CHAPTER
1
Vectors
1.1 VECTOR ANALYSIS—A LANGUAGE OF PHYSICS
Physicists have developed their own language for the description of the physical world. The beauty of
their language lies in the fact that the physical laws, when expressed in that language are concise, elegant
and transparent. Vector is a language in which the laws of mechanics and electro-magnetism have simple
appearance.
There are many physical quantities, which require magnitude and direction both for their specification.
Such quantities are displacement, velocity, force, angular momentum, torque, etc. These are called vector
quantities. On the other hand those quantities, which require only magnitude for their specifications, are
called scalars. Mass, volume, density, work, energy, pressure etc. are such quantities. In order to
differentiate vector from scalar, the former is written in bold face letters or letters with arrow mark over
it. Ordinary letter denotes the magnitude of the vector. For example a represents vector whereas a or |a|
its magnitude.
Graphically, a directed line segment represents a vector. The length of the line represents the magnitude
of the vector and the arrow direction of the vector. Two vectors are said to be equal if they have the same
magnitude and direction. If two vectors a and b have the same magnitude but opposite direction, then
one is said to be negative of the other and we write a = – b.
a
b
b
–b
a
Fig. 1.1
1.2 ADDITION OF VECTORS
Let a and b be the two vectors which are to be added. Draw a directed line segment AB such that the
length of AB represents the magnitude of vector a and the direction of AB be parallel to a. Now draw
another directed segment BC equal to vector b such that the tail of vector BC lies at the head of vector
AB as shown in Fig. (1.2). The directed line segment AC represents the vector sum of a and b.
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Introduction to Mechanics
Thus
AC = AB + BC = a + b
…(1.1)
Fig. 1.2: Addition of vectors
We may also add the vector in reverse order. In this case we draw the directed line segment AD equal
to b and then the line segment DC equal to a. Again the vector sum of b and a is the line segment AC.
Thus
AC = AD + DC = b + a
…(1.2)
From Eqns. (1.1) and (1.2), we see that
a+b=b+a
…(1.3)
Thus vector addition is commutative.
Here we see that the diagonal of the parallelogram formed by vectors a and b as its adjacent sides
represents the sum or resultant of two vectors. For this reason we say that the law of vector addition
follows the parallelogram law of addition. The addition of all vector quantities follows this law. Conversely
we define vectors as those quantities, which have magnitude and direction both and whose law of addition
follows the parallelogram law of addition.
There are quantities, which are assigned magnitude and direction both, but they are not regarded as
vectors. Such quantities do not obey the parallelogram law of addition. Finite rotation of a rigid body is
assigned magnitude and direction both but it is not regarded as vector. However, infinitesimal rotations
are regarded as vectors.
1.3 REPRESENTATION OF FINITE ROTATION
Consider a finite rotation q of a disc about an axis passing through its center and perpendicular to its
plane, as shown in the Fig. (1.3). The rotation q is represented by vector OP whose length is proportional
P
O
q
q
O
Fig. 1.3
P
Vectors
3
to the angle q and the direction of rotation is given by right hand rule. To assign direction, curl the axis
of rotation by right hand fingers such that the fingers follow the sense of rotation and the thumb stretched
outward, then the thumb gives the direction of angle of rotation q. The angular velocity w of a rigid body
about an axis is a vector quantity whose direction is assigned along the axis of rotation and the sense is
given by right hand rule.
1.4 SUBTRACTION OF VECTORS
The subtraction of vector b from a is equivalent to the addition of – b to a. Draw vector AB equal to a
and BC equal to – b. The vector AC represents the difference a – b.
b
B
a
A
a
a–b
–b
C
Fig. 1.4: Subtraction of vectors
1.5 MULTIPLICATION OF A VECTOR BY A SCALAR
Consider a vector a and a scalar l (positive or negative). The product l a is another vector whose
magnitude is l times the magnitude of a and the direction is the same as that of a if l is positive, and is
opposite to that of a if l is negative.
1.6 UNIT VECTOR n
Let a be a vector and a or | a | its magnitude, then the unit vector in the direction of a is defined by
n=
a
a
or n =
Hence a = an
a
a
…(1.4)
1.7 COMPONENT OF A VECTOR ALONG SOME SPECIFIED DIRECTION
Let OP (= a) be a vector whose component along
OQ is to be determined. Let a be the angle between
OP and OQ and n the unit vector along OQ. The
projection of OP on OQ is OP cos a or a cos a. The
component of a parallel to n is (a cos a)n.
