MATHEMATICS
424
In most sciences one generation tears down what another has built and what
one has established another undoes. In Mathematics alone each generation
builds a new story to the old structure. – HERMAN HANKEL
10.1 Introduction
In our day to day life, we come across many queries such
as – What is your height? How should a football player hit
the ball to give a pass to another player of his team? Observe
that a possible answer to the first query may be 1.6 meters,
a quantity that involves only one value (magnitude) which
is a real number. Such quantities are called scalars.
However, an answer to the second query is a quantity (called
force) which involves muscular strength (magnitude) and
direction (in which another player is positioned). Such
quantities are called vectors. In mathematics, physics and
engineering, we frequently come across with both types of
quantities, namely, scalar quantities such as length, mass,
time, distance, speed, area, volume, temperature, work,
money, voltage, density, resistance etc. and vector quantities like displacement, velocity,
acceleration, force, weight, momentum, electric field intensity etc.
In this chapter, we will study some of the basic concepts about vectors, various
operations on vectors, and their algebraic and geometric properties. These two type of
properties, when considered together give a full realisation to the concept of vectors,
and lead to their vital applicability in various areas as mentioned above.
10.2 Some Basic Concepts
Let ‘l’ be any straight line in plane or three dimensional space. This line can be given
two directions by means of arrowheads. A line with one of these directions prescribed
is called a directed line (Fig 10.1 (i), (ii)).
Chapter
10
VECTOR ALGEBRA
W.R. Hamilton
(1805-1865)
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VECTOR ALGEBRA 425
Now observe that if we restrict the line l to the line segment AB, then a magnitude
is prescribed on the line l with one of the two directions, so that we obtain a directed
line segment (Fig 10.1(iii)). Thus, a directed line segment has magnitude as well as
direction.
Definition 1 A quantity that has magnitude as well as direction is called a vector.
Notice that a directed line segment is a vector (Fig 10.1(iii)), denoted as
or
simply as
, and read as ‘vector ’ or ‘vector ’.
The point A from where the vector
starts is called its initial point, and the
point B where it ends is called its terminal point. The distance between initial and
terminal points of a vector is called the magnitude (or length) of the vector, denoted as
|
|, or | |, or a. The arrow indicates the direction of the vector.
Note Since the length is never negative, the notation | | < 0 has no meaning.
Position Vector
From Class XI, recall the three dimensional right handed rectangular coordinate
system (Fig 10.2(i)). Consider a point P in space, having coordinates (x, y, z) with
respect to the origin O(0, 0, 0). Then, the vector
having O and P as its initial and
terminal points, respectively, is called the position vector of the point P with respect
to O. Using distance formula (from Class XI), the magnitude of
(or ) is given by
|
|=
2 22
xyz
++
In practice, the position vectors of points A, B, C, etc., with respect to the origin O
are denoted by
, , , etc., respectively (Fig 10.2 (ii)).
Fig 10.1
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426
A
O
P
90°
X
Y
Z
X
A
O
B
P( )x,y,z
C
P( )x,y,z
r
x
y
z
Direction Cosines
Consider the position vector
of a point P(x, y, z) as in Fig 10.3. The angles α ,
β, γ made by the vector
with the positive directions of x, y and z-axes respectively,
are called its direction angles. The cosine values of these angles, i.e., cos α, cosβ and
cos γ are called direction cosines of the vector
, and usually denoted by l, m and n,
respectively.
Fig 10.3
From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we
have
. Similarly, from the right angled triangles OBP and
OCP, we may write
cos and cos
y z
r r
β= γ=
. Thus, the coordinates of the point P may
also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction
cosines are called as direction ratios of vector
, and denoted as a, b and c, respectively.
Fig 10.2
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Note One may note that l
2
+ m
2
+ n
2
= 1 but a
2
+ b
2
+ c
2
≠ 1, in general.
10.3 Types of Vectors
Zero Vector A vector whose initial and terminal points coincide, is called a zero vector
(or null vector), and denoted as
. Zero vector can not be assigned a definite direction
as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any
direction. The vectors
represent the zero vector,
Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The
unit vector in the direction of a given vector
is denoted by
ˆ
a
.
Coinitial Vectors Two or more vectors having the same initial point are called coinitial
vectors.
Collinear Vectors Two or more vectors are said to be collinear if they are parallel to
the same line, irrespective of their magnitudes and directions.
Equal Vectors Two vectors
are said to be equal, if they have the same
magnitude and direction regardless of the positions of their initial points, and written
as
.
Negative of a Vector A vector whose magnitude is the same as that of a given vector
(say, ), but direction is opposite to that of it, is called negative of the given vector.
For example, vector
is negative of the vector , and written as = – .
Remark The vectors defined above are such that any of them may be subject to its
parallel displacement without changing its magnitude and direction. Such vectors are
called free vectors. Throughout this chapter, we will be dealing with free vectors only.
Example 1 Represent graphically a displacement
of 40 km, 30° west of south.
Solution The vector
represents the required
displacement (Fig 10.4).
Example 2 Classify the following measures as
scalars and vectors.
(i) 5 seconds
(ii) 1000 cm
3
Fig 10.4
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428
Fig 10.5
(iii) 10 Newton (iv) 30 km/hr (v) 10 g/cm
3
(vi) 20 m/s towards north
Solution
(i) Time-scalar (ii) Volume-scalar (iii) Force-vector
(iv) Speed-scalar (v) Density-scalar (vi) Velocity-vector
Example 3 In Fig 10.5, which of the vectors are:
(i) Collinear (ii) Equal (iii) Coinitial
Solution
(i) Collinear vectors :
.
(ii) Equal vectors :
(iii) Coinitial vectors :
EXERCISE 10.1
1. Represent graphically a displacement of 40 km, 30° east of north.
2. Classify the following measures as scalars and vectors.
(i) 10 kg (ii) 2 meters north-west (iii) 40°
(iv) 40 watt (v) 10
–19
coulomb (vi) 20 m/s
2
3. Classify the following as scalar and vector quantities.
(i) time period (ii) distance
(iii) force
(iv) velocity (v) work done
4. In Fig 10.6 (a square), identify the following vectors.
(i) Coinitial (ii) Equal
(iii) Collinear but not equal
5. Answer the following as true or false.
(i)
and – are collinear.
