Vector Algebra

™

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

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 1A 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 ABJJJG or

simply as aG, and read as ‘vector AB JJJG

’ or ‘vector aG’.

The point A from where the vector ABJJJG 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

|ABJJJG|, or |aG|, or a. The arrow indicates the direction of the vector.

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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 OPJJJG 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 _OPJJJG (or rG) is given by

| OP |JJJG = _x2_+y2_+z2

In practice, the position vectors of points A, B, C, etc., with respect to the origin O

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 OPJJJG

(

)

Fig 10.3

From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we have cos x

(

)

r

α = G . Similarly, from the right angled triangles OBP and

OCP, we may write cos y and cos 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 rG, and denoted as a, b and c, respectively.

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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 0G. 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 AA, BB

JJJG JJJG

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 aG is denoted by aˆ.

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 aG and bG are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as aG =bG.

Negative of a Vector A vector whose magnitude is the same as that of a given vector (say, _ABJJJG), but direction is opposite to that of it, is called negative of the given vector. For example, vector BAJJJG is negative of the vector ABJJJG, and written as BAJJJG= −ABJJJG.

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 1Represent graphically a displacement of 40 km, 30° west of south.

Solution The vector _OPJJJG represents the required displacement (Fig 10.4).

Example 2 Classify the following measures as scalars and vectors.

(i) 5 seconds (ii) 1000 cm3

Fig 10.5 (iii) 10 Newton (iv) 30 km/hr (v) 10 g/cm3 (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 3In Fig 10.5, which of the vectors are:

(i) Collinear (ii) Equal (iii) Coinitial

Solution

(i) Collinear vectors : a cG G, and dG.

(ii) Equal vectors : aGand cG.

(iii) Coinitial vectors : b cG, G and dG.

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) aG and −aG are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

10.4 Addition of Vectors

A vector ABJJJG 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 ACJJJG and expressed as

AC JJJG

= _ABJJJG JJJG_+BC

This is known as the triangle law of vector addition.

In general, if we have two vectors aG and bG (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 10.8

For example, in Fig 10.8 (ii), we have shifted vector bG without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of aG. Then, the

vector aG+bG, represented by the third side AC of the triangle ABC, gives us the sum

(or resultant) of the vectors aG and bGi.e., in triangle ABC (Fig 10.8 (ii)), we have

AB+BC JJJG JJJG

= ACJJJG

Now again, since ACJJJG= −CAJJJG, from the above equation, we have

AB+BC+CA JJJG JJJG JJJG

= AAJJJG G=0

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)).

Fig 10.7

a

b

a b

(i) (iii)

A

C

a

b

(ii)

a +b

A

C

B B

a b

– –b

Now, construct a vector BCJJJJG′ so that its magnitude is same as the vector BCJJJG, but the direction opposite to that of it (Fig 10.8 (iii)), i.e.,

BC′ JJJJG

= −BC JJJG

Then, on applying triangle law from the Fig 10.8 (iii), we have

AC′=AB+BC′ JJJJG JJJG JJJJG

= ABJJJG+ −( BC)JJJG = −aG bG

The vector ACJJJJG′ is said to represent the difference of aG and bG.

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 aandb

G G

represented

by the two adjacent sides of a parallelogram in magnitude and direction (Fig 10.9), then their

sum aG+bG 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.

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JJJG JJJG = OCJJJG

or _OAJJJG JJJG_+OB = _OCJJJG (since ACJJJG JJJG=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 1For any two vectors aG and bG,

Proof Consider the parallelogram ABCD

(Fig 10.10). Let ABJJJG=aGand BCJJJG=bG, then using the triangle law, from triangle ABC, we have

AC JJJG

= aG+bG

Now, since the opposite sides of a parallelogram are equal and parallel, from

Fig 10.10, we have, AD = BC =JJJG JJJG bG and

DC = AB =a

JJJG JJJG _G

. Again using triangle law, from triangle ADC, we have

AC JJJJG

= AD + DC =JJJG JJJG bG+aG

Hence aG+bG = bG+aG

Property 2For any three vectors a b, andc

G

G G

(aG+bG)+cG = aG+(bG+cG) (Associative property)

Proof Let the vectors a bG, andG cGbe represented by PQ, QR and RSJJJG JJJG JJJG, respectively, as shown in Fig 10.11(i) and (ii).

