Vectors

PHYSICS - I A ELEMENTS OF VECTORS

AKASH MULTIMEDIA 65

The vector, as a mathematical object, isdefined as a directed line segment. I t should alsoobey the laws of vector addition.Examples : Displacement, velocity, acceleration,force, momentum, angular momentum , moment offorce, Torque, magnetic moment, magnetic inductionfield, Intensity of electric field, etc ....

To describe a vector quantity we require.a) The specific unit of that quantity.b) The number of times that unit is contained in

that quantity.c) The orientation of that quantity.

Ex : - A plane is flying from west to east with avelocity of 50 ms1. Here ms1 is the unit, 50 is thenumber of units of velocity and west to east is thedirection.Note 3.3: 1) A physical quantity having magnitudeand direction but not obeying laws of vector additionis treated as a scalar.

Ex : Electric current is a scalar quantityElectric current is always associated with

direction, but it is not a vector quantity. It does notobey law of vector addition for its addition.

i1

i2

(i1 + i2)q

3.1 CLASSIFICATION OF PHYSICALQUANTITIESAll measurable quantities are called physical

quantities. Most of the physical Quantities areclassified into 'Scalars' and 'Vectors'.Scalar: - Physical quant i t ies having onlymagnitude are called Scalars.Examples : Length, t ime, volume, density,temperature, mass, work, energy, electric charge,electric current, potential ,resistance, capacity, etc.....

To describe a scalar quantity we require

a) The specific unit of that quantityb) The number of times that unit is contained in

that quantityEx: A bag contains 100 kg of sugar. Here kg is theunit and 100 is the number of units of sugar presentin the bag.Note 3.1: Unit is not a compulsion to represent a scalar

Ex : - Specific gravity, Refractive indexNote3.2: Mathematical operations of scalar quantitiesyield scalar quantities and these quantities aremanipulated by ordinary algebraic rules.

Vector : - Physical quantities having bothmagnitude and direction and that obeys laws ofvector addition are called vectors (or)

VECTORS

CHAPTER

3Leonhard Euler (15 April 1707 18 September 1783) was a pioneeringSwiss mathematician and physicist. He made important discoveries in fieldsas diverse as infinitesimal calculus and graph theory. He also introducedmuch of the modern mathematical terminology and notation, particularlyfor mathematical analysis, such as the notion of a mathematical function.He is also renowned for his work in mechanics, fluid dynamics, optics, andastronomy.Euler was featured on the sixth series of the Swiss 10-franc banknote andon numerous Swiss, German, and Russian postage stamps. The asteroid2002 Euler was named in his honor. He is also commemorated by theLutheran Church on their Calendar of Saints.

Leonhard Euler1707-1783

h To differentiate between scalar and vectorquantites with examples.

h To find the resultant of vectors, and resolve avector into its rectangular components

h To explain the Relative Motion.

h To compute the product of vectors.

ELEMENTS OF VECTORS PHYSICS - I A

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The resultant of i1 and i2 is (i1 + i2) by Kirchoffscurrent law. The resultant does not depend on anglebetween currents i1 and i2.Note3.4: Equations in vector form indicate bothmathematical and geometrical relationships amongthe quantities. Physical laws in vector form are verycompact and independent of choice of coordinatesystem.

3.2 GEOMETRICAL REPRESENTATIONOF VECTORSAny vector A

can be represented geometrically

as a directed line segment (an arrow), as shown infig. The magnitude of A

is denoted by A ,

and the

direction of A

is specified by the sence of the arrowand the angle that it makes with a fixed referenceline.

A

q

When using graphical methods, the length ofthe arrow is proportional to the magnitude of thevector, and the arrow head represents the direction.

Ex:- The velocity vector v is represented byan arrow OP as shown in figure. The initial point ofthe vector is O, the final point of the vector is P. Thelength OP is the magnitude of the velocity and itsdirection is 600 north of east (or) 300 east of north .

060q

N

W

S

EO

P03 0

v

600 North of East or 300 Eash of North

04 5

NWN

NE

W

SW SES

E

3.3 TYPES OF VECTORSa) Polar Vectors : The vector whose directiondoes not change even though the co-ordinate systemin which it is defined changes is called polar vector.

Ex : Force, momentum, Acceleration.

b) Axial Vectors : The vectors whose directionchanges with the co-ordinate system in which it isdefined changes is called axial vector.

Ex:Angular velocity, torque, angular momentumc) Like vectors (or) parallel vectors : Two ormore vectors (representing same physical quantity)are called like vectors if they are parallel to each other,however their magnitudes may be different.

A

B

C

d) Unlike vectors (or) anti parallel vectors : Twovectors (representing same physical quantity) arecalled unlike vectors if they act in opposite directionhowever their magitudes can be different.

A

B

e) Equal vector s : Two or more vectors(representing same physical quantity) are called equalif their magnitudes and directions are same.

A

B

C

Ex : Suppose two trains are running on paralleltracks with same speed and direction. Then theirvelocity vectors are equal vectors.Note : If a vector is displaced parallel to itself itsmagnitude and direction does not change.

A

B

C

X

Y A B C

f) Negative Vector : A vector having the samemagnitude and opposite in direction to that of a givenvector is called negative vector of the given vector

A

A


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g) Co-initial vectors : The vectors having sameinitial point are called co-initial vectors.

CA

B

E

D

O

h) Collinear vectors : Two or more vectors aresaid to be collinear when they act along the sameline however their magnitudes may be different.

Ex : Two vectors A

and B

as shown arecollinear vectors.

A

B

(or)

A

B

i) Coplanar vectors : A number of vectors aresaid to be coplanar if they are in the same plane orparallel to the same plane. However their magnitudesmay be different.

Y

O X

A

B

C

j) Unit vector : A vector whose magnitude equalsone and used to specify a convenient direction iscalled a unit vector.

A unit vector has no units and dimensions. Itspurpose is to specify the direction of given vector.

In cartesian coordinate system, unit vectorsalong positive x, y and z axis are symbolised as i , jand k respectively. These three unit vectors aremutually perpendicular and their magnitudes i j k 1 .

If A

is a non zero vector, then the unit vector in

the direction of A

is given by A

AA

3.4 REPRESENTATION OF ANGLEBETWEEN THE TWO VECTORS

The angle between two vectors is represented by thesmaller of the two angles between the vectors whenthey are placed tail to tail by displacing either of thevectors parallel to it self.

Ex: The angle between A

and B

is correctlyrepresented in the following figures.

B

A

q q

B

A

B

A

B

q(or)

Fig.(i) Fig.(ii)

a) If the angle between A

and B

is , then theangle between A and KB

is also . Where K is a

positive constant.

qA

B

qA

K B

b) If the angle between A

and B

is , then theangle between A and KB

is (180 ) . Where

K is a positive constant.

qA

B

qA

KB

0180 q

B

c) Angle between collinear vectors is always zeroor 1800C

Q

P

P

Q

00q 0180qor

Problem : 3.1Three vectors , ,

A B C are shown in the figure. Find

angle between (i) A and B (ii) B and C (iii)A and

C.

300300

450x xx

A

B C

Sol. To find the angle between two vectors we connect thetails of the two vectors. We can shift the vectors parallelto themselves such that tails of A,B

and C

are

connected as shown in figure.


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300

300

450

A

B

C

x

y

Now we observe that angle between A

and B

is 60, B

and C

is 15 and between A

and C

is 75

Problem : 3.2I f , , A B C represents the three sides of an equilateraltriangle taken in the same order then find the anglebetween i) A and B ii) B and C iii)

A

and C .

A

A

B

C

1200

1200

1200

B

C

Sol. From the diagram the angle between thevectors A

and B

is 1200 , the angle between B

and C

is 1200, the angle between A

and C

is 1200

3.5 ADDITION OF VECTORS(GRAPHICAL METHOD)

1. To add two vectors geometrically represent thevectors by arrow head lines using the same suitablescale, with their proper directions in the choosen co-ordinate system.2. Join the initial point of the second vector withthe final point of the first vector by moving parallelto itself.3. Now, draw an arrow from the initial point ofthe first vector to the final point of the second vector.This arrow represents the resultant of the two vectors.Examples :1. When two vectors are acting in the same di-

rection.Let the two vectors P

and Q

be acting in the

same direction.

R P Q R P Q

(a) (b) (c)

P P

Q

Q

2. When two vectors are acting at some angle:First join the initial point of Q

with the final

point of P

and then, to find the resultant of thesetwo, draw a vector R

from the initial point of P

to

the final point of Q

. This single vector R

is theresultant of vectors P

and Q

.

R

P

P

QQ

R

represents the resultant of P

and Q

both inmagnitude and direction. So, R P Q.

3.6 LAWS OF VECTOR ADDITIONa) Vector addition obeys commutative law.Addition of two vectors is independent of the

order of the vectors in which they are added. If P

and Q

are two vectors then P Q Q P

asshown in fig.

Q P R

P Q R

P

P

Q

Q

O A

B C

b) Vector addition obeys associative law.While adding more than two vectors, the resultant isindependent of the order in which they are added.

If P,Q

and R

are three vectors, then P Q R P Q R

as shown in fig.

P

Q

R

P Q

Q R

P Q R

P

Q R

( )P Q R

c) Vector addition obeys distr ibutive law.If k, k1, k2 are scalars then

k P Q kP kQ

and 1 2 1 2P k k k P k P

Note 3.5 : Vector addition is possible only betweenvectors of same kind.