(ii) Two collinear vectors are always equal in
magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
Fig 10.6
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VECTOR ALGEBRA 429
10.4 Addition of Vectors
A vector simply means the displacement from a point
A to the point B. Now consider a situation that a girl
moves from A to B and then from B to C
(Fig 10.7). The net displacement made by the girl from
point A to the point C, is given by the vector
and
expressed as
=
This is known as the triangle law of vector addition.
In general, if we have two vectors
and (Fig 10.8 (i)), then to add them, they are
positioned so that the initial point of one coincides with the terminal point of the other
(Fig 10.8(ii)).
Fig 10.7
a
b
a
b
(i) (iii)
A
C
a
b
(ii)
a
b
+
A
C
B
B
a
b
–
–b
C’
Fig 10.8
For example, in Fig 10.8 (ii), we have shifted vector without changing its magnitude
and direction, so that it’s initial point coincides with the terminal point of
. Then, the
vector
+ , represented by the third side AC of the triangle ABC, gives us the sum (or
resultant) of the vectors and i.e., in triangle ABC (Fig 10.8 (ii)), we have
=
Now again, since , from the above equation, we have
This means that when the sides of a triangle are taken in order, it leads to zero
resultant as the initial and terminal points get coincided (Fig 10.8(iii)).
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Now, construct a vector so that its magnitude is same as the vector
BC
, but
the direction opposite to that of it (Fig 10.8 (iii)), i.e.,
BC
′
=
BC
−
Then, on applying triangle law from the Fig 10.8 (iii), we have
AC AB BC
′′
=+
=
AB ( BC)
+−
ab
=−
The vector
AC
′
is said to represent the difference of
and
ab
.
Now, consider a boat in a river going from one bank of the river to the other in a
direction perpendicular to the flow of the river. Then, it is acted upon by two velocity
vectors–one is the velocity imparted to the boat by its engine and other one is the
velocity of the flow of river water. Under the simultaneous influence of these two
velocities, the boat in actual starts travelling with a different velocity. To have a precise
idea about the effective speed and direction
(i.e., the resultant velocity) of the boat, we have
the following law of vector addition.
If we have two vectors
and
ab
represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10.9), then their
sum
+
ab
is represented in magnitude and
direction by the diagonal of the parallelogram
through their common point. This is known as
the parallelogram law of vector addition.
Note From Fig 10.9, using the triangle law, one may note that
OA AC
+
=
OC
or
OA OB
+
=
OC
(since
AC OB
=
)
which is parallelogram law. Thus, we may say that the two laws of vector
addition are equivalent to each other.
Properties of vector addition
Property 1 For any two vectors
and
ab
,
ab
+
=
ba
+
(Commutative property)
Fig 10.9
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VECTOR ALGEBRA 431
Proof Consider the parallelogram ABCD
(Fig 10.10). Let
then using
the triangle law, from triangle ABC, we have
Now, since the opposite sides of a
parallelogram are equal and parallel, from
Fig 10.10, we have,
and
. Again using triangle law, from
triangle ADC, we have
Hence =
Property 2 For any three vectors
, and
ab c
= (Associative property)
Proof Let the vectors
be represented by , respectively, as
shown in Fig 10.11(i) and (ii).
Fig 10.11
Then
=
and =
So =
Fig 10.10
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432
a
a
1
2
1
2
a
–2
a
a
2
and =
Hence =
Remark The associative property of vector addition enables us to write the sum of
three vectors
without using brackets.
Note that for any vector
a
, we have
=
Here, the zero vector is called the additive identity for the vector addition.
10.5 Multiplication of a Vector by a Scalar
Let be a given vector and λ a scalar. Then the product of the vector by the scalar
λ, denoted as λ
, is called the multiplication of vector by the scalar λ. Note that, λ
is also a vector, collinear to the vector . The vector λ has the direction same (or
opposite) to that of vector according as the value of λ is positive (or negative). Also,
the magnitude of vector λ
is | λ| times the magnitude of the vector , i.e.,
|λ
| = | λ || |
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10.12.
Fig 10.12
When λ = – 1, then λ = – , which is a vector having magnitude equal to the
magnitude of
and direction opposite to that of the direction of . The vector – is
called the negative (or additive inverse) of vector
and we always have
+ (– ) = (– ) + =
Also, if
1
=
||
a
λ
, provided ≠ 0 i.e. is not a null vector, then
| λ
| =|λ|| | =
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VECTOR ALGEBRA 433
So, λ represents the unit vector in the direction of . We write it as
ˆ
a
=
1
||
a
a
Note For any scalar k,
0 = 0.
k
10.5.1 Components of a vector
Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and
z-axis, respectively. Then, clearly
|
| = 1, | | = 1 and | |= 1
The vectors
, each having magnitude 1,
are called unit vectors along the axes OX, OY and OZ,
respectively, and denoted by
ˆ
ˆˆ
, and
ij k
, respectively
(Fig 10.13).
Now, consider the position vector
of a point P (x, y, z)
as in Fig 10.14. Let P
1
be the foot of the perpendicular from P on the plane XOY.
We, thus, see that P
1
P is parallel to z-axis. As
ˆ
ˆˆ
, and
ij k
are the unit vectors along the
x, y and z-axes, respectively, and by the definition of the coordinates of P, we have
. Similarly, and .
Fig 10.13
Fig 10.14
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434
Therefore, it follows that =
and =
Hence, the position vector of P with reference to O is given by
=
This form of any vector is called its component form. Here, x, y and z are called
as the scalar components of
, and
ˆ
ˆˆ
, and
xi yj zk
are called the vector components
of
r
along the respective axes. Sometimes x, y and z are also termed as rectangular
components.