Fig 10.11

Then aG+bG = PQ + QR = PRJJJG JJJG JJJG

and _bG G_+c = QR + RS = QSJJJG JJJG JJJG

So (aG+bG)+ cG = PR + RS = PSJJJG JJJG JJG

a _{1 2}a _{1 2}

a

–2

a

a 2

and aG+(bG+cG) = PQ + QS = PSJJJG JJJG JJG

Hence (aG+bG)+cG = aG+(bG+cG)

Remark The associative property of vector addition enables us to write the sum of

three vectors a bG , ,G cG as aG+ +bG cG without using brackets.

Note that for any vector aG, we have

0

aG+G = 0G G G+ =a a

Here, the zero vector 0G is called the additive identity for the vector addition.

10.5 Multiplication of a Vector by a Scalar

Let aG be a given vector and λ a scalar. Then the product of the vector aG by the scalar λ, denoted as λaG, is called the multiplication of vector aG by the scalar λ. Note that, λaG is also a vector, collinear to the vector aG. The vector λaG has the direction same (or opposite) to that of vector aG according as the value of λ is positive (or negative). Also, the magnitude of vector λaG is |λ| times the magnitude of the vector aG, i.e.,

|λaG| = | | | |λ aG

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 = −aG aG, which is a vector having magnitude equal to the

magnitude of aG and direction opposite to that of the direction of aG. The vector –aG is called the negative (or additive inverse) of vector aG and we always have

(– )

aG+ aG = (– )aG + =aG 0G

Also, if = 1 |a|

λ _G , provided aG≠0, i.e. aG is not a null vector, then

|λ = λaG| | | | |aG = 1 | | 1 |a| a =

So, λaG represents the unit vector in the direction of aG. We write it as

ˆ a = 1

|a|a G G

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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

| OA | 1, | OB |JJJG = JJJG = 1 and | OC | 1JJJG =

The vectors OA, OB and OCJJJG JJJG JJJG, each having magnitude 1, are called unit vectors along the axes OX, OY and OZ,

respectively, and denoted by _{i j}ˆ ˆ_{, and k}ˆ, respectively (Fig 10.13).

Now, consider the position vector OPJJJG of a point P (x, y, z) as in Fig 10.14. Let P₁ be the foot of the perpendicular from P on the plane XOY. We, thus, see that P₁ P is

parallel to z-axis. As i jˆ ˆ, and kˆ are the unit vectors along the x, y and z-axes,

respectively, and by the definition of the coordinates of P, we have P PJJJG JJJG₁ =OR=zkˆ.

Similarly, _QPJJJG JJJG_1=OS=yjˆ and _OQJJJG_=xiˆ.

Fig 10.13

Therefore, it follows that OPJJJG₁ = _{OQ + QP}JJJG JJJG_1=xiˆ_+yjˆ

and _OPJJJG = OP + P PJJJG JJJJG_{1 1} =xiˆ+yjˆ+zkˆ

Hence, the position vector of P with reference to O is given by

OP (or )JJJG rG = _xiˆ_+yjˆ_+zkˆ

This form of any vector is called its component form. Here, x, y and z are called

as the scalar components of rG, and xiˆ, yjˆ and zkˆ are called the vector components

of rGalong the respective axes. Sometimes x, y and z are also termed as rectangular components.

The length of any vector rG=xiˆ+yjˆ+zkˆ, is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP₁ (Fig 10.14)

1

| OP |JJJJG = 2 2 2 2

1

| OQ | +|QP |JJJJG JJJG = x +y ,

and in the right angle triangle OP₁P, we have

1

| OP | JJJJG

= 2 2 2 2 2

1 1

| OP |JJJG +| P P |JJJG = (x +y )+z

Hence, the length of any vector _rG_=xiˆ_{+yjˆ +zk}ˆ is given by

| |rG = |xiˆ+yjˆ+zkˆ| = x2+y2+z2

If aG and bG are any two vectors given in the component form a i₁ˆ+a j a k₂ˆ + ₃ˆ and