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3.7 PARALLELOGRAM LAW OF VECTORSStatement : I f two vectors are drawn from a

point so as to represent the adjacent sides of aparallelogram both in magnitude and the direc-tion, the diagonal of the parallelogram drawnfrom the same point represents the resultant ofthe two vectors both in magnitude and direction.

NA

B

O

C

aqP

Q

q

R

Let the two vectors P

and Q

, inclined at anglebe acting on a particle at the same time. Let theybe represented in magnitude and direction by twoadjacent sides OA

and OB

of parallelogram OACB,

drawn from a point O.According to parallelogram law of vectors, their

resultant vector R

will be represented by the diagonalOC

of the parallelogram.Magnitude of resultant :

The line OA is extended upto point N and drawa perpendicular from point C on to the extended lineas shown in fig.

From the parallelogram OACB.

BOA = CAN = andOA = P, OB = AC = Q,OC = R and

In CNA, AN = AC cos = Q cos

CN = AC sin = Q sin From triangle ONCOC2 = ON2 + NC2

(OC)2 = (OA+AN)2 + (NC)2

R2 = (OA)2 + 2(OA)(AN) + (AN)2 +(NC)2

R2 = P2 + 2 PQ cos + 2 2QCos QSin 2 2 2R P 2PQCos Q

R = 2 2P Q 2PQ cos ........ (3)

The magnitude of the resultant R

depends onthe magnitudes of the vectors P

and Q

and also on

the angle between them.Direction of resultant :

NA

B

O

C

aqP

Q

q

R

If the line of action of the vector P

is taken asreference line, the resultant R

makes an angle

with it. This angle indicates the direction of R

.Then from right angled triangle ONC,

tan CN CN Q sin ON OA AN P Qcos

.....(4)

Note3.6: If is the angle between the resultant R

and the vector Q,

then 1 Psintan ,Q P cos

Special Cases:a) If the magnitude of P > Q then . i.e., theresultant is closer to the vector of larger magnitude.b) When angle between two vectors increases, themagnitude of their resultant decreases.c) When t wo vector s ar e i n the samedirection(parallel).

P

Q

R P Qthen = 00, cos 00 = 1 and sin 00 = 0

from eq.(3), R = 2 2P Q 2PQ(1)

= 2(P Q) = (P + Q)

from eq.(4), tan =

Q 0

P Q(1) = 0 or = 00.

Thus for two vectors acting in the samedirection, the magnitude of the resultant vector isequal to the sum of the magnitudes of two vectorsand acts along the direction of P

and Q

.


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(d) When two vectors are acting in oppositedirections (antiparallel).

P

Q

R P Q

then = 1800, cos 1800 = 1 and sin 1800 = 0.

From eq. (3), R = 2 2P Q 2PQ( 1)= 2(P Q) = (PQ) or (QP)

From eq. (4), tan =

Q 0

P Q( 1) = 0

or = 00 or 1800.Thus for two vectors acting in opposite

directions, the magnitude of the resultant vector isequal to the difference of the magnitudes of the twovectors and its direction is along the vector of largermagnitude.(e) When two vectors are perpendicular to eachother,

= 900, sin = 1and cos = 0

R

P

Q

aFrom eq(3),

R = 2 2P Q 2PQ(0) = 2 2P Q

From eq(4), tan = Q (1)

P Q(0) = QP

or = tan1

QP

f) If the resultant of two vectors is perpendicularto any one of the vectors, theni) The angle between the two vectors is greaterthan 900

ii) The resultant vector is perpendicular to thevector having smaller magnitude.

we know Q sintan

P Q cos

0 Qsintan 90P Q cos

P + Q cos = 0

f

qP

Q R

P

cos = P

Q and

R = 2 2Q P ,P R Psin ,cos , tan .Q Q R

The angle between P

and Q

is 090

g)max

min

R P QR P Q

and max min

max min

R RP .Q R R

h) If magnitudes of P and Q are equal and theangle between them is then their resultant is

2 2R P Q 2PQ cos

q2 2 2R P P 2P cos [P=Q]

2 2R 2P 2P cosq , 2R 2P 1 c os

2 2R 2P 2cos2

, R 2P cos 2

and the resultant makes an angle with P is

Qsintan

P Q cos

P s inP P c o s

P sintan

P 1 cos

2

s in2 c o s / 2

2

2sin cos2 2tan

2cos / 2

tan2

tan tan2

, 2

a) If 060 , then R = 3P and 030

b) If 090 , then R = 2P and 045

c) If 0120 , then R = P and 060 I) The unit vector parallel to the resultant of two

vectors P & Q

is P QnP Q

.

Note 3.7 : We can add a vector to another vector ofsame kind but we cant add a vector quanity to ascalar quantity.


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Problem : 3.3The resultant of two forces whose magnitudes are inthe ratio 3 : 5 is 28 N. I f the angle of their inclination is600, then find the magnitude of each force.

Sol : Let F1 and F2 be the two forces.

Then F1 = 3x; F2 = 5x; R = 28 N and 060

2 21 2 1 2R F F 2F F cos

2 2 028 3x 5x 2 3x (5x)cos60

2 2 228 9x 25x 15x 7x

28x 4

7 ,

F1 = 3 4 = 12 N, F2 = 5 4 = 20 N. *Problem : 3.4

I f vectors A and B are 3i 4j + 5k and 2i + 3j 4krespectively then find the unit vector parallel to

A B

(Ans : 5

27i j k

) Hint : A B

nA B

*Problem : 3.5

I f magnitude of the resultant of two vectors equal ofmagnitude, is equal to the magnitude of either of thevectors, what is the angle between them ? (Ans : 1200)

Hint : 2 cos2

R P

Problem : 3.6

I f 3 4 A i j and 7 24 ,

B i j

f ind a vectorhaving the same magnitude as

B and parallel to

A

Sol. The vector parallel to A

and having magnitude of B

is^AC B B A

A

2 2B 7 24 25

and 2 2A 3i 4j 1A 3i 4jA 53 4

1C 25 3i 4j 15i 20 j5

Problem : 3. 7The resultant of two vectors A and B is perpendicularto

A and equal to half of the magnitude of B . Find

the angle between A and

B ?

Sol. Since R

is perpendicular to A

. figure shows the three

vectors A

, B

and R

.

angle between A

andB

is

A

B

R

p qqA

sin = RB

= B

2B =

12

= 30angle between A and B is 150.

Problem : 3.8What is the displacement of the point of a wheelinitial ly in contact with the ground when thewheel rolls forward half a revolution? Take theradius of the wheel as R and the x-axis as theforward direction?

Sol: From figure, during half revolution of the wheel,the point A covers AC R horizontal distance, andBC = 2R vertical distance.

qA Cx

y

B

x = R and y = 2R; Displacement,

2 2s x y 2 2 2( R) (2R) R 4

and 1 1 1y 2R 2Tan Tan Tanx R

with

x-axis.

i.e., Displacement has magnitude 2R 4 and makes

an angle 1 2Tan with x-axis.

3.8. TRIANGLE LAW OF VECTORS" If two vectors are represented in magnitude

and direction by the two sides of a tr iangle takenin one order, their resultant vector is representedin magnitude and direction by the third side ofthe tr iangle taken in reverse order" .

Let the two vectors P

and Q

, inclined at anangle be acting on a particle at the same time. Let


AKASH MULTIMEDIA 72

they be represented in magnitude and direction bytwo sides OA

and AB

of triangle OAB, taken in

the same order shown in fig.Then, according to triangle law of vector

addition, the resultant R

is represented by the thirdside

OB of triangle, taken in opposite order..

p q

g

qP A

B

NO

R

Qa

R P Q

Magnitude of resultant R

:

2 2R P Q 2PQ cos Direction of resultant R

:

QsinTan .P Qcos

Statement of tr iangle law when three forceskeep a par ticle in equilibr ium :

When three forces acting at a point can keepa par ticle in equilibr ium the three forces can berepresented as the sides of a tr iangle taken inorder both in magnitude and direction.

1F3F

1F

2F

2F

3FO P

Q

RSuppose three forece 1 2 3F ,F ,F

are

simultaneously acting at point O and the point is inequilibrium. Then the three forces can be representedas three sides of a triangle. The triangle PQR isconstructed by drawing parallel lines to the directionsin which the forces are applied. The magnitudes ofthe forces and the corresponding sides of the triangle

have equal ratio i.e., 31 2 FF FPQ PR QR

.

3.9 POLYGON LAW OF VECTORSStatement : If a number of vectors are repre-

sented by the sides of a polygon both in magni-tude and direction taken in order, their resultant

is represented by the closing side of the polygontaken in reverse order in magnitude and direction .

A B

C

DE

1F

2F

3F

4F

RF

(or) I f a number of coplanar forces are acting simul-taneously at a point keep the par ticle in equilib-r ium, these forces can be represented as the sidesof a polygon taken in order both in magnitudeand direction.