P
a
O
a
n
Q
Fig. 1.5
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Introduction to Mechanics
1.8 RECTANGULAR CO-ORDINATE SYSTEM AND CARTESIAN COMPONENTS
OF A VECTOR
Let OX, OY and OZ be three mutually perpendicular lines. Now take a right handed screw and set its
screw parallel to OZ. Now rotate the screw from OX to OY through 90º, then the direction in which the
screw advances is taken as the positive direction of Z-axis.
Similarly by setting the screw parallel to OX axis and rotating it from OY to OZ through 90º we may
find the positive direction of X-axis. The set of three mutually orthogonal lines OX, OY, and OZ. With
positive directions defined by right-handed screw rule, as described above, constitute what we call righthanded coordinate system. If the direction of any axis is reversed then the new set of axis constitute left
handed coordinate system. Fig. (1.6).
Z
R. H.
O
L. H.
O
Y
X
Y
X
Z
Fig. 1.6
Let i, j and k be unit vector along OX,OY and OZ respectively. These unit vectors are called
fundamental system of vectors.
Let OP = a be a vector whose components along the coordinate axes are to be determined. Complete
a rectangular parallelepiped with OP as diagonal and edges along the coordinate axes as shown in the
Fig. (1.7). Vectors OA, OB and OC are the components of vector OP (= a) in the directions of i, j and k
respectively. Let us denote the magnitudes of the components in the directions of i, j, k by a1, a2, a3.
Then
OA = a1 i,
OB = a2 j,
OC = a3 k
Z
C
P
a3
O
A
a1
®
a
a2
B
D
X
Fig. 1.7: Cartesian components of a vector
Y
Vectors
From the Fig. (1.7), we have
a = OP = OD + DP
= OA + AD + DP
= a1 i + a2 j + a3 k
Also
a2 = OP2 + DP2 = OA2+ AD2+ DP2 = a12 + a22 + a32
If the vector OP makes angles a, b, g with X, Y and Z-axes respectively then the ratios
cos a =
a1
, cos b =
a2
, cos g =
a3
+ +
+ +
+ a22 + a32
are called the direction cosines of the vector a. In terms of d.c we can express the vector a as
a = (a cos a)i + (a cos b)j + (a cos g)k
a12
a22
a32
a12
a22
a32
a12
5
…(1.5a)
…(1.5b)
…(1.6)
1.9 SCALAR PRODUCT OF TWO VECTORS
The scalar product of two vectors a and b is defined to be the scalar a b cos q,
b
where q is the angle between them. If q is acute, the scalar product is positive
and if q lies between p/2 and p, the scalar product is negative.
The scalar product is also denoted by a . b and is referred to as the ‘dot’ product.
We have
q
a
a . b = ab cos q
… (1.7)
Fig. 1.8
Since the scalar product of two vector is a scalar, therefore
a.b=b.a
…(1.8)
That is, the scalar product is commutative.
Self product: The scalar product of a vector with itself is called self-product.
a . a = a2
Scalar product of orthogonal vectors: If a and b are orthogonal vectors then their scalar product
vanishes.
a . b = ab cos(p/2) = 0
Unit vectors i, j and k: From the definition of scalar product we have
i.i=j.j=k.k=1
i . j = j . k = k. i = 0
…(1.9)
…(1.10)
Dot product of two vectors in component from: Let
and
a = a1i + a2j + a3k
b = b1i + b2j + b3k
…(1.11)
…(1.12)
Then using Eqns. (1.9) and (1.10), we obtain
a . b = a1b1+ a2b2 + a3b3
…(1.13)
Physical significance of scalar product: If a particle undergoes a displacement d under the influence
of a force F then F . d is defined to be the work done on the particle. Similarly if v (x, y, z, t) represents
the velocity of fluid and s an area in the fluid then v . s represents the volume of fluid crossing per sec.
through the area.
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Introduction to Mechanics
1.10 VECTOR PRODUCT OF TWO VECTORS
The vector product of two vectors a and b is defined to be another vector given by (ab sin q) n where n
is a unit vector orthogonal to both a and b and such that a, b and n form a right handed system, q is the
angle between the vectors. The vector product is denoted by a × b and hence it is also read as ‘cross’
product. Thus we have
a × b = (a b sin q)n
…(1.14)
Since b, a and –n form a right handed system, it follows that
b × a = (ab sin q)(–n) = –a × b
…(1.15)
a
×
b
=
(
ab
sin
q
)
n
Thus vector product is non-commutative.