1ˆ 2ˆ 3ˆ

b i+b j+b k, respectively, then

(i) the sum (or resultant) of the vectors aG and bG is given by

aG+bG = (a1+b i1)ˆ+(a2+b2)ˆj+(a3+b k3)ˆ

(ii) the difference of the vector aG and bG is given by

aG−bG= (a1−b i1)ˆ+(a2−b2)ˆj+(a3−b k3)ˆ

(iii) the vectors aG and bG are equal if and only if a₁ =b₁, a₂ = b₂ and a₃ = b₃

(iv) the multiplication of vector aG by any scalar λ is given by

a

The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:

Let aG and bG be any two vectors, and k and m be any scalars. Then (i) _kaG_+maG_{=(k+m a)}G

(ii) k ma( G)=(km a)G

(iii) k a( +b)=ka+kb

G G

G G

Remarks

(i) One may observe that whatever be the value of λ, the vector λaG is always

collinear to the vector aG. In fact, two vectors aG and bG are collinear if and only

if there exists a nonzero scalar λ such that bG= λaG. If the vectors aG and bG are

given in the component form, i.e. a=a i1ˆ+a j2ˆ+a k3ˆ

G

and bG=b i₁ˆ+b j₂ˆ+b k₃ˆ, then the two vectors are collinear if and only if

1ˆ 2ˆ 3ˆ

b i+b j+b k = λ(a i₁ˆ+a j₂ˆ+a k₃ˆ)

⇔ b i1ˆ+b j2ˆ+b k3ˆ = (λa i1)ˆ+ λ( a2)ˆj+ λ( a k3)ˆ

⇔ b1= λa1, b2 = λa2, b3 = λa3

⇔ 1

1

b a =

3 2

2 3

b b

a =a = λ

(ii) If aG=a i₁ˆ+a j₂ˆ+a k₃ˆ, then a₁, a₂, a₃ are also called direction ratios of aG.

(iii) In case if it is given that l, m, n are direction cosines of a vector, then liˆ+mjˆ+nkˆ

= (cos )α +iˆ (cos )β +ˆj (cos )γ kˆ 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 _aG_=xiˆ₊₂ˆ_j+zkˆ and

ˆ

ˆ ˆ

2

bG= i+yj+k are equal.

SolutionNote that two vectors are equal if and only if their corresponding components

Example 5 Let aG= +iˆ 2ˆj and _bG_=2iˆ_{+ j}ˆ. Is | |aG =| |bG ? Are the vectors aG and bG equal?

Solution We have | |aG = 12 +22 = 5 and | |bG = 22+12 = 5

So, | | | |aG = bG . But, the two vectors are not equal since their corresponding components are distinct.

Example 6Find unit vector in the direction of vector aG=2iˆ+3ˆj+kˆ

Solution The unit vector in the direction of a vector aG is given by ˆ 1 | |

a a

a

= G G.

Now |aG| = ₂2 ₊₃2 ₊₁2 _{= 14}

Therefore ˆ 1 (2ˆ 3ˆ ˆ) 14

a= i+ j+k = 2 ˆ 3 ˆ 1 ˆ

14i+ 14 j+ 14k

Example 7 Find a vector in the direction of vector _aG_{= −i}ˆ ₂ˆ_j that has magnitude 7 units.

SolutionThe unit vector in the direction of the given vector aG is

1 ˆ

| |

a a

a

= _G G = 1 (ˆ 2 )ˆ 1 ˆ 2 ˆ 5 i− j = 5i− 5 j

Therefore, the vector having magnitude equal to 7 and in the direction of aG is

7a∧ = 7 1 2 5i 5 j

∧ ∧

⎛ ₋ ⎞

⎜ ⎟

⎝ ⎠ =

7 _ˆ 14 _ˆ

5i− 5 j

Example 8 Find the unit vector in the direction of the sum of the vectors, ˆ

ˆ ˆ

2 2 – 5

aG= i+ j k and bG=2iˆ+ +ˆj 3kˆ.

Solution The sum of the given vectors is

ˆ

ˆ ˆ

( , say) = 4 3 2

+G = + −

G G

a b c i j k

Thus, the required unit vector is

1 1 _{ˆ ˆ ˆ} 4 _ˆ 3 _ˆ 2 _ˆ

ˆ (4 3 2 )

| | 29 29 29 29

c c i j k i j k

c

= G G= + − = + −

Example 9 Write the direction ratio’s of the vector _aG_{= + −i}ˆ ˆ_{j 2k}ˆ and hence calculate its direction cosines.

