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CHAPTER - 5
VECTORS
3
...


ur

ur

ur

ur

Thus A  B or B   A
uur

A


ur uur

B  A

4
...

Null vector: “A vector of zero magnitude is called zero vector or null vector”
...


It is represented by 0
...
Its
direction is indeterminate
...


A

B


C

6
...
Vectors, acting in different plane are
called non-coplanar vectors
...
The angle obtained in this way, is
the angle between the vectors
...

*Whenever angle between two vectors is to be taken we must make sure that either
their heads coincide or their tails coincide
...
If heads coincide with tail then external angle is the angle
between the two vectors as in figure (2)
...


Figure 2

Unit vector : “A vector of unit magnitude is called unit vector”
...

It is conventional to denote unit vector with a “cap” instead of “bar” over the symbol
...

IX Class - Physics

Page No : 56

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Displacement
Displacement is a shortest distance between two points
...

Displacement vector
The position of the point Q with reference to the origin is represented by the position


vector r2
...











Thus, the displacement vector  r  r2  r1 = (x2 – x1) i + (y2 – y1) i

WORK SHEET – 1

Single Answer Type
1
...

1) If two vectors have same magnitude and direction they are said to be equal vectors
...

3) A vector of zero magnitude is called zero vector or null vector
...


2
...


4
...

1) Assertion is true, and reason is true and the reason is correct explanation of assertion
...

3) Assertion is true but reason is false
...

The addition of one vector to the other result a null vector
...

4) Neither is negative vector of the other
...


If


1 
1 
k
+
+
c
is a unit vector, then the value of ‘c’ is
2 i
2 j

1) 1/ 2
12
...
Then
its displacement is














2) i  5 j 4 k







3) i  5 j 4 k

4) None of these

3) 5

4) Zero

3) 0
...
2 i + 0
...
27


18
...


4) None





1)  i  5 j 4 k
16
...


3) 19





If A = 4 i – 3 j and B = 6 i – 5 j
...


4) –1/2

Magnitude of vector 3 i – 12 j – 4 k
1) 13

13
...
4




If 2 i  y j 3 k = 5, then y =
1)

6

2)

12

15

Subjective Answer Type
19
...


If the position of a particle changes from (1,2,3) m to (5,4,2)m, then find the displacement
vectors
A vector PQ has the initial point P(1,2,–1) and terminal point Q(3,2,2)
...


IX Class - Physics

Page No : 60

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

B


A






C  AB




A

B

5
...




Q


 
 

R  PQ



P
5
...
This law is useful to find both magnitude and direction of resultant
...

D

D


A



ur
P

B

A

C

ur
R

ur
Q

ur
Q



ur
P

B

E

figure (d)

ur
ur
uuur
uuur
Explanation: P and Q are two vectors represented by AB and AD
...
If the
parallelogram ABCD is completed taking AB and AD as adjacent sides, then the diagonal
ur
uuur
AC represents their resultant R both in magnitude and direction
...
The perpendicular
drawn from ‘C’ meets the extension of AB at E
...


2
...


Arrange the vector addition so that their magnitude in increasing order are








a)

Two vectors A and B are parallel

b)

Two vectors A and B are antiparallel

c)

Two vectors A and B making an angle 600

d)

Two vectors A and B making an angle 1200

1)

b, d, c, a









2) a, b, d, c

1)

If two vectors are in the same directions the direction of resultant is same as the
individuals vectors
...


4)

None of these

The vector sum of two forces 8 N and 6 N can be
2) 20 N

ur

ur

2) 3i$ 5j$ 3k$

3) 5i$ 5j$ 3k$

4) $
i  3j$ 3k$

If two forces 12 N and 8 N at on an object in the same direction then the magnitude of
the resultant is
2) 16 N


3) 20 N

4) 50 N



If two vectors P and Q making an angle ‘  ’ between them, then the magnitude of the
resultant R of vectors is ___________
1)

R=

P 2  Q2  2PQ Sin

2) R =

P 2  Q2  2PQ Cos

3)

R=

P 2  Q2  6PQ Cos

4) R =

P 2  Q2  6PQ Sin



7
...


3) 1 N

If P  4i$ 2j$ 6k$and Q  $
i  3j$ 3k$the P  Q 
1) 5i$ 5j$ 3k$

5
...