A B

C

DE

1F

2F

3F

4

F

5

F

1 2 3 4 5F F F F F 0

Note 3. 8 : I f x is the side of a hexagon ABCDEFas shown in figure

A B

C

DE

F3x

2x

060x

3xx

x90

0

AB = x, AC = 3x , AD = 2x, AE = 3x , AF= x

Problem : 3.9ABCDEF is a regular hexagon with point O as centre.Find the value of

. AB AC AD AE AF

A B

C

DE

F O

Sol. From the diagram AB DE BC EF


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AB AC AD AE AF

= AB

AB

BC AD AD DE

AD DE

EF

3AD 3 2AO 6 AO

3.10 NULL VECTOR (OR) ZERO VECTORA vercotr whose magnitude is equal to zero is

called a null vector. Its direction is indeterminate.Examples of zero vector :1. The velocity of a particle at rest.2. The acceleration of a particle moving at uniformvelocity.3. The displacement of a stationary object over anyarbitrary interval of time.4. The position vector of a particle at the originNote 3.9 :

A 0 A

, A 0 0

3.11 EQUILIBRIANTA vector having same magnitude and

opposite in direction to that of the resultant of anumber of vectors is called the equilibr iant. (or)

Negative vector of the resultant of a numberof vectors is called the equilibr iant (E).

If 1 2 3F ,F & F

are the three forces acting on abody, then their resultant force is R 1 2 3F F F F

R 1 2 3E F , E F F F .

Note 3.10 :i) Single force cannot keep the particle in

equilibrium.ii) Minimum number of equal forces required to keep

the particle in equilibrium is two.iii) Minimum number of unequal coplanar forces

required to keep the particle in equilibrium isthree.

iv) Minimum number of equal or unequal noncoplanar forces required to keep the particle inequilibrium is four .

Application 3.1 : Lami's Theorem

" I f three coplanar forces acting at a pointkeeps i t in equi l ibr ium, then each for ce ispropor tional to the sine of the angle between theother two forces"

If 1 2 3F ,F ,F are the magnitudes of three forcesand , , are the angle between forces 2F

and 3F ,

3F

and 1F

and 1F

and 2F

respectively, as shown infig. Then according to lamis theorem

31 2FF F

sin sin sin

1

F

3

F

abg

2F

Problem 3.10 :In the given figure the tesion in the string OB is 30N.Find the weight W and the tension in the string OA.

A

B

300

O

W

Sol: Let T1 and T2 be the tensions in the strings OA and OBrespectively

A

B

300

O

W

T1

T2900

900

1500

According to lamis theorem

1 20 0 0

T T Wsin 90 sin150 sin120

(T2 = 30N)

on sloving w 30 3 N and T1= 60N

3.12 SUBTRACTION OF VECTORSThe process of subtracting one vector from

another is equivalent to adding vector ially thenegative of the vector to be subtracted.

Let P

and Q

be the two vectors as shown infigure. We want to find the difference P Q

. Let a


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b) the vector subtraction does not followassociative law ie. P Q R P Q R

P

QR

B

C

OOA

P

Q

R

P Q P Q R

AO

C

B

A B C

A

B

C

B C

O A

BC

Applications 3.2 : If iv is initial velocity of a particle

and fv is its final velocity. Then change in its

velocity is given by f iv v v .

2 2i f i fv v v 2v v cos

Where ' ' is the angle between initial and finalvelocities.Application 3.3 : When a particle is performinguniform circular motion with a constant speed v, themagnitude of change in velocity when it describes

an angle at the centre is V 2vsin .2

qO

V

Vq

Application 3.4 : If velocity of a particlechanges from iv

to fv in time t then the acceleration

of the particle is given by f iv va .

t

Problem : 3.11A particle is moving eastwards with a velocity of 5m/s.In 10s the velocity changes to 5 m/s northwards. Findthe average acceleration in this time

vector Q

be added to the vector P

by the laws ofvector addition. Their resultant gives the value of P Q

.

Q

P

P

Q

Q

A

C

O

B

S P Q

q 180 q

The magnitude of P Q

is

2 2 0S P Q 2PQ cos (180 )

S = 2 2P Q 2PQ cos and

0

0Q sin (180 - ) Qsintan .

P Q cos (180 ) P Q cos

Note 3.7 : If P Q

then P Q 2Psin .2

S 2P sin2

Note 3.8 : When two vectors P

and Q have samemagnitude and is the angle between them

Resultant Difference

00

600

900

1200

1800 0 2P

P

2P 0

P

2P 2P

3P

3P

2Pcos / 2 2Psin / 2

3.13 LAWS OF VECTOR SUBTRACTION :a) the vector subtraction does not followcommutative law i.e.

P Q Q P,

But P Q Q P

P

Q

A B

C

Q

P

A B

C

P Q

D

Q P

P

Q


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Sol. f iavv vv

at t

450

v 5 /fv m s

5 2 /m s

5 /iv m s

5 /iv m s

N

S

EW

av5j 5i 5 2 1a

10 10 2

m/s2

north west direction.

*Problem : 3.12Two vectors

A and

B have precisely equal magni-

tudes. For the magnitude of A B

to be larger thanthe magnitude of A B by a factor n, what must bethe angle between them ?

Sol. A B n A B

2A cos n2Asin2 2 (A = B)

1 11 1 1tan , tan , 2 tan2 n 2 n n

3.14RESOLUTION OF A VECTOR INTORECTANGULAR COMPONENTSResolution of a vector is the process of

obtaining the component vectors which whencombined, according to laws of vector addition,produce the given vector.

Y

Q R

y

j

Oq

ix PX

r

Consider a vector r in the xy plane and it makesan angle with x axis as shown in figure. i and jare unit vectors along X-axis and Y-axis respectivley.

From the point R, draw RP and RQperpendicular to X-axis and Y-axis respectively.From the parallelogram law of vector addition, itfollows that

OR OP OQ

............ (1)

If OP = x and OQ = y, then

OP xi and OQ yj

From equation (1) r xi y j ....... (2)

In the above equation xi

and y j

are called thex-component and the y-component of the vector r ,and x and y are called magnitudes of the twocomponent vectors.

from the triganle OPR.OP xcos x r cosOR r

......... (3)

and PR ysin y r sinOR r

......(4)

from (3) and (4) ytanx

............. (5)

The equation (3) and (4) gives the magnitudesof the rectangular component vectors in terms of themagnitude of the given vector and its inclination withX-axis.

from equation (2), r r cos i r sin j

from right angled triangle OPR

2 2OR OP PR or 2 2r x y .... (6)

The equation (6) gives the magnitude of thevector in terms of its rectangular components.Note 3.9 : Method involving resolution of vectorsinto components to find the resultant of the vectorsis known as analytical method.Note 3.10 : The components of a vector areindependent of each other and can be handledseparately.Note 3.11 : Theoretically, a given vector can be madethe diagonal of infinite number of parallelograms.Thus there can be infinite number of ways to dividea vector into components.Note 3.12 :

A

B

C Do x

y

-x

-y


AKASH MULTIMEDIA 76

a) If the vector A

is in first quadrant then it canbe written as x yA A i A j

b) If the vector B

is in second quadrant then

x yB B i B j

c) If the vector C is in third quadrant then

x yC C i C j

d) If the vector D is in fourth quardant then

x yD D i D j

3.15 ADDITION OF TWO VECTORS IN TERMSOF COMPONENTSLet us consider the addition of two vectors P

and Q

in terms of their components.We have x y x y P P i P j, Q Q i Q j

O

Q

P

R

xP xQ

yQ

yPyR

xR

x

y

Let R

be the resultant vector with component.

Rx and Ry along x and y axis respectively. Then

x xR R i

and y yR R i.

From the diagram Rx = Px + Qx and Ry = Py+Qy x x x y y y R P Q i, R P Q j

x y x x y y R R i R j; R P Q i P Q j

22x x y yR P Q P Q

y y y

x x x

R P QTan ,Tan

R P Q

y y1

x x

P QTan

P Q

Application 3.5 :

sinF q

cosF q

F

q

A block is placed on smooth horizontal surfaceand pulled by a force F making an angle withhorizontal.

The component of force along horizontal = Fcos .

The component of force along vertical = Fsin .

Application 3.6 :

q

cosmg qsinmg q qmg

m

A block of mass m is placed on inclined planeof angle then the component of weight parallel tothe inclined plane is mg sin , the component ofweight perpendicular to the inclined plane is mgcos .Application 3.7 :

q

sinT q

cosT q

mg

TF

O

q

x2 2x

A simple pendulum having a bob of mass mis suspended from a rigid support and it is pulled bya horizontal force F. The string makes an angle qwith the vertical as shown in figure.The horizontal component of tension = TsinThe vertical component of tension = Tcos


AKASH MULTIMEDIA 77

When the bob is in equilibriumTsin F, .............(1)T cos mg .............. (2)

2 2

mg mglTcos l x

From equations (1) and (2)

2 2

F xTan F mgTan mgmg l x

22T F mg

Note- 3.13 : If a vector is rotated through an angleother than integral multiple of 2 (or 3600) itchanges, but its magnitude does not change.

A B

A

B

q

Note - 3.14 : If the frame of reference is rotated thevector does not change (though its components maychange).

OS

Vecto

r

Problem : 3.13A vector 3 i j rotates about its tail through anangle 300 in clock wise direction then the new vector is

Sol: The magnitude of 3 i j is 3 1 = 2 The anglemade by the vector with x - axis is

y

x

A 1TanA 3

3i jy

o x300

030

When the given vector rotates 300 in clock wise itsdirection changes along x - axis but its magnitudedoes not change.

The new vector is 2ixo

y

2i

Problem : 3.14A weight mg is suspended from the middle of a ropewhose ends are at the same level. The rope is no longerhorizontal. What is the minimum tension required tocompletely straighten the rope is

Sol. From the diagram2T sin = mg

mg

2Tsin

T

T cosq

q qq

T cosq

T

qqmg

T2 sin

The rope will be staraight when = 0

0mgT

2sin 0

The tension required to completely straighten the ropeis infinity.

