Transcript Vector - DEP

Vectors and 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.
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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
C
Earth
C
C
C
Types of vectors
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
others it is represented as â.), where
this vector from
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:- 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
l
m = l and n = 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
B
C
Q
R
Q
α
α
O
P
A
QCosα
QSinα
D
In OBD R2 = (P+QCosα)2 + Q2Sin2α
R2 = (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α
= (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α
= P2+Q2 ( Cos2α +Sin2α) +2PQCosα
= P2+Q2 +2PQCosα
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•
On the other part current despite of having direction and
magnitude does not obey the parallelogram law of addition
R= 5
i2= 3Amp
I1 = 3Amp
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
c
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See angle between different vectors
d
c
b
b
α1
a
c α2 b
a
α3
d
Given vectors
Here we are interested
to get a+b+c+d
α3
d
See adding of vectors
R
c
d
α2
b
α1
a
Here R = a + b+ c+ d
R
c
b
a
Step wise way of addition
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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.
Here apparent weight
of the body = Mg - Sinα
RSinα
R
RCosα
α
P
Mg
If many forces are working at a point, then finding of resultant
Here these forces can be break in their
perpendicular components X & Y.
F3
4
X = F1 Cos0 + F2 Cos1 + F3Cos 2 +F4Cos 3 +
F5Cos 4
F1
Y = F1 Sin0 + F2 Sin1 + +F3Sin 2+ F4Sin 3 +
F5Sin 4
3
 2
1
F4
F2
F5
R
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
F3
4
F4
F3
3
 2
1
F4
F2
3-2
F1
4-3
F5
F5
F1
1
F2
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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
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Pulling
F Sinα
Here net weight (downward force on roller
)of roller after applying force= Mg – FSinα
The net horizontal pulling force = FCosα
α
F Cosα
Roller
Mg
It shows that here the apparent weight of
roller decreases, So it is easier to pull a
roller instead of pushing.
Click to start
Pushing
α
Here net weight (downward force on roller )of
roller = Mg + FSinα
F Cosα
The net horizontal pulling force = FCosα
F Sinα
Roller
It shows that here the apparent weight of
roller increases, So it is easier to pull a
roller instead of pushing.
Note- In pushing or pulling angle ‘α’ remains same.
Mg
Path of Projectile
C uCos = vCos1 ( Here vertical component becomes zero.)
vSin1 v

B 1
vCos1
H= It is the maximum height attained
uSin
u

A uCos
by the projectile.
vCos2
2
vSin2 v
(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
Sin
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’.)
u
uSinα
(x,y) 
uCosα
y
u

x
uCos
Position of particle
after time ‘t’
Or y = uSin(x/uCos ) – 1/2g(x/uCos)2
Or Y = 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.
Flight Time--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
Range R = 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

When  = 450 , then body covers the longest range.
450
R= longest range
A hammer thrower should throw the hammer at the angle of 450 in order to
cover maximum distance.
700
When  = 900 , then body get the greatest height.
High jumper should take lift at an angle of 900 in order to cover
maximum vertical distance.
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Vector represents in 3 dimension
Y
j
900
900
900
X
i
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
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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 ^n
^n
^n
b

a
Direction of aXb
is perpendicular
to both a & b. It
is shown by ^
n .
b

axb
a
Area covered by a & b can
be calculated as 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
If two vectors are parallel to each other
then  = 0 , so a x b = absin0 ^n = 0 .
a
b
Scalar Product
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2-Scalar product :- If a and b are two vectors then a.b = abCos
b

a
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
If two vectors are Perpendicular to each other
then  = 90 , so a . b = abCos90
=0.
a
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 I I Ĵ I Sin 900 kˆ = (1)(1)(1) kˆ = kˆ
(Î, Ĵ ,kˆ are the perpendicular vectors to each other.)
Ĵ
Similarly Ĵx kˆ = Î
kˆx Î = Ĵ
Î
kˆxĴ = - Î
Î xkˆ = - Ĵ ,
Ĵ x Î = -kˆ
kˆ
Example of scalar product: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 I I Î I Cos0 = (1)(1)(1) = 1
(Angle between same vectors is zero.)
Similarly Ĵ. Ĵ = kˆ.kˆ = 1
Î. j = I Î I I Ĵ I Cos 900 = 0
Similarly Ĵ. kˆ = kˆ. Î = 0
(Î, Ĵ ,kˆ are the perpendicular vectors to each other.)
Y
displacement
Velocity =0
C B
E
A
M
(t1, x1)
Time
Gradient = x2-x1
t2-t1
(t2, x1)
It is positive here.
X
time
displacement
N
(t2 > t1)
(t2 x2)
O
Displacement
D
O
(x2 > x1) &
B
(x2 < x1) &
M
(t2 > t1)
(t1, x1)
(t2, x2)
Gradient = x2-x1
t2-t1
N
(t1, x2)
time
It is negative here.
X
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
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
2- How vector method of doing questions are easy.
3- What is easy, pulling of trolley or pushing of trolley?
Presented by
U C Pandey
Lecturer in Mathematics
G.I.C. Dinapani Almora, Uttarakhand India
E-mail – [email protected]