3) c, d, b, a



If P = Q and if  = 120 0 between them, then select the true answer from the
following
1)

P = Q R

IX Class - Physics

2) P = Q = R

3) Both (1) & (2)

Page No : 64

4) None of these

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V  V1  V 2  V3  V 4  V5
Note: If a number of vectors simultaneously acting at a point can be represented in
magnitude and direction by the sides of a polygon taken in order, the point is in
equilibrium ie the vectors have zero resultant
...


The ratio of maximum and minimum resultant of two forces is 7 : 1
...


2) 4 : 3

2) 450

2)    /4

2) 600

2) 1000

2P act so that the resultant force is
3) 900

10 P?

4) 1200

3 N

3) 500

3 N

4) 500 N

The greatest and least resultant of two forces at a point are 29N and 5N respectively
...


4)  = 2  /3

The resultant of two forces of equal magnitude is 1414N, when they are mutually
perpendicular
...


4) 300

3)    /2

At what angle should the two forces 2P and

1) 450
5
...
Then A + B is a unit
vector if,

1)    /3
4
...
If the resultant force
39 N, the angle between the forces is
1) 600

3
...


2) 900

3) 1200

4) None of these

The resultant of two vectors (P + Q) and (P – Q) when they at right angles to each other
is

1) 2 P 2  Q2

2) 2  P 2  Q2 

3) P 2  Q2

4) (P 2  Q2 )  2

Subjective Answer Type
9
...
If their resultant is at right angles to
4N, the value of F is

IX Class - Physics

Page No : 66

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5
...
After a time interval ‘t’ its final


velocity is V 2 due north
...
The initial velocity

 
 





vector  V1  is reversed to get  V1 and added to V 2 following the usual procedure for




vector addition
...
e
...







(iii) The vector drawn from tail of V 2 to the head of  V1 represents change of velocity  V




both in magnitude and direction
...



(iv)



2
2
Magnitude of change of velocity V  V 2  V1  V1  V2  2V1V2 cos 90

 V  V12  V22

WORK SHEET - 4
Single Answer Type
1
...
Given A = 4 units and B = 3 units

1) 13 units
2
...


3) 1 unit

ur

ur

2) 4i$ 5j$ 6k$

ur ur

ur

3) 4i$ 5j$ 6k$

4) 3i$ 5j$ 6k$

If R  A  B and angle between A, B is  , then tan  is given by __ (where  is angle

ur

ur

made by R with A )
1)

A sin 
B  A cos 

IX Class - Physics

2)

A sin 
B  A cos 

3)

Bsin 
A  B cos 

Page No : 68

4)

Bsin 
A  B cos 

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ur
ur
ur
This equation confirms that A x and A y are the components of A
...
12 RESOLUTION OF A VECTOR INTO THREE RECTANGULAR COMPONENTS




(I)

Consider a right handed three dimensional coordinate system
...
The three rectangular components of A can be determined
as follows:

(ii)

Draw PP 1 perpendicular from P upon X–Z plane
...


(iii)

r


OP2 , P1P and OP3 are known as the x-component, y-component and the z-component,








respectively, of A
...



...
(3)



This gives the magnitudes of vector A
...

IX Class - Physics

Page No : 70

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8
...


2) minimum

4) none

If the component of one vector along the direction of the other vector is zero, the angle
between the two vectors is
1) 0°

10
...


One of the two rectangular components of a force is 25 N and it makes an angle of 60°
with the horizontal (force)
...


Three vectors A, B and C are of magnitudes 10 units, 5 unit and 10 unit respectively
...
13 EQUILIBRIUM
i
...


ii
...


iii
...
These
two forces are equal in magnitude and opposite in direction
...
14 APPLICATION OF TRIANGLE LAW OF VECTORS (LAMI’S THEOREM)
If three vectors, simultaneously acting at a point, have zero resultant, then these
three vectors can be represented in magnitude and direction by the three sides of a
triangle taken in order
...
Hence they are represented, in magnitude and direction,
by the sides AB, BC and CA of the triangle ABC taken in order
...




Three forces start acting simultaneously on a particle moving with velocity V
...
The particle will now move with velocity
C

A

B



5
...
Find the horizontal force in the
thread is
[Note: If mass of a body is ‘m’ kg then its weight is ‘m’ kg–wt]
1) 10

6
...
15 TANGENT OF LAW
If the bob of a pendulum is pulled to a side such that the string makes an angle  with
the vertical, F = mg tan  where F is the horizontal force applied
...