SolutionNote that the direction ratio’s a, b, c of a vector rG=xiˆ+yjˆ+zkˆ 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

1 1 2

, , as | | 6

| | 6 | | 6 | | 6

a b c

l m n r

r r r

−

= _G = = _G = = _G = G =

Thus, the direction cosines are 1 , 1 , – 2

6 6 6

⎛ ⎞

⎜ ⎟

⎝ ⎠.

10.5.2 Vector joining two points

If P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) are any two points, then the vector joining P₁ and P₂

is the vector P PJJJJG_{1 2} (Fig 10.15).

Joining the points P₁ and P₂ with the origin O, and applying triangle law, from the triangle OP₁P₂, we have

1 1 2

OP +P P JJJG JJJJG

= OP .₂ JJJJG

Using the properties of vector addition, the above equation becomes

1 2

P P JJJJG

= OP₂−OP₁ JJJJG JJJG

i.e. P P_{1 2} JJJJG

= (x i2ˆ+y j2ˆ+z k2ˆ)−(x i1ˆ+y j1ˆ+z k1ˆ)

= _{(x2−x i1)}ˆ_{+(y2−y j1)}ˆ_{+(z2−z k1)}ˆ

The magnitude of vector P PJJJJG_{1 2} is given by

1 2

P P JJJJG

= 2 2 2

2 1 2 1 2 1

(x −x) +(y −y ) +(z −z )

Example 10Find 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 JJJG , given by

PQ JJJG

= ( 1 2)− − iˆ+ − −( 2 3)ˆj+ − −( 4 0)kˆ

i.e. PQ

JJJG

= _{− −3i}ˆ ₅ˆ_{j−4 .k}ˆ

10.5.3 Section formula

Let P and Q be two points represented by the position vectorsOP and OQJJJG JJJG, 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 ORJJJGfor 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 PQJJJG such that mRQJJJG = nPRJJJG,

where m and n are positive scalars, we say that the point R divides PQ JJJG

internally in the ratio of m : n. Now from triangles ORQ and OPR, we have

RQ JJJG

= OQJJJG JJJG G G−OR= −b r

and _PRJJJG = _ORJJJG JJJG G G_{−OP= −r a},

Therefore, we have m b(G G−r) = n r(G G−a) (Why?)

or _rG = mb na

m n

+ + G _G

(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

OR JJJG

= mb na

m n

+ + G _G

Case IIWhen 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 i.e. PR QR ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ m

n is given by

OR JJJG

= mb na

m n

− − G _G

Remark If R is the midpoint of PQ , then m = n. And therefore, from Case I, the

midpoint R of PQJJJG, will have its position vector as

OR JJJG

= 2 aG+bG

Example 11Consider two points P and Q with position vectors OPJJJG=3aG−2bG and

OQJJJG= +aG bG. 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

OR JJJG

= 2( ) (3 2 ) 5

2 1 3

a+b + a− b ₌ a

+

G G

G G G

(ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2:1 is

OR JJJG

= 2( ) (3 2 ) 4 2 1

a b a b

b a

+ − − _{= −}

−

G G

G G _{G G}

Example 12 Show that the points A(2iˆ− +ˆj kˆ), B(iˆ−3ˆj−5 ), C(3kˆ iˆ−4j−4 )kˆ are the vertices of a right angled triangle.

SolutionWe have

AB JJJG

= _{(1 2)− i}ˆ_{+ − +( 3 1)j}ˆ_{+ − −( 5 1)k}ˆ _{= − −i}ˆ ₂ˆ_j−6kˆ

JJJGBC = _{(3 1)− i}ˆ_{+ − +( 4 3)}ˆ_{j+ − +( 4 5)k}ˆ _=2iˆ_{− +}ˆ_{j k}ˆ

and CAJJJG = _{(2 3)− i}ˆ_{+ − +( 1 4)}ˆ_{j+ +(1 4)k}ˆ _{= − +i}ˆ _3jˆ_+5kˆ

Further, note that

2

| AB |JJJG = _{41 6 35 | BC |= + =} JJJG 2 _{+| CA |}JJJG 2

Hence, the triangle is a right angled triangle.