Problem : 3.15The sum of magnitudes of two forces acting at a pointis 16N. If their resultant is normal to the smaller forceand has a magnitude of 8N. Then the forces are

q090

2F

1F

F

Sol. Let F

be the resultant of two forces 1F

and 2F

as shownin figure with F2 > F1.

F2 sin = F1 ...................... (i)F2 cos = F = 8 ...................... (ii)Squaring and adding Eqs (i) and (ii), we getF2

2 = F12 + 64 ...................... (iii)

Given F1 + F2 = 16 ...................... (iv)Solving Eqs. (iii) and (iv), we getF1 = 6N and F2 = 10N

* 3.16 RESOLUTION OF A VECTOR INTOTHREE RECTANGULAR COMPONENTS :Let us consider a vector A

be represented by

OP

as shown in fig. With O as origin, construct arectangular parallelopiped with three edges along thethree rectangular axes which meet at O. A

becomes

the diagonal of the parallelopiped. x yA ,A

and zA

are three vector intercepts along x, y and z axes re-spectively. These are the three rectangular compo-nents of A

.


AKASH MULTIMEDIA 78

Applying triangle law of vectors,

OP OK KP

.................... (1)

Applying parallelogram law of vectors,

yA

A

xA

zA

S

P

QO

T K

Y

Z

X

OK OT OQ

....................... (2)

from (1) and (2) OP OT OQ KP

But KP OS, OP OT OQ OS

z x yA A A A ,

or x y zA A i A j A k

Again, OP2 = OK2 + KP2 = OQ2+OT2+KP2

or A2 = Ax2 + Az

2+Ay2

yKP OS A

or 2 2 2x y zA A A A

This gives the magnitude of A

in terms of the mag-nitudes of components x yA ,A

and zA

.

* 3.17 DIRECTION COSINES :The direction cosines ,m and n of a vector

are the cosines of the angles , and which agiven vector makes with x-axis, y-axis andz-axis respectively.

A

k

aib

j

X

Z

Y

Og

If , , are angle made byA

with x, y, z axisand cos ,cos ,cos are called direction cosines.

yx zAA Acos ,m cos ,n cosA A A

Squaring and adding2 2 2cos cos cos =

2 2 2 22 2 2y x y zx z

2 2 2 2 2

A A A AA A A 1A A A A A

and 2 2 2sin sin sin 2

3.18 POSITION VECTORPosition vector of any point, with respect to

an arbitrar ily chosen or igin, is defined as thevector which connects the or igin and the pointand is directed towards the point.

Y

O X

Z

Pr

, ,p x y z

Position vector of a point helps in locating theposition of the point. Its magnitude gives thedistance between origin and the point. Consider apoint P having co-ordinates (x,y,z) as shown infigure. If O is the origin, then OP

is called the

position vector (r), r xi y j zk

Magnitude of r is, 2 2 2r x y z

The unit vector along r is given by

2 2 2

r xi yj zkrr x y z

Application 3.8 Displacement vector :

1r

2r

s

2 2 2, ,x y z

X

Y

B

O

1 1 1, ,A x y z

If 1r

is initial position vector of the particle and2r is final position vector of the particle then the dis-placement of the particle is given by 2 1s r r

. 2 1 2 1 2 1 s x x i y y j z z k


AKASH MULTIMEDIA 79

The magnitude of the displacement vector is

2 2 22 1 2 1 2 1S AB x x y y z z

Application 3.9 Condition for collision :Consider two particles, 1 and 2 move with con-

stant velocities

1v and

2v . At initial moment (t=0)their position vectors are 1r

and 2r

.If particles then collide at the point P after time

t sec.From the diagram

1 1 2 2r r s r s

x

y

1

2

P

r2r

2 2 s v t

1r

1 1 s v t

O1 1 2 2r r v t r v t

1 2 2 1r r v v t

(or) 1 2 2 1r r v v t ... (1)

1 2

2 1

r rt

v v

substitute t in equation ........(1)

1 21 2 2 12 1

r rr r v v

v v

1 2 2 1

1 2 2 1

r r v vr r v v

Problem : 3.16Find the resultant of the vectors shown in figure.

Sol.

O

AB

C

x

y

4cm3cm5 c

m

370

R OR AB BC

R 5cos37i 5sin 37 j 3i 4 j R 4i 3j 3i 4 j

R

= 7i + 7j, R 7 2 cm and045 with horizontal

Problem : 3.17Find the resultant of the vectors , , OA OB OC as shown infigure. The radius of the circle is R.

Sol : R OA OB OC

R Ri R cos45i R sin 45 j Rj

R R R R i R j2 2

45045

0

A

BC

O

R = 2R+R along OB . Problem : 3.18

A man walks 40m north, then 30m east and then 40msouth. The displacement from the starting point is

Sol. S 40 j 30i 40 j

S

N

EWS 30i

displacement is 30 m east

Problem : 3.19Vector A

is 2 cm long and is 600 above the x axis in

the first quadrant, vector B is 2cm long and is 600below the x axis in the fourth quadrant. Find A B .

Sol:

600

600

A

B

Bx

By

Ax

Ay

O X

Y

-X

-Y

R A B

0 R 2cos60 i 2sin60 j

+2cos60 i 2sin60 j

R 4 cos60i 2i

2cm, along x axis

*Problem : 3.20A vector has x component of -25.0 units and ycomponent of 40.0 units find the magnitude anddirection of the vector.

Sol. Consider a vector x y A A i A j

A 25i 40 j

2 222x yA A A 25.0 40.0

= 47.16 47.2 units.

and yx

A 40.0tan 1.6A 25.0

an (-1.6)= 58.00with ve x - axis.(This is in clock wise direction).This is equivalent to (1220 ) in anticlockwise directionwith the x axis.


AKASH MULTIMEDIA 80

y

P

S x

y1

x1S1

|psr

psr

|s sr

Now to find the velocity of a moving objectrelative to another moving object, consider a par-ticle P whose position relative to frame S is psr

and

relative to S| is |psr .

If the position of frame S1 relative to S at anytime is |s sr .

Then from the above diagram

| |ps ps s sr r r

Differentiating the above equation with respectto time

1 1ps ps s sd r d rd r

dt dt dt

1 1 1 1ps ps s s ps ps s sv v v ; v v v

...... (1)

Thus, if the velocities of two bodies (here par-ticle and frame S| ) are known with respect to a com-mon frame (s), we can find the velocity of one bodyrelative to the other.

Note 3.15 : I f Av and Bv

are velocities of twobodies A and B relative to earth, then the velocity ofA relative to B is AB A Bv v v

.a) If two bodies are moving along the same

line in the same direction with velocities vA and vBrelative to earth then the magnitude of velocity of Arelative to B is given by vAB = vA vB

If vAB is positive the direction of vAB is that ofA and if negative the direction of vAB is opposite tothat of A (or) along the direction of B.

b) If two bodies are moving towards each otheror away from each other with velocities vA and vB.Then the magnitude of velocity of A relative to B isgiven by

vAB = vA ( vB ) = vA + vBand directed towards A (or) away from A

*Problem : 3.21Find the resultant of the vectors shown in fig by thecomponent method

5

yb

x

c3

1037

0536

d

a-X

-Y

O

Sol.0

x R 1i 5cos37 i 6cos53i

x x R 1i 4i 3.6i R 6.6i

0y

R 3j 5sin37 j 6sin53 j

y y R 3j 3j 4.8 j R 1.2 j

2 22 2x yR R R 6.6 1.2 6.7

xR 1.2 R

x 6.6R

y 1.2R

010.3

xx

y

y

O

tan y 1 0x

R 1.2 0.18; tan 0.18 10.3R 6.6

with ve x axis in clock wise direction or 169.70 with+ ve xaxis in anti clock wise direction.

*Problem : 3.22Two vectors a and

b have equal magnitudes of 12.7

units. These vectors are making angles 28.20 and133.20 with the x axis respectively. Their sum is r.Find (a) the x and y components of r, (b) the magni-tude of r, (c) the angle r makes with the +x axis.Ans : 2.5, 15.3, 15.5, 6.120Hint : same as 3.21

3.19 RELATIVE VELOCITY When we consider the motion of a particle,

we assume a fixed point relative to which the givenparticle is in motion. For example, if we say that wateris flowing or wind is blowing or a person is runningwith a speed v, we mean that these all are relativeto the earth.(Which we have assumed to be fixed)


AKASH MULTIMEDIA 81

c) If two bodies A and B are moving with ve-locities vA and vB in mutually perpendicular direc-tions, then the magnitude of velocity of A relative to

B is given by 2 2AB A Bv v v .

The direction of vAB with vA is B

A

vtanv

d) If two bodies A and B are moving with ve-locities Av

and Bv , and is the angle between the

velocities, then the magnitude of velocity of A rela-tive to B is given by

2 2AB A B A Bv v v 2v v cos

and the direction of vAB with vA is given by

B

A B

v sin 180Tan

v v cos 180

B

A B

v sinTanv v cos

Differentiating equation (1) with respect to time

| |PS PS S Sd d dv v vdt dt dt

| |PS PS S Sa a a ............. (2)

Application 3.10: If two bodies A and B are mov-ing with accelerations A Ba and a

with respect to

ground the acceleration of A relative to B is AB A Ba a a

2 2AB A B A Ba a a 2a a cosq

Where is the angle between Aa and Ba

.Application 3.11:

Relative Motion on a moving train :If a boy is running with velocity

BTV

relative totrain and train is moving with velocity TGV

relative to

ground , then the velocity of the boy relative to groundBGV

will be given by:

BG BT TGV V V

So, if the boy is running along the direction oftrain

VBG = VBT + VTG

If the boy is running on the train in a directionopposite to the motion of train, then

VBG = VBT VTG

Problem : 3.23An object A is moving with 5 m/s and B is moving with20 m/s in the same direction. (Positive x-axis). Find a) velocity of B relative to A.b) Velocity of A relative to B.