IX Class - Physics

Page No : 74

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7
...
If F2  5 3 N and ‘O’ is in equilibrium, value of
F 1 and F 3 are

F2
F1
60°

30°
O

F3
1) 5N , 20 N
8
...
Then tension in that rope is
1) 1 N

2) 3 N

3) 2 N

4) 4 N

SYNOPSIS - 8
5
...
When two vectors are
multplied the product may be a scalar quantity (or) a vector quantity depending on the
physical quantity to be obtained
...

1) Cross product or vector product
2) Dot product or Scalar product
1
...













The cross product of two vectors A and B is denoted by A B and read as A cross B
...
Here 0    
...


Two vectors of magnitude 18 units and 13 units are acting at 300 to each other
...


2) 234 units

3) 117 units

4) 243 units

Two forces of magnitude 12N and 14N are acting at a point as shown in figure
...


2) 84 3N

ur

ur ur

2) 900

ur



3) A2

4) A









What is the angle between c d and c  d ?
2) 45°


3) 60°

4) 90°



If vectors A and B have angle  between them, then the magnitude of their vector
product is



2) ABsin 


3) AB



4) none of these





Given : c  a b
...


4) 21 units



2) 1

1) 0°
9
...


ur

2) 3 units

1) 0°
7
...


3) 1200

If two vectors A and B are enclosing an angle 300 and the magnitude of A is 3 units,
ur
what is the magnitude of B
...


4) 64 3N

Given that C  A  B , then the angle between C and B is
1) 00

4
...

2
...


2) iˆ  ˆj   kˆ

3) ˆj  kˆ  iˆ

4) kˆ  iˆ  ˆj







 

 i  j  k  × 2 i  3 j  k  =












1) 2 i  j  k
4
...


If P  iˆ  ˆj  2kˆ , Q  2iˆ  ˆj  kˆ then the angle enclosed by them is

6
...


2) j k  1
3) j  k  i
1) j  k  1
Choose the incorrect regarding, null vector
...




ur r







3) 300




r







4) 600


ur





r

r



What is the angle between ( A  B ) and ( A  B )?

9
...


magnitude K times of 6 i  2 j  3 k
...


2)

53

3)

4) 24

24












Area of the parallelogram formed by vectors 3 i  2 j  k and i  2 j  3 k is ____(given
vectors are diagonals)








(Note : Area of parallelogram formed by vectors A and B is A  B )
1) 3 8 units
12
...
units

3) 8 3 sq
...
The area of the parallelogram is nearly
1) 18 units
2) 4
...
0 units
IX Class - Physics

Page No : 80

4) 13units

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5
...
B  Axiˆ  Ay ˆj  Az kˆ
...
Bx iˆ  Ay ˆj
...
Bx iˆ  Ax iˆ
...
B y ˆj  Az kˆ
...
B kˆ    A ˆj
...
B kˆ 
x

z

y

z

z

z

 

 

 

 

 

 

 

 

 

 Ax
...
iˆ  Ay
...
iˆ  Az
...
iˆ  Ax
...
ˆj  Ay
...
ˆj  Az
...
ˆj 
Ax
...
kˆ  Ay
...
kˆ  Az
...
kˆ

 Ax Bx 1  Ay Bx  0   Az Bx  0   Ax B y  0   Ay B y 1  Az B y  0   Ax B y  0   Ay Bz  0   Az Bz 1

 Ax Bx  Ay By  Az Bz








A
...
B
A B









A
...
B


B






The scalar component of B in the direction of A is expressed as B cos  





A
...
If P and Q are




perpendicular, then P can not have component along Q and vice versa because




P
...


Dot product of

ur
ur
ur
ur
A and B is equal to C
...












1
4) cos 



2) 2

3) –4

4) –2


 

  
When a force  8 i + 4 j  N displaces a particle through  3 i - 3 j  m, the power is 0
...






The time of action of force is [Note : Power 
1) 10 s

14
...


 B 

 AC 

1
2) cos 

Work done
Work done
 time 
]
Time
Power

2) 20 s

3) 15 s
4) none of these
ur
ur
µ is moved from r  2i$  7$j  4k
µ to
A point of application of a force F  5i$  3$j  2k
1
uur
ur uur
µ
...
( Hint : Work done = W = F
...