The length of any vector
=
ˆ
ˆˆ
r xi yj zk
=++
, is readily determined by applying the
Pythagoras theorem twice. We note that in the right angle triangle OQP
1
(Fig 10.14)
= ,
and in the right angle triangle OP
1
P, we have
OP
=
Hence, the length of any vector =
ˆ
ˆˆ
+
r xi yj zk
=+
is given by
|
| =
If are any two vectors given in the component form
123
ˆ
ˆˆ
+
ai a j ak
+
and
12 3
ˆ
ˆˆ
bi b j bk
++
, respectively, then
(i) the sum (or resultant) of the vectors
is given by
=
11 2 2 33
ˆ
ˆˆ
( )( ) ( )
a bi a b j a bk
+ ++ ++
(ii) the difference of the vector
and
ab
is given by
=
11 2 2 3 3
ˆ
ˆˆ
( )( ) ( )
a bi a b j a bk
− +− +−
(iii) the vectors are equal if and only if
a
1
= b
1
, a
2
= b
2
and a
3
= b
3
(iv) the multiplication of vector
a
by any scalar λ is given by
λ =
12 3
ˆ
ˆˆ
()( )( )
ai a j ak
λ +λ +λ
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The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let be any two vectors, and k and m be any scalars. Then
(i)
(ii)
(iii)
Remarks
(i) One may observe that whatever be the value of λ, the vector λ
is always
collinear to the vector . In fact, two vectors are collinear if and only if
there exists a nonzero scalar λ such that
. If the vectors are given
in the component form, i.e.
=
12 3
ˆ
ˆˆ
a ai a j ak
=+ +
and , then the
two vectors are collinear if and only if
12 3
ˆ
ˆˆ
bi b j bk
++
=
12 3
ˆ
ˆˆ
( )
ai a j ak
λ++
⇔
12 3
ˆ
ˆˆ
bi b j bk
++
=
12 3
ˆ
ˆˆ
()( ) ( )
ai a j ak
λ +λ +λ
⇔
11
ba
=λ
,
2 23 3
,
b ab a
=λ =λ
⇔
1
1
b
a
=
3
2
23
b
b
aa
= =λ
(ii) If =
12 3
ˆ
ˆˆ
a ai a j ak
=+ +
, then a
1
, a
2
, a
3
are also called direction ratios of .
(iii) In case if it is given that l, m, n are direction cosines of a vector, then
ˆ
ˆˆ
li mj nk
++
=
ˆ
ˆˆ
(cos ) (cos ) (cos )
i jk
α+ β+ γ
is the unit vector in the direction of that vector,
where α, β and γ are the angles which the vector makes with x, y and z axes
respectively.
Example 4 Find the values of x, y and z so that the vectors
and
are equal.
Solution Note that two vectors are equal if and only if their corresponding components
are equal. Thus, the given vectors
will be equal if and only if
x = 2, y = 2, z = 1
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Example 5 Let
and . Is ? Are the vectors equal?
Solution We have
and
So, . But, the two vectors are not equal since their corresponding components
are distinct.
Example 6 Find unit vector in the direction of vector
Solution The unit vector in the direction of a vector is given by .
Now
=
2 22
2 3 1 14
++=
Therefore
1
ˆ
ˆˆ
ˆ
(2 3 )
14
a i jk
= ++
=
23 1
ˆ
ˆˆ
14 14 14
i jk
++
Example 7 Find a vector in the direction of vector that has magnitude
7 units.
Solution The unit vector in the direction of the given vector
is
=
1 12
ˆˆ ˆ ˆ
( 2)
5 55
ij i j
−= −
Therefore, the vector having magnitude equal to 7 and in the direction of
a
is
7
a
∧
=
12
7
55
ij
∧∧
−
=
7 14
ˆˆ
55
ij
−
Example 8 Find the unit vector in the direction of the sum of the vectors,
and .
Solution The sum of the given vectors is
and =
22 2
43(2) 29
+ +− =
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VECTOR ALGEBRA 437
Thus, the required unit vector is
1
43 2
ˆ ˆ
ˆˆ ˆˆ
(4 3 2 )
29 29 29 29
i jk i j k
= +− = + −
Example 9 Write the direction ratio’s of the vector and hence calculate
its direction cosines.
Solution Note that the direction ratio’s a, b, c of a vector
are just
the respective components x, y and z of the vector. So, for the given vector, we have
a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given
vector, then
Thus, the direction cosines are
11 2
, ,–
66 6
.
10.5.2 Vector joining two points
If P
1
(x
1
, y
1
, z
1
) and P
2
(x
2
, y
2
, z
2
) are any two
points, then the vector joining P
1
and P
2
is the
vector
(Fig 10.15).
Joining the points P
1
and P
2
with the origin
O, and applying triangle law, from the triangle
OP
1
P
2
, we have
=
Using the properties of vector addition, the
above equation becomes
=
i.e. =
2 2 2 111
ˆ ˆ
ˆ ˆ ˆˆ
( )( )
xiyjzk xiyjzk
++ −++
=
21 2 1 21
ˆ
ˆˆ
( )( ) ( )
x xi y y j z zk− +− +−
The magnitude of vector is given by
|
| =
2 2 2
21 2 1 21
( )( )( )
xx yy zz
− +− +−
Fig 10.15
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438
Example 10 Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed
from P to Q.
Solution Since the vector is to be directed from P to Q, clearly P is the initial point
and Q is the terminal point. So, the required vector joining P and Q is the vector
PQ
,
given by
=
ˆ
ˆˆ
( 1 2) ( 2 3) ( 4 0)
ijk
−− +−− +− −
i.e. =
ˆ
ˆˆ
3 5 4.
i jk
−− −
10.5.3 Section formula
Let P and Q be two points represented by the position vectors
, respectively,
with respect to the origin O. Then the line segment
joining the points P and Q may be divided by a third
point, say R, in two ways – internally (Fig 10.16)
and externally (Fig 10.17). Here, we intend to find
the position vector
for the point R with respect
to the origin O. We take the two cases one by one.
Case I When R divides PQ internally (Fig 10.16).
If R divides
such that = ,
where m and n are positive scalars, we say that the point R divides
internally in the
ratio of m : n. Now from triangles ORQ and OPR, we have
=
and = ,
Therefore, we have
= (Why?)
or
= (on simplification)
Hence, the position vector of the point R which divides P and Q internally in the
ratio of m : n is given by
=
Fig 10.16
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VECTOR ALGEBRA 439
Case II When R divides PQ externally (Fig 10.17).
