28
Vectors and Three
Dimensional Geometry UNIT 2 VECTORS – 2
Structure
2.0 Introduction
2.1 Objectives
2.2 Scalar Product of Vectors
2.3 Vector Product (or Cross Product) of two Vectors
2.4 Triple Product of Vectors
2.5 Answers to Check Your Progress
2.6 Summary
2.0 INTRODUCTION
In the previous unit, we discussed vectors and scalars. We learnt how to add and
subtract two vectors, and how to multiply a vector by a scalar. In this unit, we
shall discuss multiplication of vectors. There are two ways of defining product
of vectors. We can multiply two vectors to get a scalar or a vector. The former
is called scalar product or dot product of vectors and the latter is called vector
product or cross product of vectors. We shall learn many applications of dot
product and cross product of vectors. We shall use dot product to find angle
between two vectors. Two vectors are perpendicular if their dot product is zero.
Dot product helps in finding projection of a vector onto another vector. The cross
product of two vectors is a vector perpendicular to both the vectors.
If cross product of two vectors is zero then the two vectors are parallel (or
collinear). Cross product of vectors is also used in finding area of a triangle or a
parallelogram. Using the two kinds of products, we can also find product of
three vectors. Many of these products will not be defined. In this unit, we shall
discuss the two valid triple products, namely, the scalar triple product and the
vector triple product.
2.1 OBJECTIVES
After studying this unit, you should be able to:
define scalar product or dot product of vectors;
find angle between two vectors;
find projection of a vector on another vector;
29
Vectors - II define cross product or vector product of vectors;
use cross product to find area of a parallelogram vector product of vectors;
define scalar triple product and vector triple product of vectors.
2.2 SCALAR PRODUCT OF TWO VECTORS
Definition : The scalar product or the dot product of two vectors and ,
denoted by . is defined by
. = | | | |
where is the angle between the vectors and .
Also the scalar product of any vector with the zero vector is, by definition, the
scalar zero. It is clear from the definition that the dot product . is a scalar
quantity.
Sign of the Scalar Product
If and are two non zero vectors, then the scalar product
. = | | | | cos
is positive, negative or zero, according as the angle , between the vectors is
acute, obtuse or right. In fact,
is acute cos 0 . 0a b
is right cos 0 . 0a b
is obtuse cos 0 . 0a b
Also, note that if and and are non zero vectors then . = 0 if and only if
and are perpendicular (or orthogonal) to each other.
If . is any vector, then the dot product . ., of with itself, is given by
. = | || | cos 0= | |2
Thus, the length | | of any vector is the non negative square root , i.e., of
the scalar product , i.e.,
| | = ,
Angle between two Vectors
If is an angle between two non zero vectors and , then
. = | || | cos
30
Vectors and Three
Dimensional Geometry
So, the angle between two vectors is given by
Properties of Scalar Product
1. Scalar product is cumulative, i.e., . = . for every pair of vectors and .
2. . (– ) = – ( . ) and (– ). (– ) = . for every pair of vectors and .
3. ( λ1 ) . (λ2 ) =( λ1 λ2) ( . ) where and are vectors and λ1 , λ2 are scalars
4. (Distributivity) .( + ) = . + .
The following identities can be easily proved using above properties.
(i) ( + ) . ( – ) = a2 – b
2 (here, a
2 = . = | |
2)
(ii) ( + )2
= a2 + 2 . + b
2
(iii) ( – )
2 = a
2 – 2 . + b
2
If , and are mutually perpendicular unit vectors, then
. = . = . = 1 and
. .= . = . = 0 (1)
If and be two vectors in component
form, then their scalar product is given by
1 1 2 3 2 1 2 3 3 1 2 3
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ . ( + ) ( + ) ( + ) (using distributivity)a i bi b j b k a i bi b j b k a k bi b j b k
2 3 3 1 3 2 3 3ˆ ˆ ˆ ˆ ˆˆ= ( . ) ( . ) ( . ) ( . ) (using properties)a b j k a b k i a b k j a b k k
= a1 b1 + a2 b2 + a3 b3 (using (1))
Thus . = a1 b1 + a2 b2 + a3 b3 .