1) 28 units

15
...


2) 22 units
3) 11 units
4) 0 units
uur r uuur
A force of 3i  j  2k N displaced the body from a point (4, –3, –5)m to a point





(–1, 4, 3)m
...


ur ur
( Hint : Work done = W = F
...


1) 12 J

3) 24 J

2) 14 J

4) 36 J

ur
ur
µ and B  4i$  2$j  4k
µ are
The vector A= $i  4$j  3k

1) perpendicular

2) parallel

3) inclined at 45°

4) inclined at 60°

WORK SHEET - 11
Single Answer Type
1
...


7
7
uuur
uuur
uuur
ˆ ˆ ˆ
ˆ ˆ ˆ
If OP=2i-3j+k,OQ=i+j+k,then
PQ=

3
...


7

3) 3iˆ  2 ˆj  7 kˆ

4) none

3

3) 7

4) 14

29

3)

The length of the vector 3iˆ  6 ˆj  2kˆ is
1) 3

5
...


r
r
r r
rr
0
If a =2, b =3, a,b =120 then a
...


1) 0
2) -3
r r r r
rr
If a +b = a -b then a
...

23
...
AC 
1) 15
2) 17
3) 0
The angle between the vectors 6iˆ  2 ˆj  kˆ, 2iˆ - 9 ˆj  6kˆ is
1) 900

2) 600

4) 10
4) None of these

3) 450

4) 300

Competitive Galaxy
1
...
How far are they from each other one end a half minutes before
their collision?
(SAT-RAMAIAH-2006]

2
...
He is at
a distance of ________ from the starting point
(SAT-RAMAIAH-1995]
A man is 3m to the north of a certain point, another is 4m to the east of the same
point
...
The distance between them now is
(SAT-RAMAIAH-1993]
One of the rectangular components of a velocity 50 m/s is 30 m/s
...
A
...
Rao Awards Council, Dec-2010)
1) 20 m/s
2) 15 m/s
3) 10 m/s
4) 40 m/s

3
...


5
...
Then he turned to the right and
walked 3ft
...
How far is he from A ?
(UNIFIED CYBER OLYMPIAD - 2011)
a) 4 ft
b) 5 ft
c) 24 ft
d) 27 ft

6
...


c) 1 m / s 2

d) 3 m / s 2

KEY & HINTS

WORKSHEET – 1 (KEY)
1)

4

2)

2

3)

1

4)

3

5)

6)

3

7)

4

8)

3

9)

1

10) 1

14) 3

15) 1

11) 1

12) 1

13) 2

16) 1

17) 3

18) 2

1)

First three options are correct
...


IX Class - Physics

Page No : 86

4

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15)

ˆ
(iˆ  2jˆ  k)



Displacement vector of the particle, S =

–

ˆ
(2iˆ  3jˆ  5k)

= iˆ  5 ˆj  4kˆ
16)

Length of 2iˆ  3 ˆj  4kˆ in XY - plane =

17)

Given vector may written as (0
...
3)2 + Z2 = 1

(2)2  (3)2 =

13

or 0
...
09 + Z2 = 1
or Z2 = 1 – 0
...
87

(2)2  (y)2  (3)2 = 5 or 4 + y2 + 9 = 25 or y2 = 12 or y =

18)
19)

12

Displacement vector of the particle,








S = (x2 – x1) i +(y2 – y1) j + (z2 – z1) k






= (5 – 1) i + (4 – 2) j + (2 – 3) k






=4i + 2 j – k
20)









The initial and final position of vectors may be written as r1 = i + 2 j – k








and r2 = 3 i + 2 j + 2 k

uuur

















5)

3





Displacement vector of PQ = r2 – r1 = (3 i + 2 j + 2 k ) – ( i + 2 j – k ) = 2 i + k

uuur

Magnitude of displacement vector, PQ =

(2)2  (3)2 =

13

WORKSHEET – 2 (KEY)

1)

1)

1

2)

3

3)

4

4)

1

6)

2

7)

2

8)

4

9)

2





If two like vector A and B are parallel












a)

C  A  B for like vector = A2 + B2 + 2AB

b)

C  A  B for unlike vector = A2 + B2 – 2AB


c)