If a and b are orthogonal, then
b
a × b = ab n
…(1.16)
n
If a product (q = 0) or anti-parallel (q = p), their
q
cross product vanishes.
a
The cross product of a vector with itself is zero
–n
a×a=0
The fundamental system of vectors i, j, k satisfy the
b×a
relations.
i×i=j×j=k×k=0
Fig. 1.9: Vector product of two vectors
i × j = –j × i = k
j × k = –k × j = i
k × j = –i × k = j
…(1.17)
1.11 VECTOR PRODUCT OF TWO VECTORS IN COMPONENT FORM
Let
a = a1i + a2j + a3k
b = b1i+ b2j + b3k
Then
a × b = (a1i + a2j + a3k) × (b1i + b2j + b3k)
= 0 + a1b2k – a1b3j – a2b1k + 0 + a2b3i + a3b1j – a3b2i + 0
= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k
i
j
k
= a1
b1
a2
a3
b2
b3
…(1.18)
1.12 VECTOR AREA
Area is regarded as a vector quantity. Its direction is specified as follows. Let S be plane area and n be
unit vector normal to one side of the area. The –n is unit vector normal to the other side. Now take a right
hand screw and set its screw parallel to n. Rotate the screw in the direction in which boundary of the area
is traversed. The direction in which the screw advances gives the direction of the area. (Fig. 1.12). Thus
Vectors
7
n
S = Sn
n
b
bsinq
q
a
Fig. 1.10
Geometrical Significance of a × b: The magnitude of area of parallelogram whose sides are the
vectors a and b is ab sin q, which is the magnitude of a × b. Thus the magnitude of the cross product
a × b represents the area of the parallelogram whose adjacent sides are the vectors a and b.
Let F be the force acting on a particle located at point r referred to origin 0.Then the cross product
r × F is defined to be the torque or the moment of the force about the origin 0. Similarly if p is the linear
momentum of a particle whose position vector is r, then r × p is defined to be the angular momentum of
the particle about origin.
L=r×p
t=r×F
O
F
p
r
O
r
(b)
(a)
Fig. 1.11: (a) Moment of force; (b) Angular momentum
F = q(v × B)
B
v
Fig. 1.12: Lorentz force
1.13 LORENTZ FORCE
When a charged particle carrying charge q is fired into a magnetic field of strength B with velocity v, it
is acted upon by a force F whose magnitude and direction is given by equation
F = q (v × B)
…(1.19)
This force is called the Lorentz force.
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Introduction to Mechanics
1.14 SCALAR TRIPLE PRODUCT
Let a, b and c three non-zero and non-coplanar vectors. A product of the form a . b × c is scalar and is
called scalar triple product. It is also written as [a b c].
If
a = a1i + a2j + a3k
b = b1i + b2j + b3k
c = c1i + c2j + c3k
Then
i
j
k
b × c = b1 b2
c1 c2
b3
c3
= (b2c3 – b3c2) i + (b3c1 – b1c3) j + (b1c2 – b2c1) k
a . b ¥ c = a1(b2c3 – b3c2) + a2(b3c1 – b1c3) + a3 (b1c2 – b2c1)
a1
a2
a3
= b1
c1
b2
b3
c2
c3
…(1.20)
…(1.21)
Similarly we can show that
b1
b . c × a = c1
a1
b2
c2
b3
c3
a2
a3
c1
c2
c3
c . a × b = a1
b1
a2
a3
b2
b3
…(1.22)
And
…(1.23)
Making use of the properties of determinants we can show that
a . b × c = b . c × a = c .a × b
Taking the first and last terms we have
a.b×c=c.a×b
or
a.b¥c=a¥b.c
The above relation shows that the dot and cross operations can be interchanged in scalar triple product.
If any vector in the scalar triple product repeats then scalar triple product vanishes. Thus
[a a c] = [b b c] = [a c c] = 0
The scalar triple product of unit vectors is unity.