EXERCISE 10.2

1. Compute the magnitude of the following vectors:

1 1 1

ˆ ˆ

ˆ ˆ _{; 2}ˆ ₇ˆ _{3 ;} ˆ ˆ

3 3 3

aG= + +i j k bG= i− j− k cG= i+ j− k

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 2iˆ+3 and ˆj xiˆ+yjˆ 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 aG= −iˆ 2ˆj+k bˆ, G= − +2iˆ 4ˆj+5kˆand cG= −iˆ 6 – 7ˆj kˆ.

7. Find the unit vector in the direction of the vector _aG_{= + +i}ˆ ˆ_{j 2k}ˆ.

8. Find the unit vector in the direction of vector PQ,JJJG where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

9. For given vectors, aG=2iˆ− +ˆj 2 kˆ and bG= − + −iˆ ˆj kˆ, find the unit vector in the

direction of the vector a bG+G.

10. Find a vector in the direction of vector _5iˆ_{− +}ˆ_{j 2k}ˆ which has magnitude 8 units.

11. Show that the vectors _2iˆ₋₃ˆ_{j+4 k}ˆ _and−4iˆ_+6jˆ_−8kˆ are collinear.

12. Find the direction cosines of the vector _iˆ₊₂ˆ_j+3kˆ.

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 _iˆ_{+ +}ˆ_{j k}ˆ 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 _iˆ_+2jˆ_−kˆ _{and –i}ˆ_{+ +}ˆ_{j k}ˆ respectively, in the ratio 2 : 1

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, _aG_{= −3i}ˆ ₄ˆ_{j−4 ,k}ˆ

ˆ ˆ ˆ 2

bG= i− +j k and _cG_{= −i}ˆ ₃ˆ_j−5kˆ, respectively form the vertices of a right angled triangle.

18. In triangle ABC (Fig 10.18), which of the following is not true:

(A) AB + BC + CA = 0JJJG JJJJG JJJG G

(B) AB BC ACJJJG JJJG JJJG+ − =0G

(C) AB BC CA+ − =0 JJJG JJJG JJJG G

(D) AB CB CAJJJG JJJG JJJG G− + =0

19. If aGandbG are two collinear vectors, then which of the following are incorrect:

(A) b= λa, for some scalar λ

G _G

(B) aG= ±bG

(C) the respective components of aGandbG are proportional

(D) both the vectors aGandbG 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

defined as a bG⋅ G = |aG| | | cos ,bG θ

where, θ is the angle between aGand , 0bG ≤ θ ≤ π (Fig 10.19).

If either aG=0 orG bG=0,G then θ is not defined, and in this case,

we define a bG⋅ =G 0

Observations

1. a bG⋅ G is a real number.

2. Let aGandbGbe two nonzero vectors, then a bG⋅ =G 0 if and only if aGandbG are perpendicular to each other. i.e.

0

a bG⋅ = ⇔G aG⊥bG

3. If θ = 0, then a bG⋅ =G | | |aG bG|

In particular, _{a a}G G_{⋅ =| | ,a}G 2 _as θ _{in this case is 0.}

4. If θ = π, then a bG⋅ = −G | | |aG bG|

In particular, _aG_{⋅ −( a}G_{)= −| |a}G 2_{, as} θ _{in this case is} π_.

5. In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆ

ˆ ˆ_{, and ,}

i j k we have

ˆ ˆ ˆ ˆ

i i⋅ = ⋅j j = k kˆ ˆ 1,⋅ =

ˆ ˆ ˆ ˆ

i j⋅ = ⋅j k = k iˆ ˆ⋅ =0

6. The angle between two nonzero vectors

a

G

b

G

cos ,

| || | a b a b

⋅ θ =

G G

G

G or cos–1 . | || |

a b a b

⎛ ⎞

θ = _{⎜ ⎟}

⎝ ⎠

G G

G G

7. The scalar product is commutative. i.e.

a bG⋅G= b aG G⋅ (Why?) Two important properties of scalar product

Property 1 (Distributivity of scalar product over addition) Let a bG, and G cG be any three vectors, then