 2

 2





C  A  B  2 A B Cos60 0 = A2 + B2 +

IX Class - Physics

Page No : 88

2AB
= A2 + B2 + AB
2

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11)

tan  =

Q Sin 
Q Sin 
=
Q + Q Cos  Q [1 + Cos ]

Sin 


= 2 Cos2   = Tan

2
2 



[Q Sin  =2 sin



cos ]
2
2

Tan  = Tan


2



 =


2

WORKSHEET – 3 (KEY)

1)

1)

2

2)

1

3)

4

6)

2

7)

2

8)

2

4)

1

5)

1

PQ 7

PQ 1
P  Q  7P  7Q

6P  8Q 
2)

R 2  P 2  Q 2  2PQ cos 

 39 

2

  20    25   2  20  25  cos 

cos  

2

2

62 1
;    600
125 2


3)

P 4

Q 3



Given, A  B = A2 + B2 + 2AB cos  and A = 1, B = 1




The required condition is A  B = 1

  = 1200 or

IX Class - Physics

 1 = 2 + 2 cos  = cos  =

2
3

Page No : 90

1
2

V e cto rs
2
2
 F1  F2 = 25

F12  F22  2F1F2 cos1200 =

13

2
2
 F1  F2  2F1F2 × 1/2 = 13

 F1F2 = 12
(F1 – F2)2 = F12 + F22 – 2F1F2
= 25 – 2 × 12 = 1
 F1 – F2 = 1  (1) and F1 + F2 = 7

 (2) and from (1) and (2) we get
F1 = 4N and F2 = 3N

WORKSHEET – 4 (KEY)

1
...


ur ur
A  B 3i$ 2$j  8k$ $i  3j$ 2k$



 



 3i$ 2j$ 8k$ $i  3j$ 2k$
 4i$ 5j$ 6k$
3
...



v

B

O

A



v

Change in velocity


 
 v    v   2 v = 2 × 10 = 20ms–1



IX Class - Physics

Page No : 92

5)

4

V e cto rs

9
...

uur
Initial velocity; v1 = 5 ms–1 along
east = 5 ˆi
uur
Final velocity, v 2  5 ms 1 along
north = 5 ˆj
N

uuur
 v1
uur

45° v 2
W

E

O

S

uur uur uur
Change in velocity, v  v 2  v1  5 ˆj  5 ˆi


uur
v 

52   5   5 2
2

uur
The direction of v is north - west
...

N
V2

W

V1 = 50
kmph

V
-V1 = 50
kmph
S

uur
v1 = 50 km h –1 due north
uur
v 2 = 50 km h–1 due west
...


IX Class - Physics

Page No :

94

V e cto rs

5
...


Let P cos  =

 sin  

3P
3
or cos  =
5
5

4
5
4P
5

Other component of force P sin  =
7
...


Let v the vector, vertical component

9
...






The given condition is Ab = Acos  = 0

 cos  = 0 or  = 90°
10
...


Given that, P cos  = 25 or P =



25
cos 60

25
 50N
1/2

Other component of force = Psin 
= 50 sin 60° = 50 ×
IX Class - Physics

3
= 25 3N
2
Page No : 96

V e cto rs

1
...
These two
forces are equal in magnitude and opposite in direction
...


The magnitude of P, Q and R are proportional to the lengths of AB, BC and CA
respectively
...
e
...


Whenever a triangle is formed to represent three forces keeping a body in equilibrium,
then the length of each side of the triangle to proportional to the sin of the angle opposite
to it
...


From triangle law of vectors, the resultant force acting on a particle is zero and hence
there is no change in its velocity
...


P = 10 Kg - wt,  = 60°
The body is in equilibrium at B under the action of three forces

(i)

weight of the body

(ii)

horizontal force and

(iii) tension ‘T’ in the string
...

0

60°

T

B

A

150°

120°

According to Lamis’ theorem
...


2F
T
3

3 P=

3 × 10 Kg – wt

Tension in the rope, T =

× 9
...