[i j k] = i . j × k = 1
Vectors
9
Geometrical meaning of scalar triple product
Taking a, b, and c as co-terminus edges, construct a
parallelepiped as shown in the figure. Let n be unit
vector normal to the plane containing b and c.
a
Now
n
a . b ¥ c = a . n bc sin q
= (a . n) (bc sin q)
b
= (a cos a) (bc sin q)
a
= Volume of the parallelepiped
q
c
…(1.24)
Thus the scalar triple product a . b ¥ c equals
Fig. 1.13
the volume of the parallelepiped formed by vectors
a, b and c. If the three vectors a, b and c are coplanar, no solid figure can be formed from these vectors
and hence the volume of the parallelepiped is zero. Conversely, if the scalar triple product of three nonvanishing vectors is zero, then the vectors will be coplanar.
1.15 VECTOR TRIPLE PRODUCT
The vector product b ¥ c is orthogonal to b and c. Then a ¥ (b ¥ c) lies in the plane containing b and c.
Therefore a ¥ (b ¥ c) may be expressed as a linear combination of b and c. Hence
a × (b × c) = a b + b c
where a and b are unknown constants. Multiplying the above equation with a scalarly we have
a . a × (b × c) = a a . b + b a . c
Left hand side is zero by virtue of property of scalar triple product. Therefore
aa.b+ba.c=0
a
b
= g ( say )
or
a .c = a . b
Then
a = g (a . c) and b = –g (a . b)
Substituting the value of a and b in eqn. (1.15), we obtain
a ¥ (b ¥ c) = g (a . c) b – g (a . b) c
For special case a = c = j and b = i
we have
j ¥ (i ¥ j) = g (j . j)i – g (j . i)j
g=1
Hence
a ¥ (b ¥ c) = (a . c) b – (a . b) c
…(1.25)
1.16 SCALAR PRODUCT OF FOUR VECTORS: (a × b) . (c × d)
(a × b) . (c × d) = m . (c × d) where m = a × b
= (m × c) . d = {(a × b) × c} . d = – {c × (a × b)} . d = –[(c . b) a – (c . a) b] . d
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Introduction to Mechanics
= (c . a) (b . d) – (c . b) (a . d)
=
a .c c .b
a.d b .d
…(1.26)
1.17 VECTOR PRODUCT OF FOUR VECTORS (a × b) × (c × d)
(a × b) × (c × d) = m × (c × d), where m = a × b
= (m . d) c – (m . c) d
= (a × b . d) c – (a × b . c) d
= [a b d] c – [a b c] d
Putting c × d = m, we have
(a × b) × (c ×d) = (a × b) × m
= – m × (a × b)
= (m . a) b – (m . b) a
= (c × d . a) b – (c × d . b) a
= [c d a] b – [c d b] a
Equating the two expressions obtained in two ways we have
[a b d] c – [a b c] d = [c d a] b – [c d b] a
[c d b] a – [c d a] b + [a bd] c – [a b c] d = 0
The above equation can be written as
d=
[dbc] a + [dca] b + [dab]c
[abc]
…(1.27)
…(1.28)
This relation expresses the fact that any vector such as d can be written as linear combination of any
three non-coplanar vectors a, b and c.
1.18 RECIPROCAL SYSTEM OF VECTORS
If a, b and c are three non-coplanar vectors, then the vectors a¢¢, b¢¢ and c¢¢ defined by
b¥c
c¥a
a¥b
, b¢ =
, c¢ =
a¢¢ =
[a, b, c]
[a, b, c]
[a, b, c]
are called the reciprocal system of vectors.
Properties of reciprocal system of vectors
If a, b, c and a¢¢, b¢¢, c¢¢ are two sets of reciprocal vectors then
(1) a . a¢¢ = b . b¢¢ = c . c¢¢ = 1
(2) a . b¢¢ = a . c¢¢ = b . a¢¢ = b . c¢¢ = c . a¢¢ = c . b¢¢ = 0
(3) a × a¢¢ + b × b¢¢ + c × c¢¢ = 0
(4) [a, b, c] = 1/[a¢¢, b¢¢, c¢¢]
…(1.29)
Vectors
11
(5) If a¢¢ = (b × c)/[a, b, c], b¢¢ = (c × a)/[a, b, c] and
c¢¢ = (a × b)/[a, b, c] then
a = (b¢¢ × c¢¢)/[a¢¢, b¢¢, c¢¢], b = (c¢¢ × a¢¢)/[a¢¢, b¢¢, c¢¢] and
c = (a¢¢ × b¢¢)/[a¢¢, b¢¢, c¢¢]
Proof
a.b ¥ c
=1
[ a, b , c ]
b . b¢¢ = 1, c . c¢¢ = 1
a . a¢¢ =
(1)
Similarly
(2) Now
a . b¢¢ =
a .c ¥ a
=0
[a, b, c]
Similarly other relations can be proved.
