Vector analysis

Basic concept of vector

1. VECTOR
ANALYSIS
Introduction Of Vector
Analysis

2. VECTOR
A vector is a quantity which has both magnitude
and direction.
We will denote vectors by bold-faced letters or
letters with an arrow over them. Thus the
vectors
may be denoted by A or as in figure. The
magnitude or length of the vector is then denoted
by


PQ A
AAAPQ or ,,,

3. VECTOR
we introduce the concept of a vector as a directed
line segment from one point P to another Q.
Here P is called the initial point or origin of ,
and Q is called the terminal point , end, or
terminus of the vector.
Q
Q
P P
PQ
PQ
B
A

4. SCALARS
A scalar is a quantity which has only magnitude
but no direction.
Such as mass, length and temperature.

5. VECTOR ALGEBRA:
There are two basic operations with vectors:
a)Vector addition and b) Scalar multiplication.
Vector addition:
Consider vectors , pictured in figure- 2(a).
The sum or resultant of , is a vector
formed by placing the initial point of on the
terminal point of and then joining the initial
point of to the terminal point of , pictured in
fig-2(b). The sum is written by
Fig-2(a) fig- 2(b)
BA

and
BA

and
C

B

A

A

B

BAC

+=
A
 B

A

B

BAC

+=

6. SCALAR MULTIPLICATION
Let be any vector and m be any given scalar,m
denotes the m times of the vector in the
direction of .
The length or modulus is given by,

a

a

a

a

am

maamam ==


7. LAWS OF VECTOR ALGEBRA:
ion]multipicat[Unit ,)(1 ][M
law] ve[Associati ,)()(]M[
]law veDistributi[ ,)( ][M
law] ive[Distribut ,)( ][M
addition]for law ve[Commutati , ][A
negative] of [Existence ,0)()(
such that, vector a exits there, every For ][A
element] zero of [Existence,00
or every vectfor such that, 0 vector zero aexist There ][A
addition]for law ve[Associati ),()( ][A
:hold laws following Then the scalars. aren and m and vectorsare ,, Suppose
4
3
2
1
4
3
2
1
AA
AmnnAm
AnAmAnm
BmAmBAm
ABBA
AAAA
AA
AAA
A
CBACBA
CBA











=
=
+=+
+=+
+=+
=+-=-+
-
=+=+
++=++

8. UNIT VECTOR
Unit vectors are vectors having unit length.
Suppose is any vector with length

Null Vector
A null vector is a vector whose module or
magnitude is zero and is denoted by 0
A

0>A

r.unit vecto a is
A
A
a


=
vector.zero a as define we,0 aa

=

9. EQUAL VECTOR
Two vectors are equal if they have the same
magnitude and direction.
Collinear Vector
Vectors which have the same support are said to be
collinear and the line is known as the line of
vectors.
Coplanar Vector
A system of vectors is said to be coplanar if their
supports are parallel to the same plane.

10. MULTIPLE PRODUCT OF VECTORS
(a)Dot or Scalar Product
The dot or scalar product of two vectors
denoted by and is defined as the product of
the magnitudes of and the cosine of the
angle between them. In symbols,

BA

and
BA

.
BA

and
q
pqq ££= 0,cos. BABA


11. PROPOSITION
0... and 1...)(
).().().().( )(
law] ive[Distribut , ..).( )(
product]dot for law ve[Commutati , .. )(
:hold laws following
Then the scalar. a is and vectorsare ,, Suppose
======
===
+=+
=
ikkjjikkjjiiiv
mBABmABAmBAmiii
CABACBAii
ABBAi
mCBA






12. CROSS PRODUCT
The cross product of vectors is a vector as
follows:
The magnitude of is equal to the product
of the magnitudes of and the sine of the
angle between them.
The direction of is perpendicular to the
plane of . So that form a right
handed system. In symbols,

BA

and
BAC

´=
BA

and
q
BAC

´=
BA

and CBA

,,
BAn
nBABA


´
££=´
ofdirection theindicatingr unit vecto a is ˆ where
0,ˆsin pqq

13. PROPOSITION
jikikjkji
kkjjiiiv
mBABmABAmBAmiii
CABACBAii
ABBAi
mC B,A






=´=´=´
=´=´=´
´=´=´=´
´+´=+´
´-=´
;;
0)(
)()()()()(
law] ive[Distribut )()()()(
law] ve[Commutati , )()(
Then, scalar. a is and vectorsare and Suppose

14. TRIPLE PRODUCT
Dot and cross multiplication of three vectors
may produce meaningful products, called triple
products of the form
Proposition
CBA

,,
)( and ).(,).( CBACBACBA

´´´
ACBBCACBA
CBABCACBAii
BACACBCBAi



).().()(
).().()()(
)()().()(
-=´´
-=´´
´=´=´

15. PROBLEM-1
21
5
6.7
.)442)(623(
cos (1) From
6)4(42and
7623 Here,
)1(
.
coscos.
Then, em.between th angle thebe let :
.442 and 623
vectorsebetween th angle theFind
222
222
-=
-+++
=Þ
=-++=
=++=
=Þ=
-+=++=
kjikji
b
a
ab
ba
abba
Sol
kjibkjia
q
qq
q


16. PROBLEM-2
.
2914
59
2914
405
sin
405)10(716
10716)64()29()124(
243
321
292)4(3 and 14321 Here,
sinˆsin
Then em.between th angle thebe Let :Sol
.243 and 32
vectorsebetween th angle theof sine theFind
222
222222
==
=-++=´
-+=--+-++=
-
=´
=+-+==++=
´
=Þ=´
+-++
q
qq
q
ba
kjikji
kji
ba
ba
ab
ba
nabba
kjikji

17. PROBLEM-3
. lar toperpendicu is So . 0. since
06612)34).(623(.
vectors, twoofproduct dot takeusLet :Sol
other.each lar toperpendicu are 34
and 623 vectors that theShow
baba
kjikjiba
kjib
kjia
=
=--=+--+=
+-=
-+=

18. PROBLEM-4
c
C
b
B
a
A
b
B
a
A
c
C
BcaAbcCab
BcaAbcCabei
cbaccbac
cbbacbab
acbacbacbaa
cba
cba
ABCcba
c
C
b
B
sinsinsin
sinsinsin
sinsinsin
)sin()sin()sin(..
0)(
0)(
0)(0)(
then, (1) with seperately
,, ofproduct cross theTake
)1(0Then
. of sides therepresent ,,Let :Sol
.
sinsin
a
sinA
leany triang that Prove
==Þ
==Þ
==Þ
-=-=-
´=´Þ=++´
´=´Þ=++´
´=´Þ=+´Þ=++´
=++
D
==
ppp

a
A
C
B

19. PROBLEM-5
2
2
][
]][[)]([
}0]){[(
}])).(()).).{(([(
)}]()).{([(
],,[L.S. :Sol
][],,[ that Prove
cba
cbacbacbacba
ccbaba
accbcacbba
accbba
accbba
cbaaccbba
=
=´=
-´=
´-´´=
´´´´=
´´´=
=´´´

20. PROBLEM-6
]2[][0000][
).().().(
).().().(
)0).((
)).((
)}()().{(
)}()).{((
][L.S. :Sol
]2[][ that Prove
c,b,ac,b,ac,b,a
acbabbcbb
acaabacba
acabcbba
acccabcbba
accacbba
accbba
ac,cb,ba
c,b,aac,cb,ba
=+++++=
´+´+´+
´+´+´=
´++´+´+=
´+´+´+´+=
+´++´+=
+´++=
+++=
=+++