Projection of a Vectors on another Vector
Let = be two vectors
A
Figure 1
O L B
31
Vectors - II Drop a perpendicular form A on OB as shown in Figure 1. The projection of
on is the vector . If is the angle between and , then projection of
and has length | cos and direction along unit vector . Thus, projection
of on = (| | cos )
2
. .
| | | |
a b a bb b
b b
The scalar component of project of on is Similarly, the scalar
component of projection of on
a . ba b .
|b |
a . bb a
| a |
Example 1: Find the angle between the following pair of vectors.
(a) +
(b)
Solution: (a) . = (2 + 3 ) . ( + 4 )
= 2.1 + 0.4 + 3 ( 1)
= 2 3 = 1
= . = (2 + 3 ) . (2 + 3 )
= 2.2 + 0.0 + 3.3
= 13
|a| =
= . = ( +4 – ) . ( +4 )
= 1.1 + 4.4 + ( 3) ( 3)
= 26
|a| =
(b) Here, . = ( + 2 + ). (2 + 2 + )
= 1.2 + 2.(–2) + 2.1
= 0
= = 3
32
Vectors and Three
Dimensional Geometry = = 3
.
Example 2 : Show that is perpendicular to – , for any
two non zero vectors .
Solution : We know that two vectors are perpendicular if their scalar product is
zero.
( ) . ( )
= ( ) + . ( )
(using distributivity)
( ) ( ) + ( ) ( )
( ) + ( ) = 0
so, the given vectors are perpendicular.
Example 3 : Find the scalar component of projection of the vector
ˆ ˆˆ ˆ ˆ 2 3 5 on the vector 2 2 . a i j k b i j k
Here,
2 22and |b| = = 3( 2) ( 1)2
Example 4 : Show that the diagonals of a rhombus are at right angles.
Solution : Let A B C D be a rhombus (Figure 2 )
D C
A B (Figure 2)
33
Vectors - II ABCD being a rhombus, we have
=
= ( + ) . ( )
=
= – = 0
2 2
( )AB AD
Example 5 : For any vectors and , prove the triangle inequality
| + | ≤ | | + | |
Solution : If = or = , then the inequality holds trivially.
So let | | ≠ 0 ≠ | |. Then,
= ( + )2
= ( + ). ( + )
=
= | |2 2 + | |
2 ( = )
= | |2
2| | |cos | |2
where is the angle between and
| |2
2| | | | |2
(cos 1 )
= (| | | |) 2
Hence | | | | | |
Remark : Let = and = , then + =
C
+
A B
Figure 3
As shown in figure 3 inequality says that the sum of two sides of triangle is
greater than the third side. If the equality holds in triangle inequality, i.e.,
| + | | | +| |
Then | AC
| = | AB
| + | BC
|
showing that the points A, B and C are collinear.
Check Your Progress – 1
1. If = 5 –3 –3 and = + –5 then show that + and – are
perpendicular.
34
Vectors and Three
Dimensional Geometry
2. Find the angle between the vectors
(a) = 3 – +2 = – 4 +2
(b) = – –2 = –2 +2 +4
3. Find the vector projection of on where =3 –5 +2 and = 7 + –2 .
Also find the scalar component of projection of vector on .
4. Prove the Cauchy – Schwarz Inequality
| . | ≤ | | | |
5. Prove that | – |2 = | |
2 + | |
2 – 2 .a b
2.3 VECTOR PRODUCT (OR CROSS PRODUCT) OF TWO
VECTORS
Definition : If and are two non zero and non parallel (or equivalently non
collinear) vectors, then their vector product × is defined as
= | | | |sin
where is the angle between and (0 < < ) and is a unit vector
perpendicular to both and such that , and form a right handed system.
If and are parallel (or collinear) i.e., when = 0 or , then we define the
vector product of and to be the zero vector i.e., = . Also note that if
either or ,then
Properties of the Vector Product
1. = since = 0
2. Vector product is not commutative i.e., ≠ .
However, = –
We have = | || |sin where , and form a right handed system and
= | || | |sin | where , and ight handed system. So the
direction of | is opposite to that of
Hence, = | || |sin
= –| || |sin
= – .
3. Let from a right–handed system of mutually perpendicular unit vectors
in three dimensional space. Then
= = = and
= =
Also, = – = –
a
Figure 4
35
Vectors - II 4. Two non zero vectors and are parallel if and only if
=
5. Vector product is distributors over addition i.e., if , and
are three vectors, then
(i) ( ) = .