We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
m : n
PR
i.e.
QR
=
m
n
is given by
=
Remark If R is the midpoint of PQ , then m = n. And therefore, from Case I, the
midpoint R of
PQ
, will have its position vector as
=
Example 11 Consider two points P and Q with position vectors and
. Find the position vector of a point R which divides the line joining P and Q
in the ratio 2:1, (i) internally, and (ii) externally.
Solution
(i) The position vector of the point R dividing the join of P and Q internally in the
ratio 2:1 is
=
(ii) The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
=
Example 12 Show that the points
ˆ ˆ ˆ
ˆˆ ˆ ˆ ˆ
A(2 ), B( 3 5 ), C(3 4 4 )
i jk i j k i j k
−+ − − − −
are
the vertices of a right angled triangle.
Solution We have
=
ˆ
ˆˆ
(1 2) ( 3 1) ( 5 1)
i jk
− +−+ +−−
ˆ
ˆˆ
26
i jk
=− − −
=
ˆ
ˆˆ
(3 1) ( 4 3) ( 4 5)
i jk
− +−+ +−+
ˆ
ˆˆ
2
i jk
= −+
and =
ˆ
ˆˆ
(2 3) ( 1 4) (1 4)
i jk
− +−+ + +
ˆ
ˆˆ
35
i jk
=− + +
Fig 10.17
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440
Further, note that
=
Hence, the triangle is a right angled triangle.
EXERCISE 10.2
1. Compute the magnitude of the following vectors:
=
ˆˆ
;
i jk
++
=
ˆ
ˆˆ
2 7 3;
i jk
−−
=
11 1
ˆ
ˆˆ
33 3
i jk
+−
2. Write two different vectors having same magnitude.
3. Write two different vectors having same direction.
4. Find the values of x and y so that the vectors
ˆˆ ˆˆ
2 3 and
i j xi yj
++
are equal.
5. Find the scalar and vector components of the vector with initial point (2, 1) and
terminal point (– 5, 7).
6. Find the sum of the vectors
=
ˆ
ˆˆ
2,
i jk
−+
=
ˆ
ˆˆ
245
i jk
−+ +
and =
ˆ
ˆˆ
6 –7
ci j k
=−
.
7. Find the unit vector in the direction of the vector
=
ˆ
ˆˆ
2
ai j k
=++
.
8. Find the unit vector in the direction of vector
PQ
, where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively.
9. For given vectors,
=
ˆ
ˆˆ
22
ij k
−+
and =
ˆ
ˆˆ
i jk
−+ −
, find the unit vector in the
direction of the vector
.
10. Find a vector in the direction of vector
ˆ
ˆˆ
52
ij k
−+
which has magnitude 8 units.
11. Show that the vectors
ˆ ˆ
ˆˆ ˆˆ
2 3 4 and 4 6 8
i jk i jk
−+ −+−
are collinear.
12. Find the direction cosines of the vector
ˆ
ˆˆ
23
i jk
++
.
13. Find the direction cosines of the vector joining the points A (1, 2, –3) and
B(–1, –2, 1), directed from A to B.
14. Show that the vector
ˆ
ˆˆ
i jk
++
is equally inclined to the axes OX, OY and OZ.
15. Find the position vector of a point R which divides the line joining two points P
and Q whose position vectors are
ˆ ˆ
ˆ ˆ ˆˆ
2 and –
i jk i jk
+ − ++
respectively, in the
ratio 2 : 1
(i) internally (ii) externally
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VECTOR ALGEBRA 441
16. Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2).
17. Show that the points A, B and C with position vectors, =
ˆ
ˆˆ
3 4 4,
i jk
−−
=
ˆ
ˆˆ
2
i jk
−+
and =
ˆ
ˆˆ
35
i jk
−−
, respectively form the vertices of a right angled
triangle.
18. In triangle ABC (Fig 10.18), which of the following is not true:
(A)
(B)
(C)
(D)
19. If are two collinear vectors, then which of the following are incorrect:
(A)
(B)
(C) the respective components of are not proportional
(D) both the vectors
have same direction, but different magnitudes.
10.6 Product of Two Vectors
So far we have studied about addition and subtraction of vectors. An other algebraic
operation which we intend to discuss regarding vectors is their product. We may
recall that product of two numbers is a number, product of two matrices is again a
matrix. But in case of functions, we may multiply them in two ways, namely,
multiplication of two functions pointwise and composition of two functions. Similarly,
multiplication of two vectors is also defined in two ways, namely, scalar (or dot)
product where the result is a scalar, and vector (or cross) product where the
result is a vector. Based upon these two types of products for vectors, they have
found various applications in geometry, mechanics and engineering. In this section,
we will discuss these two types of products.
10.6.1 Scalar (or dot) product of two vectors
Definition 2 The scalar product of two nonzero vectors
, denoted by , is
Fig 10.18
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442
defined as =
where, θ is the angle between (Fig 10.19).
If either
then θ is not defined, and in this case, we
define
Observations
1.
is a real number.
2. Let
be two nonzero vectors, then if and only if are
perpendicular to each other. i.e.
3. If θ = 0, then
In particular, as θ in this case is 0.
4. If θ = π, then
||||
ab a b
⋅ =−
In particular, , as θ in this case is π.
5. In view of the Observations 2 and 3, for mutually perpendicular unit vectors
ˆ
ˆˆ
, and ,
ij k
we have
ˆˆ ˆ ˆ
ii jj
⋅=⋅
=
ˆˆ
1,
kk
⋅=
ˆ
ˆˆ ˆ
i j jk
⋅=⋅
=
ˆ
ˆ
0
ki
⋅=
6. The angle between two nonzero vectors is given by
or
7. The scalar product is commutative. i.e.
(Why?)