(3)
\
a × a¢¢ =
a ¥ (b ¥ c) (a . c)b - (a . b)c
=
[a, b, c]
[a, b, c]
b × b¢¢ =
(b .a)c - (b .c)a
[a, b, c]
c × c¢¢ =
(c . b )a - (c .a)b
[a, b, c]
a ¥ a¢¢ + b ¥ b¢¢ + c ¥ c¢¢ = 0
b¥c c¥a
a¥b
(b ¥ c).(c ¥ a) ¥ (a ¥ b) [a, b, c]
¥
=
=
.
[a, b, c] [a, b, c] [a, b, c]
[a, b, c]3
[a, b, c]3
2
(4)
a¢¢. b¢¢ ¥ c¢¢ =
=
(5)
b¢¢ × c¢¢ =
\
a=
1
a
,
b
[ , c]
(c ¥ a) ¥ (a ¥ b )
[a, b, c]
2
=
[a, b, c] a = a = a a ¢, b ¢, c ¢
[
]
[a, b, c]2 [a, b, c]
b¢ ¥ c¢
a
[ ¢, b ¢, c ¢ ]
Similarly other relations can be proved.
SOLVED EXAMPLES
Example 1: If a = 2i + j – k, b = –3i – 2j + 2k, find (1) a . b (2) angle between the vectors (3) a × b
(4) Unit vector normal to the plane of a and b (5) magnitude of the area of the parallelogram formed by
vectors a and b.
Solution:
(1) a . b = (2) (–3) + (1) (–2) + (–1) (2) = –10
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Introduction to Mechanics
(2) cos q = (a . b)/a b = – (10/÷102)
i
j k
(3) a ¥ b = 2 1 -1 = - j - k
-3 -2 2
a¥b = 2
(4) Unit vector normal to the plane of a and b is
n=
a ¥ b -j - k
=
a¥b
2
Area of parallelogram a ¥ b = 2 units
Example 2: A particle is displaced from point (1, 1, 2) to point (2, 4, 5) under the influence of a constant
force i – 3j + 8k. Calculate the work done on the particle.
Solution:
Displacement
d = (2i + 4j +5k) – (i + j +2k)
= i + 3j + 3k
Work done = F . d = (i – 3j +8k) . (i +3j +3k) = 16 units
Example 3: A particle is located at point (3, 4, 5) and is acted upon by a force 4 i – j – k. Calculate the
torque exerted on the particle about the origin.
Solution:
Position vector of the particle r = 3i + 4j + 5k
i j k
Torque t = r × F = 3 4 5 = i + 23j - 19k.
4 -1 -1
Example 4: A particle at point r = i – j + k has linear momentum 2i + 5j – 3k. What is its angular
momentum about the origin?
Solution:
i j k
Angular momentum J = r ¥ p = 1 -1 1 = -3i + 5 j + 7k
2 5 -3
Example 5: If a = i + 2j + k, b = –2i + 3j + 4k, c = 3i + k. Find a . b ¥ c, a ¥ (b ¥ c) and (a ¥ b) ¥ c.
Solution:
1 2 1
a . b × c = -2 3 4 = 22
3 0 1
Vectors
13
a × (b × c) = (a . c) b – (a . b) c
=4b–8c
= –32 i + 12j + 8k
(a × b) × c = – c × (a × b)
= (c . a) b – (c . b) a
= 4b + 2a
= – 6i + 16j + 18k.
Example 6: Show that
a × (b × c) + b × (c × a) + c × (a × b) = 0.
Solution:
a × (b × c) = (a . c) b – (a . b) c
b × (c × a) = (b . a) c – (b . c) a
c × (a × b) = (c . b) a – (c . a) b
Adding we get the desired result.
Example 7: Prove that [(a + b), (b + c), (c + a)] = 2 [a, b, c].
Solution:
(a + b) . (b + c) × (c + a) = (a + b) . (b × c + b × a + c × a)
=a.b×c+b.c×a
= 2 a . b × c = 2 [ a, b, c].
Example 8: Prove that (a × b).(b × c) × (c × a) = [a,b,c]2
Solution:
(b × c) × (c × a) = m × (c × a) where m = b × c
= (m . a) c – (m . c)a
= (b × c . a) c – (b× c . c)a
= [b, c, a]c – 0
= [a, b, c]c
\
(a × b).(b × c) × (c × a) = (a × b).c[a, b, c]
= [a, b, c] [a, b, c] = [a, b, c]2.