(ii) ( ) = .
6. If is a scalar and and are vectors, then
( = (
Vector Product in the component form
Let = a1 a2 a3 and
= b1 b2 b3
Then
= (a1 a2 a3 ) (b1 b2 b3
= a1 b1 ( ) a1 b2 ( ) a1 b3 ( ) a2 b1 ( ) a2 b3
( ) a2 b3 ( ) a3 b1 ) a3 b2 ( ) a3 b3
( )
= a1 b2 a1 b3 (– a2 b1 (– a2 b3 a3 b1 a3 b2 (–
= (a2 b3 a3 b2) (a1 b3 a3 b1) (a1 b2 a2 b1)
=
Example 6: Find if = and = 3
Solution : We have
=
= (3 – 2) – (3 – 1) + (2 – 1)
= – 2
Example 7: Find a unit vector perpendicular to both the vectors
ˆ ˆˆ ˆ ˆ 4 3 and 2 2 a i j k b i j k
Solution : A vector perpendicular to both and is
=
= (2 – 3) – (–8 6) (4 – 2)
= – 2
| | = = = 3.
So the desired unit vector is
36
Vectors and Three
Dimensional Geometry
Example 8 : Prove the distributive law
( ) a b c a b a c
using component form of vectors.
Solution : Let = + +
= + + and
= + + .
So, + = + + ( +
Now,
× ( + ) =
=
= + .
Example 9: Show that
| |2
= (
Solution : | |2
= | || | sin
| |2
= (| || | sin )2
= | |2| |
2 (1– cos
2)
= | |2| b
|2 – ( | |
2| |
2 cos
2)
2 2 ( ) ( ) ( . )aa bb a b
Example 10 : Find | | if
| | = 10, | | = 2, = 12.
Solution : Here = 12
a
a
37
Vectors - II Area of Triangle
Let , be two vectors and let be the angle between them ( O < < ). Let O
be the origin for the vectors as shown in figures (Fig. 5) below and let
= , =
B D
O C A Figure 5
Draw BC OA.
Then BC = OB Sin = |b| sin
Thus, if and represent the adjacent sides of a triangle, then its area is given as
Area of a Parallelogram
In above figure, if D is the fourth vertex of the parallelogram formed by O, B, A
then its area is twice the area of the triangle OBA.
Hence, area of a parallelogram with and as adjacent sides = |
Example 11: Find the area of ABC with vertices A (1,3,2), B (2, –1,1) and
C(–1, 2, 3).
Solution : We have
= –
= (2 – 1) (–1 – 3) (1 – 2)
= – 4 –
and = ( –1 – 1) (2 – 3) (3 – 2)
= – 2 –
θ
38
Vectors and Three
Dimensional Geometry
Example 12 : Show that ( – ) ( ) = 2( ). Interpret the result
geometrically.
Solution : ( – ) ( )
= ( – ) ( – )
= – –
= 0 0
= 2 ( )
Let ABCD be a parallelogram with = and = – .
Then area of parallelogram = =
D C
b
A a
B Figure 6
Also, diagonal = +
and diagonal = –
( ) ( ) =
= area of parallelogram formed by and .
Thus, the above result shows that the area of a parallelogram formed by diagonals
of a parallelogram is twice the area of the parallelogram.
Check Your Progress – 2
1. Find a unit vector perpendicular to each of the vector ( + ) and ( – ),
where = 2 – 4 and = – 2
2. Show that
( ) ( ) ( ) =
39
Vectors - II 3. For Vectors
= – 2
Compute ( and ( ).
is ( × = ( ).
4. If the vectors , and satisfy + + = 0, then prove that
= =
5. Find the area of a parallelogram whose diagonals are
= 3 + –2
2.4 TRIPLE PRODUCT OF VECTORS
Product of three vectors may or may not have a meaning. For example,
( . as is a scalar and dot product is defined only for
vectors. Similarly, ( . has no meaning. The products of the type
( and are meaningful and called triple products. The
former is a vector while the latter is a scalar.
Scalar Triple Product
Definition : Let , and be any three vectors. The scalar product of . )
is called scalar triple product of , and and is denoted by [ , ]. Thus
[ , ] = .( .