Two important properties of scalar product
Property 1 (Distributivity of scalar product over addition) Let
be
any three vectors, then
Fig 10.19
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VECTOR ALGEBRA 443
(i)
B
C
A
l
B
l
A
C
(ii)
A
B
C
l
(iv)
l
C
B
A
(iii)
θ
θ
θ
θ
p
p
p p
a
a
a
a
(90<<180)
00
θ
(0<<90)
00
θ
(270<<360)
00
θ
(180<<270)
00
θ
Property 2 Let be any two vectors, and l be any scalar. Then
If two vectors are given in component form as
12 3
ˆ
ˆˆ
ai a j ak
++
and
12 3
ˆ
ˆˆ
bi b j bk
++
, then their scalar product is given as
=
123123
ˆ ˆ
ˆˆ ˆˆ
( )( )
ai a j a k bi b j bk
++ ⋅++
=
1123 2123
ˆ
ˆ
ˆˆ ˆ ˆˆ ˆ
(
)( )
ai bi b j bk a j bi b j bk
⋅++ +⋅++
+
312 3
ˆ ˆ
ˆˆ
( )
ak bi b j bk
⋅++
=
11 12 13 21 22 23
ˆ ˆ
ˆˆ ˆˆ ˆ ˆˆ ˆˆ ˆ
() () () () ( ) ( )
ab i i ab i j ab i k ab j i ab j j ab j k
⋅+ ⋅+ ⋅ + ⋅+ ⋅+ ⋅
+
31 32 33
ˆ ˆ ˆˆ
ˆˆ
() ( ) ( )
ab k i ab k j ab k k
⋅+ ⋅+ ⋅
(Using the above Properties 1 and 2)
= a
1
b
1
+ a
2
b
2
+ a
3
b
3
(Using Observation 5)
Thus =
11 2 2 3 3
ab ab ab
++
10.6.2 Projection of a vector on a line
Suppose a vector
makes an angle θ with a given directed line l (say), in the
anticlockwise direction (Fig 10.20). Then the projection of on l is a vector
p
(say) with magnitude , and the direction of being the same (or opposite)
to that of the line l, depending upon whether cosθ is positive or negative. The vector
Fig 10.20
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444
is called the projection vector, and its magnitude | | is simply called as the projection
of the vector
on the directed line l.
For example, in each of the following figures (Fig 10.20 (i) to (iv)), projection vector
of
along the line l is vector .
Observations
1. If
ˆ
p
is the unit vector along a line l, then the projection of a vector on the line
l is given by
ˆ
ap
⋅
.
2. Projection of a vector
on other vector
b
, is given by
ˆ
,
ab
⋅
or
3. If θ = 0, then the projection vector of will be itself and if θ = π, then the
projection vector of
will be .
4. If
=
2
π
θ
or
3
=
2
π
θ
, then the projection vector of will be zero vector.
Remark If α, β and γ are the direction angles of vector
12 3
ˆ
ˆˆ
ai a j ak
=+ +
, then its
direction cosines may be given as
Also, note that are respectively the projections of
along OX, OY and OZ. i.e., the scalar components a
1
, a
2
and a
3
of the vector , are
precisely the projections of
along x-axis, y-axis and z-axis, respectively. Further, if
is a unit vector, then it may be expressed in terms of its direction cosines as
ˆ
ˆˆ
cos cos cos
ijk
=α+β+γ
Example 13 Find the angle between two vectors with magnitudes 1 and 2
respectively and when
=1.
Solution Given
. We have
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VECTOR ALGEBRA 445
Example 14 Find angle ‘θ’ between the vectors .
Solution The angle θ between two vectors
is given by
cosθ =
Now =
ˆˆ
ˆˆ ˆˆ
( )( ) 1 1 1 1
i jk i jk
+ − ⋅ − + =−−=−
.
Therefore, we have cosθ =
1
3
−
hence the required angle is θ =
Example 15 If , then show that the vectors
are perpendicular.
Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero.
Here
=
ˆ ˆ ˆ
ˆˆ ˆ ˆ ˆ ˆ
(5 3) ( 3 5) 6 2 8
ij k i j k i j k
−− + + − = + −
and =
ˆ ˆ ˆ
ˆˆ ˆ ˆ ˆ ˆ
(5 3 ) ( 3 5 ) 4 4 2
ij k i j k i j k
−− −+− =−+
So ( ) . ( )
ˆ ˆ
ˆˆ ˆˆ
(6 2 8 ) (4 4 2 ) 24 8 16 0.
ijkijk
= +− ⋅−+ =−−=
Hence are perpendicular vectors.
Example 16 Find the projection of the vector
ˆ
ˆˆ
23 2
i jk
=++
on the vector
ˆ
ˆˆ
2
i jk
=+ +
.
Solution The projection of vector
a
on the vector
b
is given by
=
2 22
(2 1 3 2 2 1) 10 5
6
3
6
(1) (2) (1)
×+× +×
==
++
Example 17 Find , if two vectors are such that
and .
Solution We have
=
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446
B
C
A
a
b
+
a
b
=
=
2 2
(2) 2(4) (3)
−+
Therefore =
5
Example 18 If is a unit vector and , then find .
Solution Since
is a unit vector, . Also,
or = 8
or
Therefore = 3 (as magnitude of a vector is non negative).
Example 19 For any two vectors
, we always have (Cauchy-
Schwartz inequality).
Solution The inequality holds trivially when either
0 or 0
ab
==
. Actually, in such a
situation we have
. So, let us assume that .
Then, we have
=
| cos | 1
θ≤
Therefore
Example 20 For any two vectors , we always
have
(triangle inequality).
Solution The inequality holds trivially in case either
(How?). So, let . Then,
=
= (scalar product is commutative)
≤
(since
||
xxx
≤ ∀∈
R
)
≤
(from Example 19)
=
Fig 10.21
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VECTOR ALGEBRA 447
Hence
Remark If the equality holds in triangle inequality (in the above Example 20), i.e.
= ,
then
=
showing that the points A, B and C are collinear.
Example 21 Show that the points
ˆ
ˆ
ˆˆ ˆˆ
A ( 2 3 5 ), B( 2 3 )
i jk i jk
−+ + + +
and
ˆ
ˆ
C(7 )
ik
−
are collinear.