Sol. a) B V 20i

m/s ; A V 5i

m/s ;

B AV V 15i

m/s

b) B A V 20im / s,V 5i

m/s ; AB A BV V V

= 15 i m/s

Problem : 3.24Two object A and B are moving each with velocities10 m/s. A is moving towards East and B is movingtowards North from the same point as shown. Findvelocity of A relative to B AB(V )

.

Sol:

45045 0 ABV E

N

W

SVAO

VB

AB A BV V V = 10i 10j2 2 110 10 10 2ABV ms

along SE

Problem : 3.25Find the speed of two objects if when they move unifomlytowards each other, they get 4.0 metres closer eachsecond and when they move uniformly in the same di-rection with original speeds, they get 4.0 metres closereach 10 sec.

Sol. Let their speeds be v1 and v2 and let v1 > v2 .In First Case:Relative velocity, v1 + v2 = 4 ...... (1)In Second Case:Relative velocity, v1 v2 =

410

= 0.4 ...... (2)

Solving eqns. (1) and (2), we getv1 = 2.2 m s

1, v2 = 1.8 m s1

Problem : 3.26A person walks up a stationary escalator in time t1. I fhe remains stationary on the escalator, then it can takehim up in time t2. How much time would it take him towalk up the moving escalator ?


AKASH MULTIMEDIA 82

Sol. Let L be the length of escalator. Speed of man w.r.t.

escalator is me1

Lvt

.

Speed of escalator ve = 2

Lt

.

Speed of man with respect to ground would be

m me e1 2

1 1v v v Lt t

The desired time is 1 2m 1 2

t tLtv t t

.

Problem : 3.27

Two ships A and B are 10km apart on a line runningsouth to north. Ship A farther north is streaming westat 20 km/h and ship B is streaming north at 20 km/h.What is their distance of closest approach and howlong do they take to reach it ?

Sol. N

E

AV

BV

W

S

450

20 2 /BAV km h

20 /AV km h

20 /

BV km h

450

B

A

C

ABV

Velocity of B relative to A is BA B AV V V

2 2BAV 20 20

= 20 2 km/h

i.e., BAV

is 20 2 km/h at an angle of 450 from easttowards north.

Condition A is at rest and B is moving with BAV

in the

direction shown in figure.

Therefore, the minimum distance between the two is

Smin = AC = AB sin 450 =

1102

km = 5 2 km

and the time isBA

BC 5 2 1t420 2V

h= 15 min

3.20MOTION OF A BOAT IN THE LINE OFRIVER FLOWSuppose a boat moves relative to water with

velocity BWV

and water is following relative toground with velocity

WGV , velocity of boat relative to

ground

BGV will be given by : BG BW WGV V V

a) If the boat is moving along the direction offlow of water (down stream)

A Bd

VBWVWG

VBG = VBW + VWGThe time taken for the boat to move to a dis-

tance d along the direction of flow of water is

1BW WG

dtV V

.................(1)

b) If the boat is moving opposite to the direc-tion of flow of water (up stream)

A Bd

VBWVWG

VBG = VBW VWGThe time taken for the boat to move to a

distance d opposite to the direction of flow ofwater is

1BW WG

dtV V

.........(2)

From equations (1) and (2) BW WG1

2 BW WG

V Vtt V V

c) If the engine of boat is switched off , thenthe boat will be carried by the river current.

VBG = VWG (VBW = 0)

The time taken for travel a distance d is

3WG

dtV

time taken by person to go down streams a dis-tance d and comes back

d uT t t

BW WG BW WG

d dTV V V V

1 22

1Gmm2 d


AKASH MULTIMEDIA 83

b) Shortest Time :

d

WGV

BWV

BGV

A

CB

q

x

To cross the river in shortest time the boatshould be rowed along the normal to flow of water(or) by making an angle of 900 with the flow of water

The minimum time taken to cross the river is

minBW

dtV

.... (5) [from (4) 0 ]

The velocity of boat with respect to ground is2 2

BG BW WGV V V ... (6)

In this case the boat reaches the other bank atpoint C, due to the flow of water.

The displacement of boat along the directionof flow of water (or) drift is given by

x = VWG (tmin) , WGBW

dx VV

.... (7)

The displacement of the boat is2 2s d x

Note 3.16 :The time can be obtained by dividing the

distance in a particular direction by velocity in thatdirection.

BW WG BG

AB BC ACtV V V

c) Suppose the boat starts at point A on one bankwith velocity BWV

and reaches the other bank at

point D as shown in the diagram.

d BWV

BEV

A

DB

q

C x

BWV cosq

BWV sinq WGV

3.21 MOTION OF A BOAT ACROSS A RIVERa) To cross the r iver over the shotest distance:

d

WGV

BWV BGV

A

B

900q

C

Suppose a boat starts from a point A on onebank of the river of width d. To cross the river overthe shortest distance, the boat should move by mak-ing an angle q with the normal to the flow of wateras shown in above figure.

Let WGV

is the velocity of the water with re-spect to ground, BWV

is the velocity of boat in still

water (or) velocity of boat relative to water. Velocity of boat with respect to ground is

given by BG BW WGV V V

from the triangle ABC2 2

BG BW WGV V V .......... (1)

and BGBW

VcosV

......... (2)

WG

BW

VsinV

......... (3)

To cross the river in shortest path the anglemade by the velocity of boat with the flow of wateris 90 +

= 90 + sin1 WGBW

VV

The component of velocity of boat normal tothe flow of water is = VBW cos.

The time taken for the boat to cross the river is

BW

dtV cos

......... (4)

BGBGBW

BW

d dtVVV

V

2 2BW BG

dtV V

........... (5) [ from (1)]


AKASH MULTIMEDIA 84

The component of velocity of boat parallel tothe flow of water is BWV

The component of velocity of boat normal tothe flow of water is of BWV Cos

The time taken for the boat to cross the river is

BW

dtV Cos

The horizontal displacement of the boat (or)drift is

WG BWx V V sin t

WG BWBW

dx V V sinV Cos

Note 3.17 : i) If VWG > VBWsin , the boat reachesthe other end of the river to the right of B.

ii) If VWG < VBWsin , the boat reaches theother end of the river to the left of B.

iii) If VWG = VBWsin , the boat reaches theexactly opposite point on the other bank. i.e, at B.

Problem : 3.28A boat is moving with a velocity vbw = 5 km/hr relative towater. At time t = 0, the boat passes through a piece ofcork floating in water while moving downstream. I f itturns back at time t1=30 min,a) when the boat meet the cork again ?b) The distance travelled by the boat during this time.

Sol.

t = 0

BA

C

vw vw

Liet AB=d is the distance travelled by boat along downstream, in t1 sec and it returns back and it meets thecork at point C after t2 sec.

Let AC=x is the distance travelled by the cork dur--ing (t1 + t2) sec.

B w 1d V V t .................. (1)

B w 2d x V V t ........... (2)and w 1 2x V t t ............. (3)

Substitute (1) and (3) in (2) we get t1 = t2 The boat meets the cork again afterT = 2t1 = 60 minand the distance (AB+BC) travelled by the boat beforemeets the cork isD = 2d x

B W 1 W 1D 2 V V t V 2t

B 1 W 1 W 1D 2V t 2V t 2V t

B 130

D 2V t 2 5 5km60

Problem : 3.29A man wishe 00000s to cross a river flowing withvelocity u jumps at an angle with the river flow. Findthe velocity of the man with respect to ground if he canswim with speed v. Also find how far from the pointdirectly opposite to the starting point does the boatreach the opposite bank. In what direction does theboat actually move. I f the width of the river is d.

Sol.

qfA

B C

v

u

d

x

Cv

MG MW WGV V V

and is the angle between MGV

and WGV

2 2MGV V u 2Vucos

drift = du V cosVsin

Vsintanu V cos

Problem : 3.30A swimmer crosses a flowing stream of width d to andfro in time t1. The time taken to cover the same dis-tance up and down the stream is t2. If t3 is the time theswimmer would take to swim a distance 2d in still wa-ter, then

Sol. Let v be the river velocity and u the velocity of swim-mer in still water. Then

1 2 2

dt 2u v

..... (i)


AKASH MULTIMEDIA 85

2 2 2d d 2udt

u v u v u v

.... (ii)

and t3 =2du

........ (iii)

from equations (i), (ii) and (iii)21 2 3 1 2 3t t t t t t

Problem : 3.31Two persons P and Q crosses the river starting frompoint A on one side to exactly opposite point B on theother bank of the river. The person P crosses the riverin the shortest path. The person Q crosses the river inshortest time and walks back to point B. Velocity ofriver is 3 kmph and speed of each boat is 5 kmph w.r.triver. I f the two persons reach the point B in the sametime, then the speed of walk of Q is

Sol : For person (P) : For person (Q) :

C B

A

VB

WV

CB

A

VB

x

WV

P 2 2 2 2B W

d d dt45 3V V

Q P QB

d dt , t t tV 5

man

d d x ,4 5 V But W

B

dx VV

W

B man

V dd d ,4 5 V V man

d d 3d4 5 5 V

man

1 1 3 ,4 5 5V

man

1 320 5V

man

3 20V 12kmph

5

*Problem : 3.32A man in a boat attempts to cross a river from point A.I f he rows the boat perpendicular to the banks, hereaches point C at a distance s = 120m down stream,from point B, in 10 minutes shown in figure. I f the manrows his boat at an angle to the straight line AB hereaches the point B in 12.5 minutes. Find a) the width ofthe river, (b) velocity of the boat relative to the water, (c)velocity of the curent and (d) the angle . Assume thevelocity of the boat relative to the water to be constnatand of the same magnitude in both cases.