It is clear from the definition that [ , ]. Is a scalar quantity.
Geometrically, the scalar triple product gives the volume of a parallelepiped
formed by vectors , as adjacent sides.
Scalar Triple Product as a determinant
Let =
and =
Then
.(
= ( –
= –
40
Vectors and Three
Dimensional Geometry =
Thus, .(
Note : We can omit the brackets in .( because
( . is meaningless.
Properties of Scalar Triple Product
1. [ , = [
= – [ ] = – [ ,
This is clear if we note the properties of a determinant as [ , can be
expressed as a determinant.
2. In scalar triple product .( , the dot and cross can be interchanged.
Indeed, .( = [ = [
= =
3. [ , , ] [ , , ]pa qb rc pqr a b c
where p, q and r are scalars. Again it is clear
using properties of determinants.
4. If any two of are the same then [ For example,
[ ,
Coplanarity of three vectors
Theorem : Three vectors are coplanar if and only if [
Proof : First suppose that are coplanar . If are parallel
vectors, then If are not parallel,
then being perpendicular to plane containing is also
perpendicular to because
Conversely, suppose that [ , = 0. If are both non zero, then
are perpendicular as their dot product is zero. But is
perpendicular to both and hence must lie in a plane, i.e.,
they are coplanar. If = then are coplanar as zero vector is
coplanar with any two vectors. If = , then
, must be coplanar.
Note : Four points A, B , C and D are coplanar if the vectors , and are
coplanar.
41
Vectors - II Example 13 : If = 7 – 2 +3 = – 2 +2 , and = 2 + 8
find [ , ,
Solution : [ , , = . ( ×
= 7(0 – 16) + 2(0 – 4) + 3(8 + 4)
= –112 – 8 + 36
= – 84
Example 14 : Find the value of λ for which the vectors
= – 4 + = λ – 2 + , and = 2 + 3 + 3 are
coplanar,
Solution : If , and are coplanar, we have
= . ( = 0 i.e.,
1(–6 –3) + 4(3 – 2) + (3 + 4) = 0
Example 15 : If , , are coplanar then prove that , and are also
coplanar.
Solution : Since , , are coplanar,
[ , ,
Now [ , ] = ( ) .( )
= ( + ) .[( + ( ]
= ( + ) .[ × ] ( as 0
= .( a
.( ) + .( ) + ( + ( ) +
= [ , , [ , , +
= 2 [ , , (using property (4)
= 0
42
Vectors and Three
Dimensional Geometry
Vector Triple Product
Definition : Let , , and any three vectors. Then, the vectors ( ) and
( ) are called vector triple products.
It is clear that, in general
( ) ( ) ×
Note that ( ) is a vector perpendicular to both and . And also
is perpendicular to both and . Thus ( ) lies in a plane containing
the vectors and , i.e., ( ), and are coplanar vectors.
We now show that for any three vectors ,
( ) = ( – (
Let =
and = Then
=
= ( – + ( – + ( –
(
– – –
= [ ( – + – (
+ [ (
Also, (1)
= (
(
= [ (
[ (
[ (
= [ (
[ (
[ ( (2)
43
Vectors - II From (1) and (2), we have
( = ( – (
It is also clear from this expression that ( is a vector in the plane of
and
Now = – (
= – [ – ( )
= ( )
= ( )
So, ( is a vector in the plane of .
Theorem : Let be any three vectors. Then
if and only if are collinear.
Proof : First suppose
Now, (
and ( (
, ( . ) ( . )So a b c b c a
Conversely, suppose that
. Then
– ( –
and – ( –
So, – (
( ( =(
i.e., (
Example 16 : Show that
Solution :
= ( – ( . ) – ( ) + ) – (
= since dot product is commutative.
Example 17 : Let be any four vectors. Then prove that
(i) ( ) = [ ] c
[
(ii) ( ) = [ ] [
44
Vectors and Three
Dimensional Geometry Solution : (i) Let = Then
( ) = )
= (
= (
= [ –
= [
= [
(ii) Let = . Then
( ) = (
(
= – [( r
– (
= (
= (
= ( ) – ( )
= [ , ] – [ ]
Example 18 : Prove that
[ , , a
=
Solution : [ , , a
= ). [ × ) (
= )[ ) )
= ). [ ) a
c
] ( )
= ) [ ) ] = [ a
)][ )]
= [ ]2.