Solution We have
=
ˆˆ
ˆˆ ˆˆ
(1 2) (2 3) (3 5) 3 2
i j k ij k
+ + − + − = −−
,
=
ˆ ˆ
ˆˆ ˆˆ
(71) (02) (13) 6 2 4
i j ki jk
−+−+−−=−−
,
=
ˆ ˆ
ˆˆ ˆˆ
(7 2) (0 3) ( 1 5) 9 3 6
i j ki jk
+ + − +−− = − −
=
Therefore
Hence the points A, B and C are collinear.
Note In Example 21, one may note that although but the
points A, B and C do not form the vertices of a triangle.
EXERCISE 10.3
1. Find the angle between two vectors with magnitudes
3 and 2
,
respectively having
.
2. Find the angle between the vectors
ˆ ˆ
ˆ ˆ ˆ ˆ
2 3 and 3 2
i j k i j k
− + − +
and
ˆ ˆ
ˆ ˆ ˆ ˆ
2 3 and 3 2
i j k i j k
− + − +
3. Find the projection of the vector
ˆˆ
ij
−
on the vector
ˆˆ
ij
+
.
4. Find the projection of the vector
ˆ
ˆˆ
37
i jk
++
on the vector
ˆ
ˆˆ
78
ij k
−+
.
5. Show that each of the given three vectors is a unit vector:
1 1 1
ˆ ˆ ˆ
ˆˆ ˆˆ ˆˆ
(2 3 6 ), (3 6 2 ), (6 2 3 )
7 7 7
i jk i jk i jk
++ −+ +−
Also, show that they are mutually perpendicular to each other.
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448
6. Find
, if .
7. Evaluate the product
.
8. Find the magnitude of two vectors
, having the same magnitude and
such that the angle between them is 60
o
and their scalar product is
1
2
.
9. Find
, if for a unit vector , .
10. If
are such that is
perpendicular to
, then find the value of λ.
11. Show that
is perpendicular to , for any two nonzero
vectors
.
12. If
, then what can be concluded about the vector ?
13. If
are unit vectors such that , find the value of
.
14. If either vector
. But the converse need not be
true. Justify your answer with an example.
15. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2),
respectively, then find ∠ABC. [∠ABC is the angle between the vectors
and
].
16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.
17. Show that the vectors
ˆˆ ˆ
ˆˆ ˆ ˆ ˆ ˆ
2 , 3 5 and 3 4 4
i j ki j k i j k
−+ − − − −
form the vertices
of a right angled triangle.
18. If
is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit
vector if
(A) λ = 1 (B) λ = – 1 (C) a = | λ| (D) a = 1/| λ|
10.6.3 Vector (or cross) product of two vectors
In Section 10.2, we have discussed on the three dimensional right handed rectangular
coordinate system. In this system, when the positive x-axis is rotated counterclockwise
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VECTOR ALGEBRA 449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10.22(i)).
In a right handed coordinate system, the thumb of the right hand points in the
direction of the positive z-axis when the fingers are curled in the direction away from
the positive x-axis toward the positive y-axis (Fig 10.22(ii)).
Fig 10.22 (i), (ii)
Definition 3 The vector product of two nonzero vectors
, is denoted by
and defined as
= ,
where, θ is the angle between
,
0
≤θ≤π
and
ˆ
n
is a
unit vector perpendicular to both
, such that
form a right handed system (Fig 10.23). i.e., the
right handed system rotated from
moves in the direction
of
ˆ
n
.
If either
, then θ is not defined and in this case, we define .
Observations
1.
is a vector.
2. Let
and
ab
be two nonzero vectors. Then if and only if are
parallel (or collinear) to each other, i.e.,
=
Fig 10.23
2019-20
MATHEMATICS
450
In particular,
and , since in the first situation, θ = 0 and
in the second one, θ = π, making the value of sin θ to be 0.
3. If
2
π
θ=
then .
4. In view of the Observations 2 and 3, for mutually perpendicular
unit vectors
ˆ
ˆˆ
, and
ij k
(Fig 10.24), we have
ˆˆ
ii
×
=
ˆˆ
ij
×
=
ˆˆˆ
ˆ ˆ ˆˆ
, ,
k jkiki j
× = ×=
5. In terms of vector product, the angle between two vectors may be
given as
sin θ =
6. It is always true that the vector product is not commutative, as = .
Indeed,
, where form a right handed system,
i.e., θ is traversed from
, Fig 10.25 (i). While, , where
form a right handed system i.e. θ is traversed from ,
Fig 10.25(ii).
Fig 10.25 (i), (ii)
Thus, if we assume
and
ab
to lie in the plane of the paper, then
1
ˆˆ
and
nn
both
will be perpendicular to the plane of the paper. But,
ˆ
n
being directed above the
paper while
1
ˆ
n
directed below the paper. i.e.
1
ˆˆ
nn
=−
.
Fig 10.24
2019-20
VECTOR ALGEBRA 451
Hence
=
=
7. In view of the Observations 4 and 6, we have
ˆˆ ˆ
ˆˆ ˆ ˆ ˆ ˆ
, and .
ji k k j i ik j
× =− × =− × =−
8. If represent the adjacent sides of a triangle then its area is given as
.
By definition of the area of a triangle, we have from
Fig 10.26,
Area of triangle ABC =
1
AB CD.
2
⋅
But (as given), and CD = sin θ.
Thus, Area of triangle ABC =
9. If represent the adjacent sides of a parallelogram, then its area is given
by
.
From Fig 10.27, we have
Area of parallelogram ABCD = AB. DE.
But
(as given), and
.
Thus,
Area of parallelogram ABCD =
We now state two important properties of vector product.