( )

a bG⋅ G+cG = a bG⋅G + ⋅a cG G

(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 )0 _θ 0 (0 < < 90 )0 _θ 0

(270 < < 360 )0 _θ 0 (180 < < 270 )0 _θ 0

Property 2 Let aG and bG be any two vectors, and λ be any scalar. Then

(λ ⋅a bG) G = (λ ⋅ = λ ⋅a bG) G (a bG G)= ⋅ λaG ( bG)

If two vectors aG and bG are given in component form as a i₁ˆ+a j₂ˆ+a k₃ˆ and

1ˆ 2ˆ 3ˆ

b i+b j+b k, then their scalar product is given as

a bG⋅G = (a i₁ˆ+a j₂ˆ+a k₃ˆ) (⋅ b i₁ˆ+b j₂ˆ+b k₃ˆ)

= a i₁ˆ⋅(b i₁ˆ+b j₂ˆ+b k₃ˆ)+a j b i₂ˆ⋅( ₁ˆ+b j₂ˆ+b k₃ˆ) +a k3ˆ⋅(b i1ˆ+b j2ˆ+b k3ˆ)

= a b i i1 1(ˆ ˆ⋅ +) a b i j1 2(ˆ ˆ⋅ +) a b i k1 3(ˆ⋅ +ˆ) a b j i2 1(ˆ ˆ⋅ +) a b2 2(ˆ ˆj j⋅ +) a b j k2 3(ˆ⋅ ˆ)

+ a b k i3 1(ˆ⋅ +ˆ) a b k j3 2(ˆ⋅ +ˆ) a b k k3 3(ˆ ˆ⋅ )(Using the above Properties 1 and 2)

= a₁b₁ + a₂b₂ + a₃b₃ (Using Observation 5)

Thus a bG⋅G = a b1 1+a b2 2+a b3 3

10.6.2 Projection of a vector on a line

Suppose a vector ABJJJG makes an angle θ with a given directed line l (say), in the

anticlockwise direction (Fig 10.20). Then the projection of ABJJJG on l is a vector pG

(say) with magnitude | AB | cosθ JJJG

, and the direction of pG being the same (or opposite)

to that of the line l, depending upon whether cosθ is positive or negative. The vector pG

is called the projection vector, and its magnitude |pG| is simply called as the projection

of the vector ABJJJG on the directed line l.

For example, in each of the following figures (Fig 10.20 (i) to (iv)), projection vector

of ABJJJG along the line l is vector

AC

JJJG

1. If pˆ is the unit vector along a line l, then the projection of a vector aG on the line l is given by a pG⋅ ˆ .

2. Projection of a vector aG on other vector bG, is given by

ˆ,

a bG⋅ or , or 1 ( )

| | | |

b

a a b

b b ⎛ ⎞ ⋅⎜ ⎟ ⋅ ⎝ ⎠ G G G G G G

3. If θ = 0, then the projection vector of ABJJJG will be ABJJJG itself and if θ = π, then the

projection vector of ABJJJG will be BAJJJG.

4. If = 2 π

θ or =3 2

π

θ , then the projection vector of ABJJJG will be zero vector.

RemarkIf α, β and γ are the direction angles of vector aG=a i₁ˆ+a j₂ˆ+a k₃ˆ, then its direction cosines may be given as

3

1 2

ˆ

cos , cos , and cos

ˆ _{| | | | | |}

| || |

a

a a

a i

a a a

a i

⋅

α = = β = γ =

G

G G G

G

Also, note that | | cos , | |cos and | |cosaG α aG β aG γ are respectively the projections of aG along OX, OY and OZ. i.e., the scalar components a₁, a₂ and a₃ of the vector aG, are precisely the projections of aG along x-axis, y-axis and z-axis, respectively. Further, if aG is a unit vector, then it may be expressed in terms of its direction cosines as

ˆ

ˆ ˆ

cos cos cos

aG= α +i β +j γk

Example 13Find the angle between two vectors aG and bG with magnitudes 1 and 2

respectively and when a bG⋅ =G 1.