Example 19 : For vectors =
and = verify that
) )
Solution : We have
× =
= 3 + 3 + 3
45
Vectors - II So, (
=
= – 9 –6 – 3 (1)
Also,
= ( –2 + .( + 2 +
= 1 – 4 – 1 = – 4
( ) = – 4 ( + + = – 8 – 4 – 4
and = ( +2 + .( + + = 1
( ) = + 2 –
Thus, ( ) – ( ) = – 6 – (2)
Hence, from (1) and (2)
) = ( ) – ( )
Check Your Progress – 3
1. For vectors = + , = +2 = + +2 find [ , ].
2. Find the volume of the parallelepiped whose edges are represented by
= – 3 , = + = – +2
3. Show that the four points having position vectors
+ , 6 +11 +2 + 6 i + + 4 are coplanar.
4. Prove that
( ). ( ) = ( . )( . ) – ( )( . )
5. For any vector prove that
( ) + ( ) + ( ) = 2
6. Prove that
[ ( )] = ( . ) ( × )
2.5 ANSWERS TO CHECK YOUR PROGRESS
Check Your Progress – 1
1. Here, + = + 2 and
– = – 4 + 2
So, ( + ). ( – ) = ( + 2 – 4 + 2
= 24 – 8 – 16 = 0
Hence + and + are perpendicular vectors.
46
Vectors and Three
Dimensional Geometry 2. (a) Here, . = 3 (– 4) + (– 1). 0 +2.2
= – 12 + 4 = – 8
| | = =
| | = =
(b) Here, . = 1(– 2) + (– 1) 2 + (– 2) 4
= – 12
| | = =
| | = =
θ =
3. Vector Projection of on
2
.
| |
a bb
b
4. The inequality holds trivially if ( ) ( ) 0, 0a b c a b c b
Now, = | | | cos
Hence ≤ | |b
|
5. =
= –
= – 2 +
( dot product is commulative)
47
Vectors - II Check Your Progress – 2
1. Here = 2 +
= 3
Let = . Then is vector perpendicular to both
Now =
= (–6 + 6)
= 12
A unit vector in the direction of is
So, is a unit vector perpendicular to both
2. ) a
( )
= (using distributivity)
= – – (
= – and = – = 0
3. = –
= – 5
Also,
=
= –
Clearly,
(
48
Vectors and Three
Dimensional Geometry 4.
i.e., +
i.e., + 0
(
i.e., = –
i.e., = ……………..(1)
Similarly,
=
i.e., +
i.e., = –
i.e., = ……………. (2)
From (1) and (2), we have
=
5. Let ABCD be the parallelogram
D C
A B
Figure 7
=
So area of parallelogram ABCD
= |
49
Vectors - II
Now, | =
= (2 – 3) – (1+ 6) + (–1 – 4)
= –1 –7 – 5
| ADAB
| =
= 5
Hence, the area of parallelogram ABCD = 5
Check Your Progress – 3
1. = = 3 +5 –7
( = (2 ). (3 )
= 6 + 5 –21
= –10
2. Volume of Parallelopiped = | ( |
Now, ( =
= – 7
Volume = | –7| = 7
3. Let = − + 2
= + + 2
= +2 + 6
So, = – = +12
= OC
– = – 9 + 4
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Vectors and Three
Dimensional Geometry
, and are coplanar.AB BC CD
Hence A, B, C and D are coplanar.
4. Let × = Then
( ). ( ) = r
.( )
= ( r
).
= (( ) ).
= [(
= ( ) – ( . )
= ( ) – ( )
5. L.H.S. = ( ) ( ) ( )
= [ ( ) i [ ( )
= )
= i )
Let =
So, –
Similarly, =
L.H.S. = 3 –
= 3
6.
= .
= (
= – ( ( 0
= ( (
51
Vectors - II 2.6 SUMMARY
This unit discusses various operations on vectors. The binary operation of scalar
product is discussed in section 2.2. In the next section, the binary operation of
vector product (also, called cross product) is illustrated. Finally in section 2.4, the
ternary operation of tripe product of vectors is explained.
Answers/Solutions to questions/problems/exercises given in various sections of
the unit are available in section 2.5.
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