Property 3 (Distributivity of vector product over addition): If
are any three vectors and λ be a scalar, then
(i)
(ii)
Fig 10.26
Fig 10.27
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MATHEMATICS
452
Let
be two vectors given in component form as
12 3
ˆ
ˆˆ
ai a j ak
++
and
12 3
ˆ
ˆˆ
bi b j bk
++
, respectively. Then their cross product may be given by
=
123
123
ˆ
ˆˆ
i jk
aaa
bbb
Explanation We have
=
12 3 12 3
ˆ
ˆ
ˆˆ ˆˆ
(
)( )
ai a j a k bi b j bk
++ ×++
=
11 12 13 21
ˆ
ˆˆ ˆˆ ˆ ˆˆ
() () () ()
ab i i ab i j ab i k ab j i
×+ ×+ × + ×
+
22 23
ˆ
ˆˆ ˆ
() ()
ab j j ab j k
×+ ×
+
31 32 33
ˆ ˆ ˆˆ
ˆˆ
() ( ) ( )
ab k i ab k j ab k k
×+ ×+ ×
(by Property 1)
=
12 13 21
ˆ
ˆˆ ˆ ˆˆ
() () ()
abijabki abij
×− ×− ×
+
23 31 32
ˆˆ
ˆ
ˆ
ˆˆ
() () ()
ab j k ab k i ab j k
× + ×− ×
ˆˆ ˆˆ ˆˆ
ˆˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ
(as
0 and , and )
ii j j kk ik kiji i j k j jk
× = × = × = × =− × × =− × × =− ×
=
12 13 21 23 31 32
ˆ ˆ
ˆ ˆˆˆ
abk ab j abk abi abj abi
−−++−
ˆˆ ˆ
ˆˆ ˆ ˆ ˆˆ
(as , and )
i j kjki ki j
×= ×= ×=
=
23 3 2 13 31 12 21
ˆ
ˆ ˆ
( )( ) ( )
ab ab i ab ab j ab ab k
−−−+−
=
123
123
ˆ
ˆˆ
i jk
aaa
bbb
Example 22 Find
Solution We have
=
ˆ
ˆˆ
21 3
35 2
ijk
−
=
ˆ
ˆ ˆ
( 2 15) ( 4 9) (10 – 3)
i jk
−− −− − +
ˆ
ˆˆ
17 13 7
i jk
=− + +
Hence =
2 22
( 17) (13) (7) 507
−+ + =
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VECTOR ALGEBRA 453
Example 23 Find a unit vector perpendicular to each of the vectors and
where .
Solution We have
A vector which is perpendicular to both
and
ab ab
+−
is given by
=
Now =
4 16 4 24 2 6
+ += =
Therefore, the required unit vector is
||
c
c
=
12 1
ˆ
ˆˆ
66 6
i jk
−
+−
Note There are two perpendicular directions to any plane. Thus, another unit
vector perpendicular to
will be
12 1
ˆ
ˆˆ
.
66 6
i jk
−+
But that will
be a consequence of .
Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3)
and C(2, 3, 1) as its vertices.
Solution We have
. The area of the given triangle
is
.
Now,
=
ˆ
ˆˆ
ˆ
ˆˆ
012 4 2
120
i jk
i jk
=− + −
Therefore =
16 4 1 21
++=
Thus, the required area is
1
21
2
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Example 25 Find the area of a parallelogram whose adjacent sides are given
by the vectors
Solution The area of a parallelogram with as its adjacent sides is given
by
.
Now
=
ˆ
ˆˆ
ˆ
ˆˆ
3145 4
1 11
i jk
ij k
= +−
−
Therefore =
25 1 16 42
++ =
and hence, the required area is
42
.
EXERCISE 10.4
1. Find .
2. Find a unit vector perpendicular to each of the vector
, where
.
3. If a unit vector
makes angles
ˆˆ
with , with
34
ij
ππ
and an acute angle θ with
ˆ
k
, then find θ and hence, the components of .
4. Show that
5. Find λ and µ if .
6. Given that
and . What can you conclude about the vectors
?
7. Let the vectors
be given as
1 2 31 2 3
ˆ
ˆ
ˆˆ ˆˆ
,
,
ai a j ak bi b j bk
++ ++
12 3
ˆ
ˆˆ
ci c j ck
++
. Then show that .
8. If either
then . Is the converse true? Justify your
answer with an example.
9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
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VECTOR ALGEBRA 455
10. Find the area of the parallelogram whose adjacent sides are determined by the
vectors
and .
11. Let the vectors
be such that , then is a unit
vector, if the angle between
is
(A) π/6 (B) π/4 (C) π/3 (D) π/2
12. Area of a rectangle having vertices A, B, C and D with position vectors
11
ˆˆ
ˆˆ ˆˆ
– 4, 4
22
i jki jk
++ ++
,
1
ˆ
ˆˆ
4
2
i jk
−+
and
1
ˆ
ˆˆ
–4
2
i jk
−+
, respectively is
(A)
1
2
(B) 1
(C) 2 (D) 4
Miscellaneous Examples
Example 26 Write all the unit vectors in XY-plane.
Solution Let
be a unit vector in XY-plane (Fig 10.28). Then, from the
figure, we have x = cos θ and y = sin θ (since |
| = 1). So, we may write the vector
r
as
=
ˆˆ
cos sin
ij
θ+ θ
... (1)
Clearly, |
| =
22
cos sin 1
θ+ θ=
Fig 10.28
Also, as θ varies from 0 to 2π, the point P (Fig 10.28) traces the circle x
2
+ y
2
= 1
counterclockwise, and this covers all possible directions. So, (1) gives every unit vector
in the XY-plane.
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Example 27 If
ˆ ˆ ˆ
ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ
, 2 5 , 3 2 3 and 6
++ + + − − −
ijkijijk ijk
are the position
vectors of points A, B, C and D respectively, then find the angle between
and .
Deduce that
and are collinear.
Solution Note that if θ is the angle between AB and CD, then θ is also the angle
between
and .
Now
= Position vector of B – Position vector of A
=
ˆˆ
ˆ ˆ ˆˆ ˆ ˆ
(2 5 ) ( ) 4
i j i jk i jk
+ − ++ =+ −
Therefore | | =
22 2
(1)(4)(1) 32
+ +− =
Similarly =
ˆ
ˆˆ
2 8 2 and |CD | 6 2
i jk−−+ =
Thus cos θ =
=
1( 2) 4( 8) ( 1)(2) 36
1
36
(3 2)(6 2)
− + − +− −
= =−
Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that and are collinear.
Alternatively,
which implies that and are collinear vectors.
Example 28 Let be three vectors such that and
each one of them being perpendicular to the sum of the other two, find
.