SolutionGiven a b⋅ =1, | | 1 and | |a = b =2

G G

G G

. We have

1 1 1

Example 14Find angle ‘θ’ between the vectors aG= + −iˆ ˆj kˆ and b iG= − +ˆ ˆj kˆ.

SolutionThe angle θ between two vectors aG and bG is given by

cosθ = | || |

a b a b

⋅G G

G G

Now a bG⋅G = (iˆ+ − ⋅ − +ˆj kˆ) (iˆ ˆj kˆ) 1 1 1= − − = −1.

Therefore, we have cosθ = 1 3 −

hence the required angle is θ = cos 1 1 3

− ⎛₋ ⎞

⎜ ⎟ ⎝ ⎠

Example 15 If _aG_{= − −5i}ˆ ˆ_{j 3 k}ˆ _{and b}G_{= +i}ˆ ₃ˆ_j−5kˆ, then show that the vectors

and

aG+bG aG−bG are perpendicular.

SolutionWe know that two nonzero vectors are perpendicular if their scalar product is zero.

Here aG+bG = (5iˆ− −ˆj 3 )kˆ +(iˆ+3ˆj−5 )kˆ =6iˆ+2ˆj−8kˆ

and aG−bG = (5iˆ− −ˆj 3 )kˆ −(iˆ+3ˆj−5 )kˆ =4iˆ−4ˆj+2kˆ

So (aG+bG) (⋅ aG−bG)=(6iˆ+2ˆj−8 ) (4kˆ ⋅ iˆ−4ˆj+2 )kˆ =24 8 16− − =0.

Hence aG+bG and aG−bG are perpendicular vectors.

Example 16 Find the projection of the vector aG=2iˆ+3ˆj+2kˆ on the vector

ˆ ˆ ₂ˆ bG= +i j+k .

SolutionThe projection of vector aG on the vector bG is given by

1

( )

| |b a b⋅

G G

G =

2 2 2

(2 1 3 2 2 1) 10 5

6 3 6 (1) (2) (1)

× + × + ×

= =

+ +

Example 17Find |aG−bG|, if two vectors aGandbG are such that | |a =2, | |b =3 G G

and a bG⋅ =G 4.

SolutionWe have

2

|aG−bG| = (aG− ⋅bG) (aG−bG)

B

C

A

a + b

a

b

= | |aG 2−2(a bG⋅ +G) | |bG 2

= (2)2 −2(4)+(3)2

Therefore |aG−bG| = ₅

Example 18 If aG is a unit vector and (xG G−a) (⋅ +G Gx a)=8, then find | |xG .

SolutionSince aG is a unit vector, |aG| 1= . Also, (xG G− ⋅ +a) (xG Ga) = 8

or x xG G⋅ + ⋅ − ⋅ − ⋅x aG G G Ga x a aG G = 8

or | |xG 2−1 = 8 i.e. |xG|2 _{= 9}

Therefore | |xG = 3 (as magnitude of a vector is non negative).

Example 19For any two vectors aGandbG, we always have |a bG⋅G| | | | |≤ a bG G (Cauchy-Schwartz inequality).

SolutionThe inequality holds trivially when either aG=0 or G bG=0G. Actually, in such a situation we have |a bG⋅G|= =0 | | | |a bG G . So, let us assume that | |aG ≠ ≠0 | |bG . Then, we have

| |

| || | a b a b

⋅G G

G

G = | cos | 1θ ≤

Therefore |a bG⋅G|≤ | | | |a bG G

Example 20 For any two vectors aGandbG, we always

have |aG+bG| | |≤ aG +| |bG (triangle inequality).

SolutionThe inequality holds trivially in case either

0 or 0

aG=G bG=G (How?). So, let | |aG ≠ ≠0G | |bG . Then,

2

|aG+bG| = (aG+bG)2 =(aG+bG) (⋅ +aG bG)

= a aG G⋅ + ⋅ + ⋅ + ⋅a bG G b aG G b bG G

= | |aG 2 + ⋅ +2a bG G | |bG 2 (scalar product is commutative)

≤ 2 2

| |aG +2 |a bG⋅G|+| |bG (since x≤|x|∀ ∈x R) ≤ _{| |a}G 2_{+2 | ||a b}G G_{| |+ b}G_|2

(from Example 19)

= (| |aG +|bG|)2

Hence |aG+bG|≤ | |aG +|bG|

RemarkIf the equality holds in triangle inequality (in the above Example 20), i.e.

|aG+bG| = | |aG +| |bG ,

then | AC |JJJG = | AB |JJJG +| BC |JJJG showing that the points A, B and C are collinear.