Sol : Let the velocity of the boat with respect to water be u,the velocity of the water current with respect to thebanks is v, and the width of the river ' '.

A

B C

u a

v

When boat is rowed perpendicular to the banks the re-sultant velocity of the boat with respect to the banks isalong AC. The boat moves along the river with velocityv. Suppose the boat moves a distnace BC = s along theriver in time t1.

Then BC = s = vt1 = 120m............(1)

Exactly in this time t1 boat moves across the river andtravels a distnace' ', (width of the river), with velocityu.

Then = ut1 ........(2)Now we consider the case when the boat moving withvelocity u makeing an angle ' ' with the line AB.

The relative velocity of the boat with respect to thebank along the river is v u sin = 0,

or v = u sin ........(3)

The relative velocity of the boat with respect to thebank and perpendicular to it is ucos a . If the time takento reach the point B in the case is t2then (ucos )t2 = ......... (4)From (1), (2), (3) & (4) t1 = s/v; t2 = /ucossin = s/ ; cos = t1/ t2We know that sin2 + cos2 = 1. From this identity

2 21 2s / t / t 1

Solving for ' ' we get

= st2 2 22 1t t and substituting the valuess = 120 m, t1 = 10 minutes, t2 = 12.5 minutes,we get = 200m.using (1) and (2) we getu = 19.2 mmin; v = 12 m/minand sin1(v/u) = sin1(12/19.2)= sin1(0.6667) = 410.49


AKASH MULTIMEDIA 86

3.22 A MAN IN RAINLet us consider the rain is falling with a veloc-

ity RV

and a man moves with a velocity MV

rela-tive to ground, he will observe the rain falling witha velocity RM R MV V V .

Since RV

is assumed

as a constant (for the rain falling at a constant rate),the man will record different velocities of the rain ifhe moves with different velocities relative to ground.i) If rain is falling vertically with a velocity RV

and an observer is moving horizontally with velocityMV

, the velocity of rain relative to observer will be :

RV

MV

MV

RV

RMV

MV

a

b

RM R MV V V

The magnitude of velocity of rain relative toman is 2 2RM R MV V V

If is the angle made by the umbrella with

horizontal, then RM

VtanV

(or) RRM

VsinV

(or)

M

RM

Vcos .V

If is the angle made by the umbrella with

vertical, then MR

VtanV

(or) MRM

VsinV

(or)

R

RM

VcosV

ii) When the man is moving with a velocity1M

V relative to ground towards east, and the rain isfalling with a velocity RV

relative to ground by

making an angle with vertical.Then the velocity

of rain relative to ground 1RM

V is as shown in figure.

1MV

1RMV

1MV

1 MV

RV

1RMV

yq

b

x

NESW

1 1

RM R MV V V

From the diagram xtany

......... (1)

and 1Mx v

tany

.......... (2)

iii) If the man speeds up, at a particular velocity

2MV

, the rain will appear to fall vertically with

2 2RM R MV V V

as shown in figure.

2RMV

2MV

2MV

2 MV

RV2RMV

iv) If the man increases his speed further, he will seethe rain falling with a velocity as shown in figure.

3MV

3RMV

3MV3

MV

RV3RMV

bq

x

y

33 3

MRM R M

v xxV V V tan ; tany y

(where x and y are assumed unknowns involvedin the problem and their value can be either obtainedor eliminated).

From the above threee cases we understand thatsometimes the man obseves the rain falling forwards,sometimes vertically downwards, and sometimes therain appears to fall backwards. Hence the velocityof rain relative to man depends upon the velocity ofman relative to ground.

*Problem : 3.33A pipe is fixed to a moving cart such that the axis ofthe pipe makes an angle with the floor of the cart.The cart moves uniformly along a horizontal path withvelocity v1 = 2ms-1. I f rain drops which are falling witha velocity v2 = 6ms-1 vertically down reach the bottomof the pipe without touching the walls of the pipe findthe value of . Assume that the air resistance isnegligible and velocity v of the drops is constant.


AKASH MULTIMEDIA 87

Sol. From the fig .2

1

v 6tan 3v 2

. 21v

1v

2v

1v

1v

atan (3)

Problem : 3.34To a man in car which is moving due east with a speed of20 m/s, the rain appears to fall at an angle of 300 withvertical. When he reduces the speed to 8 m/s, again therain appears to fall at the same angle with vertical. Findthe velocity of rain.

Sol:y

300r mV

RV

x20mV

0 20 xTan30y

.... (i)

RV

RMV

030

x

y

-VM = 8

Tan 300 = x 8

y

...............(ii)

Solving eq. (i) and (ii), we have

x = 14 and 6 3y

Hence rgV 14i 6 3 j

Problem : 3.35Rain is falling vertically with a speed of 20ms1. Aperson is running in the rain with a velocity of 5 ms1and a wind is also blowing with a speed of 15ms1(both from the west). The angle with the vertical atwhich the person should hold his umbrella so that hemay not get drenched is:

Sol. Rain RV V 20( K)

Man MV V 5i

wind WV V 15i

Resultant velocity of rain and wind =

RMV 20K 15i

Now,Velocity of Rain relative to man = RM MV V

( 20K 15i) (5i)

a

a20K

10i

Vertical

20K 10i 1Tan2

11Tan2

Problem : 3.36To a man walking at the rate of 3km/h the rain appearsto fall vertically. When he increases his speed to 6 km/h it appears to meet him at an angle og 450 with vertical.Find the angle made by the velocity of rain with thevartical and its value.

Sol.

RMV

RV

MV 3kmph

RMV

RV

3 36

y

045 q

From the diagram 0 3Tan45y

....... (1)

and 3

Tany

.............. (2)

From (1) and (2) 045

and 0R R

3 1 3sin 45 , V V2

RV 3 2 kmph

3.23 MULTIPLICATION OFA VECTOR BY A SCALAR

Consider a vector P

as shown in figure (a).When it is multiplied by a scalar m, the result is

another vector R

such that

R mP

So, whenever a vector is multiplied by a scalarm the result is another vector having magnitude mtimes that of the first and directed along the first vec-tor.


AKASH MULTIMEDIA 88

2B A

2C A

a b c

P

An illustration of this for m = + 2 and m= 2 isshown in figure (b) and figure (c )respectively.

If m = 0, the result is a null vector.

If 1mP

then R is a unit vector..

3.24 SCALAR PRODUCT ORDOT PRODUCT

Dot product of two vectors is defined as theproduct of their magnitudes and the cosine of theangle between the two.

(or)The dot product is the product of the mag-

nitude of one of the vector quantities and the sca-lar component of second vector in the directionof the first vector

(i)q

Q

P

The Scalar product of the two vectors, P

andQ

is defined as P.Q P Q cos

. Here is theangle between P

and Q

.

The angle between the two vectors P and Q

is

given by P.QcosP Q

1 P.Qcos

P Q

PROPERTIES OF SCALAR PRODUCT :(a) The Dot product between two vectors obeyscommutative law. i.e., P.Q Q.P

(b) The Dot product obeys distributive law,i.e., P.(Q R) P.Q P.R

(c) It is always a scalar which is positive if anglebetween the vectors is acute (i.e. < 90) and negativeif angle between them is obtuse (i.e. 90 < 180)

(d) Scalar product of two vectors will be maximumwhen cos = max = 1, i.e., = 0, i.e., vectors areparallel.

max(P.Q) PQ

(e) P.Q

P (Q cos ) = Q (P cos )(f) If the scalar product of two non zero vectorsvanishes, then the vectors are orthogonal.(perpendicular).

i.e., = 900, 0P.Q PQ cos90 0

(g) The scalar product of a vector by itself istermed as self dot product and is given by

P.P

= PP cos = P20( 0 )

(h) In case of unit vector n ,n . n = 1 x 1 x cos = 1n . n = i . i = j . j = k . k = 1

(i) In case of orthogonal unit vectors i , j and

k ; i . j = j . k = k . i = 0

(j) If y z P P i P j P k

and

y z Q Q i Q j Q k

y z y z P.Q P i P j P k . Q i Q j Q k

= (PxQx + PyQy + PzQz)

(k) The product of vectors is possible betweenvectors of different kind.

Application 3.12 :Geometrically, Q cos is the projection of

Q onto P and P cos is the projection of

P onto

Q

as shown. So P .

Q is the product of the magnitude of

P and the component of Q along

P and vice versa.

q

P

Q

q

q cosQ

Q

P

Q

qQ c

osq

P

(a) Component of Q

along P

= Q cos =

P.QP

P.Q


AKASH MULTIMEDIA 89

(b) The vector component of Q

along P

=

P.Q Bcos P P P.Q PP

(c) Component of P

along Q

= P cos =

P.QQ

P.Q

(d) The vector component of P

along Q

=

P.Q P cos Q Q P.Q QQ

e) The component of P

perpendicular to Q

andin the same plane is C P P.Q Q

f) The component of Q

perpendicular to P

and

in the same plane is D Q Q.P P Examples of Scalar Producta) The scalar product of force and displacement,is called workdone by that force.