Solution Given
Now
=
+
=
= 9 + 16 + 25 = 50
Therefore
=
50 5 2
=
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Example 29 Three vectors
satisfy the condition . Evaluate
the quantity
.
Solution Since
, we have
= 0
or
= 0
Therefore
... (1)
Again,
= 0
or
... (2)
Similarly
= – 4. ... (3)
Adding (1), (2) and (3), we have
= – 29
or 2µ = – 29, i.e., µ =
29
2
−
Example 30 If with reference to the right handed system of mutually perpendicular
unit vectors
, then express in the form
is parallel to is perpendicular to
Solution Let is a scalar, i.e., .
Now
=
ˆ
ˆˆ
(2 3 ) (1 ) 3
i jk
− λ + +λ −
.
Now, since
2
β
is to be perpendicular to
α
, we should have . i.e.,
3 (2 3 ) (1 )
− λ − +λ
= 0
or λ =
1
2
Therefore
1
=
31
ˆˆ
22
ij
−
and
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Miscellaneous Exercise on Chapter 10
1. Write down a unit vector in XY-plane, making an angle of 30° with the positive
direction of x-axis.
2. Find the scalar components and magnitude of the vector joining the points
P(x
1
, y
1
, z
1
) and Q (x
2
, y
2
, z
2
).
3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of
north and stops. Determine the girl’s displacement from her initial point of
departure.
4. If
, then is it true that ? Justify your answer.
5. Find the value of x for which
ˆ
ˆˆ
()
xi j k
++
is a unit vector.
6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
.
7. If
, find a unit vector parallel
to the vector
.
8. Show that the points A (1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and
find the ratio in which B divides AC.
9. Find the position vector of a point R which divides the line joining two points
P and Q whose position vectors are
externally in the ratio
1 : 2. Also, show that P is the mid point of the line segment RQ.
10. The two adjacent sides of a parallelogram are
ˆ ˆ
ˆˆ ˆˆ
2 4 5 and 2 3
ijk ijk
−+ −−
.
Find the unit vector parallel to its diagonal. Also, find its area.
11. Show that the direction cosines of a vector equally inclined to the axes OX, OY
and OZ are
±
1
3
1
3
1
3
,,
.
12. Let
. Find a vector which
is perpendicular to both
and , and .
13. The scalar product of the vector
ˆ
ˆˆ
i jk
++
with a unit vector along the sum of
vectors
ˆ
ˆˆ
245
i jk
+−
and
ˆ
ˆˆ
23
i jk
λ+ +
is equal to one. Find the value of λ.
14. If
are mutually perpendicular vectors of equal magnitudes, show that the
vector
is equally inclined to .
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VECTOR ALGEBRA 459
15. Prove that , if and only if are perpendicular, given
.
Choose the correct answer in Exercises 16 to 19.
16. If θ is the angle between two vectors
, then only when
(A)
0
2
π
<θ<
(B)
0
2
π
≤θ≤
(C) 0 < θ < π (D) 0 ≤ θ ≤ π
17. Let
be two unit vectors and θ is the angle between them. Then is
a unit vector if
(A)
4
π
θ=
(B)
3
π
θ=
(C)
2
π
θ=
(D)
2
3
π
θ=
18. The value of
ˆ ˆˆ
ˆˆ ˆ ˆ ˆ ˆ
.( ) ( ) ( )
ijk ji k ki j
× +⋅× +⋅×
is
(A) 0 (B) –1 (C) 1 (D) 3
19. If θ is the angle between any two vectors
, then when θ
is equal to
(A) 0 (B)
4
π
(C)
2
π
(D) π
Summary
Position vector of a point P(x, y, z) is given as , and its
magnitude by
2 22
xyz
++
.
The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes.
The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of
any vector are related as:
,,
a bc
l mn
r rr
= ==
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460
The vector sum of the three sides of a triangle taken in order is .
The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors.
The multiplication of a given vector by a scalar λ, changes the magnitude of
the vector by the multiple | λ|, and keeps the direction same (or makes it
opposite) according as the value of λ is positive (or negative).
For a given vector
, the vector gives the unit vector in the direction
of
.
The position vector of a point R dividing a line segment joining the points
P and Q whose position vectors are
respectively, in the ratio m : n
(i) internally, is given by
.
(ii) externally, is given by
.
The scalar product of two given vectors having angle θ between
them is defined as
.
Also, when
is given, the angle ‘θ’ between the vectors may be
determined by
cos θ =
If θ is the angle between two vectors , then their cross product is
given as
where
ˆ
n
is a unit vector perpendicular to the plane containing . Such
that
form right handed system of coordinate axes.
If we have two vectors , given in component form as
and λ any scalar,
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VECTOR ALGEBRA 461
then =
11 2 2 3 3
ˆ
ˆˆ
( )( ) ( )
a bi a b j a bk
+ ++ ++
;
λ
=
12 3
ˆ
ˆˆ
()( ) ( )
ai a j ak
λ +λ +λ
;
=
11 2 2 3 3
ab ab ab
++
;
and
=
111
222
ˆ
ˆˆ
.
i jk
abc
abc
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry”. The germinal ideas of modern vector theory date from around 1800
when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described
that how a complex number a + ib could be given a geometric interpretation with
the help of a directed line segment in a coordinate plane. William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853). Hamilton’s
method of quaternions (an ordered set of four real numbers given as:
ˆˆ
ˆ ˆ ˆˆ
,, ,
a bi cj dk i j k
+++
following certain algebraic rules) was a solution to the
problem of multiplying vectors in three dimensional space. Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B.C.), a Greek
philosopher, and pupil of Plato (427-348 B.C.). That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law. The correct law for the composition of
forces, that forces add vectorially, had been discovered in the case of perpendicular
forces by Stevin-Simon (1548-1620). In 1586 A.D., he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics. But it took another 200 years for the general concept
of vectors to form.
In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist and
mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
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462
part of quaternion from its imaginary (vector) part. In 1881 and 1884, Gibbs
printed a treatise entitled Element of Vector Analysis. This book gave a systematic
and concise account of vectors. However, much of the credit for demonstrating
the applications of vectors is due to the D. Heaviside and P.G. Tait (1831-1901)
who contributed significantly to this subject.
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