102  10 3



2

= 20 Kg – wt
...


sin150
1
 80  =40 kg - wt
sin 90
2

Horizontal force or applying force, (F) = mg tan = 2 × 10 × tan60 = 20 3 N
Tension in the string, T 






 20 3 

2

F 

2

  mg 

2

  2  10   40N
2

Horizontal force
F 20 3
 
Tension in t he string T
40

3
2

Alternative method

T
30°

T sin30°
F

T cos30°
2 kgwt

F = T Cos30° =

T 3
--(1)
2

2 = Tsin30°  T = 4 kg – wt --(2)
Substitute (2) in (1) we get, F = 2 3 kg – wt

IX Class - Physics

Page No : 100

V e cto rs

WORKSHEET – 8 (KEY)
1)

3

2)

2

3)

2

4)

1

5)

1

6)

4

7)

2

8)

3

9)

2

10) 1

11) 4
1
...


1
 117 units
2

A  B  AB sin 
 AB sin 600  14  12 

3
 84 3 N
2

3
...


24  A  B  sin 300  24  3  B 

1
2

 B  16 Units
5
...


Cross product of equal vectors is zero
...


According to definition of cross product
...


The result follows from the definition of cross product
...


A  B and B  A are oppositely directed
...


10
...






c  d is in the plane of c and d
...
So,







A B = (15)(8) sin 60° = 60 3 units
...


















So, A  B is perpendicular to A  B
...





radian
...










A  B is perpendicular to the plane of A and B
...


r ur

Torque,   r  F



r

Here r  5iˆ  2 ˆj  kˆ



 iˆ  ˆj  kˆ =  6iˆ  ˆj  0kˆ
ˆj kˆ
iˆ
  6 1 0
2 1 1

 iˆ  1  0   ˆj  6  0   kˆ  6  2 
 iˆ  6 ˆj  4kˆ
Magnitude of torque

IX Class - Physics

Page No : 104

V e cto rs



ˆi



ˆj

A  B 3

ˆ
k

1 2

2 2 4






 i  4  4   j 12  4   k  6  2






 8( i  j k)




AB 

1   1   1  8
2

2

2

2

8 3

   
8
 i  j k 



 Unit vector perpendicular to plane containing A and B  
8 3






i  j k

3

WORKSHEET – 10 (KEY)
1)

4

2)

1

3)

2

4)

1

5)

6)

1

7)

3

8)

2

9)

4

10) 3

14) 1

15) 3

11) 2

12) 3

13) 2

1

16) 1

SOLUTIONS
2
...
B  5iˆ  2 ˆj  3kˆ
...










P
...




tan  = 1   =450



A
...










P
...


The dot product of two equal vectors is equal to the square of the magnitude of either of

IX Class - Physics

Page No : 106

V e cto rs


14
...
6
6








 
  

 r  r2  r1   5 i  2 j  3 k    2 i  7 j  4 k 

 








 3 i  5 j k




 
 
  

 W  F
...
 3 i  5 j  k 

 


= 15 + 15 – 2 = 28 units
15
...
S



  
  

=  3 i  j  2k 
...


= 15 – 7 + 16 = 24 joule
...
B = i + 4k + 3k
...
B = 4 + 8 – 12 = 0
...


WORK SHEET -11 (KEY)

1
...






r
r
Required unit vector =
7
9  36  4
ab
2
...


If a,b,c are the sides of the triangle respectively then
r
a  2iˆ  3 ˆj  6kˆ  4  9  36  49  7

r
b  iˆ  2 ˆj  3kˆ  36  4  9  7,
r
c  3iˆ  6 ˆj  2kˆ  9  36  4 49  7
...

18
...

20
...

22
...


 Perimeter = 7+7+7=21
r
r r
c  5iˆ  9 ˆj  3 iˆ  4 ˆj  2iˆ  3 ˆj  3a  b
uuur
AB=iˆ  2 ˆj  2kˆ  D
...
's are 1,-2,2   D
...
's are 1/3,-2/3,2/3
rr
a
...

rr r r
r r
a
...
b  0  a
...

uuur uuur uuur
uuur uuur uuur
uuur uuur
AB  OB  OA  2iˆ  ˆj  5kˆ
...
AC  2  5  10  17
...
b  12  18  6  0  a  b  a, b  900



 



 

Competitive Galaxy
60
 0
...


Time taken t meet t 

2
...
5 km
Two given displacement form two sides of equilateral triangle, then closing side equal
to one of the side = 100 m
...

4
...
6
Let 50sin   30

sin   3 / 5

52  32  4
The other component is given by 50cos 

IX Class - Physics

Page No : 110



---