If a force F

acts on a body and displaces itthrough a displacement S

, the work done by the force

W F.S F(Scos ) (F cos )S

Workdone is maximum when force anddisplacement are in the same direction (the anglebetween them = 0) and is equal to FS.

If force and displacement are perpendicular toeach other, the workdone by the force is zero.b) Dot product between force and velcity vecotors

gives power i.e., power = F.S F.Vt

c) A block of mass 'm' is lifted to a vertical height'h' from the surface of the earth. The work done (W)against the gravitational force (F=mg) is

W F.h Fh cos

0 W = Fh

W = mgh

mg

h

If h decreases, potential energy of the body willdecreases.

d) The dot product of the magnetic fieldinduction and area of the coil gives the magnetic flux.d B.ds (or) B.ds.

Bds

n

O

q

e) The dot product of current density J

and areads gives electric curret i.e., dI J.ds

(or) I J.ds

f) Electric potential is the scalar product of electric

field intencity E and position vector dr .

dV = E.dr

or V E.dr.

g) If a magnetic dipole of moment of M

is in amagnetic field of induction B,

the potential energy

of the dipole is given by BU M.B.

3.25 CROSS PRODUCT OF TWO VETORSThe vector product of two vetors is a vector

whose magnitude is equal to the product ofmagnitudes of the two vectors and sine of theangle between them.

The pr oduct vector i s in a planeperpendicular to the plane of two vectors. Thedirection of product vector can be known eitherwith r ight hand screw rule or with r ight handthumb rule.

n

PQ

P Q

q

According to vector product P Q PQ sin n

n is a unit vector perpendicular to plane of P

and Q

as shown in fig.The direction of P Q

is given by the following

rules.


AKASH MULTIMEDIA 90

(a) Right handed screw rule : Imagine a righhanded screw to be placed along the normal of theplane (ABCD) containing P

and Q

. Rotate the

cap of the screw from P

to Q

through the smallerangle between them. The direction of motion of thetip of the screw gives the direction of P Q

.

A

D

B

C

O

P Q

P

Q

Q P

q

It is clear from fig. that tip of the screw movesup when cap is rotated from P

to Q

. Hence P Q

directed upwards. Similarly, it can be proved thatQ P

is directed downward along the normal.(b) Right hand thumb rule : Imagine the

normal to the plane (ABCD) containing P

and Q

to be held in the right hand with the thumb erect. Ifthe fingers curl in the direction from P

to Q

, through

smaller angle between them, then the direction of

thumb gives the direction of P Q

.

P

Q

P Q

q

Q P

O

A

D C

B

Fig. shown as that P Q

is directed along thenormal in the up ward direction while Q P

is

downward.Properties of vector product : Following are

the properties of vector product :a) The vector product is not commutative

P Q Q P

.b) The vector product is distributive

P Q R P Q P R .

c) The vector product of two parallel vectors isa null vector i.e.,

P Q

PQ sin n = PQ isn 0 n = 0

d) The vector product of a vector by itself is anull vector (zero)

P P

= P P sin n = PP sin 0 n = 0

e) The magnitude of the vector product of twovectors mutually at right angle is equal to the productof the magnitude of the vectors.

P Q

= P Q sin n = PQsin900 n = PQ nf) The vector product of unit orthogonal

vectors, i, j and k have the following relations. i i j j k k 0 i j j i k, j k k j i, and

k i i k j They follow the right hand screw rule as shown

in fig.

i

j

k

g) The vector product of two vectors intermsof their x, y and z components can be expressed inthe form of determinant.

If x y z P P i P j P k

x y z Q Q i Q j Q k

x y z

x y z

i j kP Q P P P

Q Q Q

y z z y x z z x x y y x P Q P Q i P Q P Q j P Q P Q k h) The unit vector perpendicular to both P

and

Q

isP Q

nP Q


AKASH MULTIMEDIA 91

i) If x y z P P i P j P k,

x y z Q Q i Q j Q k

are parallel then yx zx y z

PP PQ Q Q

= constant

Note 3.18 :

1) The angle between P and P Q

is 900

2) The angle between Q and P Q is 900

3) The angle between (P Q) and (P Q) is 900

4) The angle between (P Q) and (P Q) is 900

5) The angle between P Q and Q P

is 1800

Application 3.13 : Area of prarallelogram :

If P

and Q

are the two adjacent sides of aparallelogram OABC as shown in figure.

BN is the perpendicular drawn from B on OA.

Q

CB

qO AP

N

The area of parallelogram= OA X perpendicular distance BN

= OA X OB sin

= PQ sin

= P Q

Application 3.14 : Area of parallelogram in terms of diagonals.If 1 2d and d

are the diagonals of a parallelogramof OABC as shown in fig.

A

B C

O1d

2d

Q

P

From the diagram

1d P Q

and 2d P Q

1 2d d P Q P Q

= P P P Q Q P Q Q

= O P Q P Q O

= 2(P Q)

1 2d d 2 P Q

Area of the pralleogram = 1 21 d d .2

Application 3.15 : Area of tr iangle : Let OAB bea triangle formed with P

and Q

as its two sides as

shown in figure.

Q

Oq

P

B

AC

Area of the triangle = 12 (base)(height)

= 12 (OA)(BC)

From the triangle OBC BC Q sin

Area of the triangle = 1 P Qsin2

1 1PQsin P Q2 2

Area of triangle is equal to half the magnitudeof the cross product of two vectors representing twosides of the triangle .

Examples of Vector Product :a) Relat ion between L inear Veloci ty V,

Angular Velocity and Position Vector r . a) Consider a particel is moving in a circular

path about an axis passing through OO1,perpendicular to plane of the path, with aconstant angular velocity

. P is the position of the


AKASH MULTIMEDIA 92

body at any time given by the position vector r

described with 'O' as the origin. The radius of thecircular path is O1 P = r sin . The linear velocity v= (rsin ). The direction of the velocity is tangentialto the path at 'P' and it is perpendicular to the planecontained by

and r . So vectorially v r .

O

v

p

O

w

q

z

r

r sinq

b) Torque (Moment of force) :Turning effect of a force is called torque,

and is equal to the product of the force and theperpendicular distance between axis of rotationand the line of action of force.

Let F

be the force acting on a particle at apoint P in XY plane. The position vector of theparticle be r . The torque acting on the particlewith respect to the origin O is given by ON F

r sin Fn ON rsin

The torque acts perpendicular to XY plane i.e.,along Z axis.

Magnitude of torque is r sin F

q

X

Y

Z

qF

O

r

r

NP

c) Angular momentum :The moment of momentum is cal ledangular momentum.

O

L r p

pr

N

X

Y

Z

qQ

Consider a particle of mass m at a point Q inXY plane. The position vector of Q is r . If it ismoving with a velocity v in that plane, itsmomentum P mv.

Angular momentum = perpendicular distancefrom origin to the line of action of momentum xmomentum

L = ON(P) = (rsin )P

L r p, L rpsin n

The direction of angular momentum is alongZaxis.d) Force on a charged par ticle q moving withvelocity v in a magnetic field B

is given by

F q(v B)

e) Tor que on a magnet ic moment M

inmagnetic field B

is given by (M B)

f) The force on a current carrying conductorof length

in a magnetic field B

is given by

F i( B)

g) If ' i ' denotes a unit vector along incident rayr a unit vector along refracted ray into a mediumof refractive index and n is unit vector normalto boundary of medium directed towards theincident medium, the law of refraction is

i n (r n)

*Problem : 3.37Find the angle between two vectors 2

A i j k

and B = i k.

Hint : A.B

cosA B

, Ans : 300

Problem : 3.38Show that the vectors

3 2 , 3 5

a i j k b i j kand 2 4 c i j k from a right angled triangle.

Sol.

a

b

c

a 9 4 1 14, b 1 9 25 35 and

c 4 1 16 21.


AKASH MULTIMEDIA 93

which gives b2 = a2 + c2

b c (i 3j 5k) (2i j 4k) 3i 2 j k a

Hence, a,b,c are coplanar..

So they form a right angled triangle.

Problem : 3.39I f 1

a and 2

a are two non collinear unit vectors and if

1 2 3, a a then the value of 1 2 1 2( ).(2 )

a a a a

isSol. a1 = a2 = 1 and

a12 +a2

2 + 2a1a2cos 2

3 3 or

1 + 1 + 2 cos = 3 or cos = 1/2

Now 2 21 2 1 2 1 2 1 2(a a ).(2a a ) 2a a a a cos

= 1 12 12 2

Problem : 3.40If a and b are two unit vectors such that 2

a b and

5 4

a b are perpendicular to each other, then the anglebetween a and

b is

Sol. (a 2b).(5a 4b) 0

2 25a 4abcos 10abcos 8b 0 = 5 4 cos + 10 cos 8 = 0 [a = b = 1]= 3 + 6 cos = 0cos = 1/2, = 600

Problem : 3.41Find the component of 3 i + 4 j along i j ?

Sol. Componant of A

along B

is given by A.BB

(3i 4 j).(i j) 7

2 2

*Problem : 3.42

I f 2 3 F i j k and r = ij+6k find r F..

Sol.i j k

r F 1 1 62 3 1

= i(118)j(112)+k(3+2)

= 17i+13j+5k

*Problem : 3.43

Find the uni t vector perpendi cular to2 3

A i j k and .

