vectors then axb = abSin

axb

Force is the vector quantity, displacement is the vector quantity , because their product which give rise to moment is also a vector quantity , so here we take vector product.

T(Torque) = Fxd = FdSin


Scalar Product quantity

a. b = abCos

b

a

If two vectors are Perpendicular to each other

then  = 90 , so a.b = abCos90 = 0 .

a

b

Click to start

2-Scalar product :- If a and b are two vectors then a.b = abCos

Here products of two vectors give rise to a scalar quantity.

Force is the vector quantity, displacement is the vector quantity ,but their product which give rise to work is a scalar quantity.

W = F.d = FdCos


Ĵ quantity

Î

Example of vector product:-

If a = (2î+3Ĵ+4kˆ) and b = (4î+5Ĵ+6kˆ) then

axb = (2î+3Ĵ+4kˆ)x(4î+5Ĵ+6kˆ)

= 8îx î+ 10 îxj + 12 Îxkˆ + 12 Ĵxi + 15Ĵx Ĵ +18Ĵxkˆ +16 kˆx î +20kˆx Ĵ +24kˆxkˆ

= 0 +10kˆ +12(-Ĵ) +12(-kˆ)+0 +18î+16(- Ĵ)+20(-î)+0

= 10kˆ -12 Ĵ – 12k +18 î -16 Ĵ – 20 î

= -2 î – 28 Ĵ -2kˆ

Important to know-: Î x Î = I Î I I Î I Sin00 kˆ = (1)(1)(0) kˆ = 0

(Angle between same vectors is zero.)

Similarly Ĵ. Ĵ = kˆ.kˆ = 0

Îxj = I Î II Ĵ I Sin 900 kˆ= (1)(1)(1) kˆ = kˆ (Î, Ĵ ,kˆ are the perpendicular vectors to each other.)

Similarly Ĵx kˆ = Î

kˆxÎ = Ĵ

kˆxĴ = - Î

Î xkˆ = - Ĵ , Ĵ x Î = -kˆ


Example of scalar product:- quantity

If a = (2î+3Ĵ+4kˆ) and b = (4î+5Ĵ+6kˆ)

a.b = (2î+3Ĵ+4kˆ). (4î+5Ĵ+6kˆ)

= 8î. î+ 10 î.j + 12 Î.kˆ + 12 Ĵ.i + 15Ĵ. Ĵ +18Ĵ.kˆ +16 kˆ. î +20kˆ. Ĵ +24kˆ.kˆ

= 8 +0 +0 +0+15 +0+0+0+24

= 47

Important to know-:

Î . Î = I Î II Î I Cos0 = (1)(1)(1) = 1 (Angle between same vectors is zero.)

Similarly Ĵ. Ĵ = kˆ.kˆ = 1

Î. j = I Î II Ĵ I Cos 900 = 0 (Î, Ĵ ,kˆ are the perpendicular vectors to each other.)

Similarly Ĵ. kˆ = kˆ.Î = 0


(x quantity2 < x1) &

(t2 > t1)

Gradient = x2-x1

t2-t1

It is negative here.

(x2 > x1) &

(t2 > t1)

Gradient = x2-x1

t2-t1

It is positive here.

B

(t2 x2)

Velocity =0

E

D

displacement

C

M

(t1, x1)

B

(t2, x1)

A

O

X

Velocity decreasing

time

Displacement

Velocity constant

M

Velocity is constant

Velocity increasing

(t1, x1)

displacement

(t2, x2)

N

(t1, x2)

time

X

Time

Y

N

O

1- Graph OB and MN is showing that the body is moving with constant velocity. In graph OB the gradient is positive so it is positive velocity. In graph MN the gradient is negative so it is positive velocity.

2- In graph OC the displacement is decreasing with time .It shows that velocity is decreasing.

3- In graph OA the displacement is increasing with time .It shows that velocity is increasing.

4- Graph DE shows that displacement is zero so the velocity is also zero.


Application of vectors quantity

With the help of vectors it is easy to comprehended the complicated chapters in higher sciences.

Today vectors are used to explain higher mathematics and physics


1-Explain vectors and scalars quantity

2- How vector method of doing questions are easy.

3- What is easy, pulling of trolley or pushing of trolley?


Presented by quantity

U C Pandey

Lecturer in Mathematics

G.I.C. Dinapani Almora, Uttarakhand India

E-mail – [email protected]


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