Example 9: Show that the vectors a = 2i + j – k, b = –3i – 2j + 2k, c = i + 2j – 2k are coplanar.
Solution: Scalar triple product of vectors
2 1 -1
a . b × c = -3 -2 2 = 0
-1 2 -2
Hence the given vectors are coplanar.
PROBLEMS
1. If a = 2i – j + k, b = i + 2j – 3k, c = – 3i + 2j + k, find a . b, b . c, c . a,(a + b) . (a – b), (b + c) .
(b – c), (a + 2b – 3c) . (a –3b + 2c)
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Introduction to Mechanics
2. Find the angle between vectors a = 5i – 4j + 2k and 4i + 3j + 2÷5k.
3. For what value of a the vectors 5i + 6j – k and 2i + (3 + a)j – ak are orthogonal.
4. A particle moves from point (1,2,3) to point (3,2,5) under the action of forces 2i + j + 3k and
3i – 4j – 2k. Find the work done by the force.
5. If a = 2i +3j – 4k, b = i +2j + 3k, find a ¥ b, b ¥ a, (a + b) ¥ (a – b), unit vector normal to the plane
of a and b, area of the parallelogram whose adjacent sides are a and b.
6. The position vector of a particle is 4i – 3j + k, It is acted upon by a force 2i – 4j + 8k. Find the
moment of the force about origin.
7. A particle of mass 2 kg is moving with velocity 5i + 4j – 2k m/s. What is its angular momentum
about origin at the instant its position vector is (i + j + k) m?
8. Find the volume of the parallelepiped whose adjacent sides are 3i – j – k, – i + j + k, 2i +2j + 3k.
9. For what value a the vectors i + 2j – 3k, -aj + k, i + k are coplanar?
10. Show that
i × (a × i) + j × (a × j) + k × (a × k) = 2a.
11. Obtain a set of vectors reciprocal to the three vectors –i + j + k, i – j + k, and i + j – k
12. Show that i, j, k are self-reciprocal system of vectors.
VECTOR CALCULUS
1.19 DIFFERENTIATION OF VECTORS
If to each value of some scalar t there corresponds a vector A then A is said to be vector function of scalar
variable t and is denoted by A (t). Let A (t) and A (t + dt) represent the vector function corresponding to
the scalar variables t and t + dt respectively. Let A (t + d t) – A (t) = dA.
The limit of the quotient
lim
dt Æ 0
dA
A(t + dt ) - A(t )
= lim
dt Æ 0 dt
dt
provided it exists, is called the derivative of vector A with respect to t and is denoted by, dA/dt. The
derivative of a vector is also a vector.
dA = A(t + dt) – A(t)
A(t + dt)
A(t)
O
Fig. 1.14
Thus
dA
dA
A(t + dt ) - A(t )
= lim
= lim
dt Æ 0 d t
dt
dt dt Æ0
(1.30)
Vectors
15
If A1(t), A2(t) and A3(t) are the components of vector A then
dA
dA1
dA
dA
i + 2 j+ 3 k
=
dt
dt
dt
dt
…(1.31)
A(t) = A1(t) i +A2(t) j +A3(t) k
Time derivative of position r(t) of a moving particle
Let r(t) and r(t + dt) be the position vectors of the particle at time t and t + dt. The displacement dr of the
particle is given by
dr = r (t + dt) – r(t)
The limit of the quotient dr/dt as dt tends to zero is called the velocity v of the particle
v = lim dt Æ0
and is denoted by
v=
dr
dt
dr
dx
dy
dz
= i + j +k
dt
dt
dt
dt
…(1.32)
The acceleration of the particle is
a=
dv
r
= dt
…(1.33)
1.20 RULES OF VECTOR DIFFERENTIATION
In the following A and B are vector functions of scalar variable t, c is a constant vector, and j is a scalar
function.
dc
=0
dt
…(1.34)
dkA
dA
= k
dt
dt
…(1.35)
d (A ± B)
dA dB
±
=
dt
dt
dt
…(1.36)
d jA
dA
dj
+A
= j
dt
dt
dt
…(1.37)
dB dA
d
( A ◊ B) = A ◊
+
◊B
dt
dt
dt
…(1.38)
dB dA
d
( A ¥ B) = A ¥
+
¥B
dt
dt
dt
…(1.39)