B i j k

Hint : A Bn ,A B

Ans : 4i j 5k42

Problem : 3.44A particle of mass m is moving with velocity 'v' parallelto x axis along line y = b. I ts angular momentum w . rto origin after time (t) will be

Sol: Shortest distance between line of momentum and originis 'b'.

mv(0,b)

(0,0)

Hence angular momentum of the particle with respetto origin is L mvb( k)

.

Problem : 3.45I f A = ( )

i j and B = ( i j + 5

k ). Find angle

betweenA and

B .

Sol. A B = 2 1 0

1 1 5

i j k

= i (5 + 0)

j (10 0) +

k (2 1)

A B = 5

i 10

j 3

k

sin =

| A B |AB

2 22

2

5 10 3

2 1 1 1 25

sin = 1345 27

, = sin1 134135

.

Problem : 3.46I f a rigid body is rotating about an axis passingthrough the point 2i j k and paral l el to

2 2i j k with an angular velocity of 3rad/sec.

Find the velocity of the point of the rigid body whoseposition vector is 2 3 4i j k

Sol. Unit vector in the direction of

2 2i j k is 2 2 1 2 2

39i J k

n i J k

Angular velocity of rigid body

1 3 2 23 i j k

, 2 2i j k

and position vector 2 3 4 2r i j k i j k , 4 3r J k

Linear velocity of the point 'P' is

1 2 20 4 3

i j kv r

2 3 4v i j k


AKASH MULTIMEDIA 94

Problem : 3.47

Find the vector components of vectorA

= 2 i + 3 j along the directions of B = i + j .

Sol. A.B BC= Acosq B=B B

2+3 i+j 5 C= = i+j .21+1 1+1

*Problem : 3.48A particle is in equilibrium under the simultaneousaction of three forces. Prove that each force bears aconstant ratio with the sine of the angle between theother two.

Sol : Let, P+Q and R

be the three forces acting at a point Oshown in figure. Since the particle is in equilibrium.

P+Q +R=0

............... (1)

(or) (P+Q)=-R

................ (2)

Taking cross product, with P

on both sides of (2),

P P+Q =-PR

(or) PP+P Q=RP

(or) PQ=RP

............... (3)

Taking cross product, with Q

on both sides of (2)

Q P+Q =- QR

QP+Q Q=-QR

QP= QR

P Q

gb a

R(or) PQ=QR

From (3) and (4)

PQ=QR=RP

Taking their magnitudes

PQ sin = QR sin RP sin

Dividing throughout by PQR

PQsin QRsin RP sinPQR PQR PQR

(or) P Q R

sin sin sin

*Problem : 3.49Find the angle between the diagonals of a cube withedges of length a.

Sol.

zF

G

E

D

B

O

A

y

xc

From the diagram

1OG d

ai+aj+ak, 2CF d

ai+ aj + ak

The angle between the daigonals is

Hint : 1 21 2

d dcos

d d

Ans : 1

1cos3

Long Answer Questions1. Define scalar and vector products and explain

their properties. Give two examples for each

Short Answer Questions1. Define scalar product. Give its properties and

two examples.

2. Define vector product. Give its properties andtwo examples

3. State parallelogram law of vectors. Derive anexpression for the magnitude and direction ofthe resultant vector.

4. State triangle law of vectors and explain it.

5. What is relative motion. Explain it?

6. Show that a boat must at an angle of 900 withrespect to river water in order to cross the riverin minimum time ?

7. Define unit vector, null vector and positionvector.

8. If a b a b prove that the angle between

a and b

is 900.


AKASH MULTIMEDIA 95

13. The mass of a body is 5kg and its velocity is 4i 2 j k ms1. What is the momentum of the

body?

14. Find the values of 2 2

a.b a b and

2 2a.b a b .

15. Does vector subtraction obey commutativelaw?

Assess Yourself1) Identify the vector and scalar quantities among

the following.a) temperature b) force c) volume of the milkin a bottle d) bus fare to Hyderabad from yourvillage/town e) height of Charminar f) velocityof horse - cart you travel, the age of the Earth.

Ans : 1) Force, velocity of horse - cart are vectors.2) Temperature, volume of the milk, bus fare,height of charminar, the age of the earth areScalars.

2. Can we add a vector quantity to a scalar quantity?

Ans : We can add a vector to another vector of samekind but we cant add a vector quantity to ascalar quantity.

3. The magnitudes of two vectors A

and B

are8 units and 5 units respectively. What are themaximum and minimum possible values for themagnitude of the resultant vecor R A B

?

Ans. is maximum when A and B are in the samedirections. The magnitude of R is equal to themagnitude of A and B.

Rmax = A + B = (8+5) unit = 13 units.R has minimum magnitude when A and B areopposite (antiparallel) to each other.Rmin = A B =(85) units = 3 units.

4. If A

and B

are two vectors, under what

conditions A B A B

. Under what

condition the resultant vector equals to zero ?

Very Short Answer Questions1. The vertical component of a vector is equal to

its horizontal component. What is the anglemade by the vector with xaxis?

2. A vector V make an angle with the hori-

zontal. The vector is rotated through an angle . Does this rotation change the vector V

?

3. Two forces of magnitudes 3 units and 5 unitsact at 600 with each other. What is the magni-tude of their resultant?

4. A i j.

What is the angle between the vec-tor and xaxis ?

5. When two right angled vectors of magnitude7 units and 24 units combine, what is the mag-nitude of their resultant?

6. If P

= 2i + 4j + 14k and Q

= 4i +4j + 10k findthe magnitude of P Q

.

7. A force 2i + j k newton acts on a body whichis initially at rest. At the end of 20 seconds thevelocity of the body is 4i + 2j + 2k ms1. Whatis the mass of the body?

8. Can a vector of magnitude zero have a non-zero components?

9. What are the units of a unit vector?10. Can two vectors of unequal magnitude add up

to give the zero vector? Can three unequal vec-tors and upto give the zero vector?

11. During practice a player throws a ball high into the air, and then runs in a straight line andcatches it. Which had and greater displacement,the player or the ball ?

12. In the below fig. there are four forces that areequal in magnitude. Identify any three of themwhich act on a body simultaneously and keepit with unchanged velocity.

C

AB

D


AKASH MULTIMEDIA 96

Ans. when these vectors have same direction. A +B = 0 when they are oppositely directed.

5. A displacement vector in x y plane hasmagnitude 25m and direction 300 with x axis.What are its x and y components?

Ans :

Y

Xx

y

P

O

25m

30y

3x 25cos30 25 21.65N2

1y 25sin30 25 12.5N2

6. A certain force has magnitude of 200N, its xcomponent is 80N. What is its y componentand its direction. Two possible answers existfind them.

Ans :

X

Y

80N

200N

O

in first case

-Y

-X

a) 2 2x yF F F

2 2200 80 Fy yF 183N and x

y

F 80tanF 183

= 240

The angle made by the force with Xaxis is90 + 24 = 1140 in anticlock wise direction.

b) 2 2 2 2x y yF F F 200 80 F

yF 183N

y

x

F 183tanF 80

in second case

Y

X-X

Fy

O-80

-Y

200N

066

The angle made by this vector with Xaxis is180 + 66 = 2400 in anti clock wise

7. If A B , what can you conclude about the

components of A and B.Ans. All their components may or may not be equal.8. If one component of a vector is not zero, can

the magnitude of the vector be zero ?

Ans. The magnitude of vector A is given by

2 2 2x y zA +A +A . Hence if any component is

non zero, A can never be zero.9. Under what circumstances can the component

of a vector be equal to the magnitude of thevector ?

Ans : If vector is directed along positive direction ofone of the three axes x, y or z then themagnitude of the vector is the same as thecomponent associated with the direction, othercomponents are zero.

10. Write the following vectors in their unit vectorform in cartesion coordinate system in twodimensions (i.e., consider x y plane only.)a) A velocity vector of 10 ms1 at an angle ofelevation of 600;

Ans :

Y

R P

Q XO

10 m

s-1

Vy

Vx600

V=10cos60i+10sin60j

5 8.66 V i j

b) A vector A of magnitude A = 5m and angle = 2250

Ans :Y

X-XVx

Vy450

2250

10ms

-1

-Y


AKASH MULTIMEDIA 97

10 cos 45 10sin 45 V i j

3.54 3.54 V i j

c) A displacement from the origin to the pointx = 14m, y = 6m

Ans : 1 2 S S i 5 j

, S 14i 6 j

11. If the resultant of three vectors is zero, mustthey all be in the same plane ?

Ans. Yes, the vectors must be in the same plane.Resultant of two vectors lies in the plane inwhich the vectors lie. Hence the third vectorwhich is equal to the resultant of the two andopposite in direction should lie in the sameplane.

12. Two boats travelling at the same speed set offacross a river at the same time. One headsstraight across and is pulled downstreamsomewhat by the currentThe other one headsupstream at an angle so as to arrive at a pointopposite the starting point. Which boat reachesthe opposite side first ?

Ans : The boat which heads straight across reachesthe opposite side first because it takes where

minimum time to cross i.e.,2 2BW WG

dt

V V

is width of the river and VBW is boat velocityand VWG is river velocity. The boat which headsupstream at an angle crosses the river followingthe shortest path. The condition for that is

= sin1 WGBW

VV

.

13. Prove that P.P

= Px2+Py

2+Pz2 = P2.

Ans. Consider x y z P P i P j P k.

x y z x y z P.P P i P j P k . P i P j P k

2 2 2 2x y zP.P P P P P

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