Example 21Show that the points _{A ( 2− +i}ˆ ₃ˆ_{j+5 ), B(k}ˆ _iˆ₊₂ˆ_{j+3 )k}ˆ and _C(7iˆ_−kˆ₎ are collinear.

SolutionWe have

AB JJJG

=(1 2)+ iˆ+(2−3)ˆj+ −(3 5)kˆ=3iˆ− −jˆ 2kˆ,

BC JJJG

= (7 1)− iˆ+(0−2)ˆj+ − −( 1 3)kˆ=6iˆ−2ˆj−4kˆ,

AC JJJG

= (7+2)iˆ+(0−3)ˆj+ − −( 1 5)kˆ=9iˆ−3ˆj−6kˆ

| AB |JJJG = 14, | BC |JJJG =2 14 and | AC | 3 14JJJG =

Therefore AC

JJJG

=| AB |+| BC | JJJG JJJG

Hence the points A, B and C are collinear.

$

EXERCISE 10.3

1. Find the angle between two vectors aG and bGwith magnitudes

3 and 2

respectively having _{a b}G⋅ =G ₆.

2. Find the angle between the vectors

i

ˆ

− +

2

ˆ

j

3

k

ˆ

3

i

ˆ

− +

2

ˆ

j k

ˆ

3. Find the projection of the vector _iˆ₋ ˆ_j on the vector _iˆ₊ˆ_j.

4. Find the projection of the vector _iˆ_+3jˆ_+7kˆ on the vector _7iˆ_{− +}ˆ_{j 8k}ˆ. 5. Show that each of the given three vectors is a unit vector:

1 _{ˆ ˆ ˆ} 1 _{ˆ ˆ ˆ} 1 _{ˆ ˆ ˆ}

(2 3 6 ), (3 6 2 ), (6 2 3 )

7 i+ j+ k 7 i− j+ k 7 i+ j− k

6. Find | | and |aG bG|, if (aG+bG) (⋅ aG−bG)=8 and | aG| 8 | |= bG .

7. Evaluate the product (3aG−5 ) (2bG ⋅ aG+7 )bG .

8. Find the magnitude of two vectors aG and bG, having the same magnitude and

such that the angle between them is 60o _{and their scalar product is} 1

2.

9. Find | |xG , if for a unit vector aG, (G Gx−a) (⋅ +xG Ga) 12= .

10. If aG=2iˆ+2ˆj+3 ,kˆ bG= − +iˆ 2ˆj+kˆ and cG=3iˆ+ ˆjare such that aG+ λbG is

perpendicular to cG, then find the value of λ.

11. Show that | |a bG G+|b aG|G is perpendicular to | |a bG G−| |b aG G, for any two nonzero

vectors aG and bG.

12. If a aG G⋅ =0 and a bG⋅ =G 0, then what can be concluded about the vector bG?

13. If a b cG, ,G G are unit vectors such that aG+ + =bG cG 0G, find the value of

a bG⋅ + ⋅ + ⋅G Gb cG G Gc a.

14. If either vector aG=0 or G bG=G0, then a bG⋅ =G 0. 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 BAJJJG

and BCJJJG].

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 2iˆ− +ˆj k iˆ, ˆ−3ˆj−5kˆ and 3iˆ−4ˆj−4kˆ form the vertices of a right angled triangle.

18. If aG is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λaG 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

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 3The vector product of two nonzero vectors aGandbG, is denoted by aG×bG and defined as

a bG×G = | || | sina bG G θnˆ,

where, θ is the angle between aGandbG, 0≤ θ ≤ π and nˆ is

a unit vector perpendicular to both aG and bG, such that

ˆ , and

a bG G n form a right handed system (Fig 10.23). i.e., the

right handed system rotated from aG tobG moves in the

direction of nˆ.

If either aG=G0 orbG=0G, then θ is not defined and in this case, we define a bG× =G G0.

Observations

1. a bG×G is a vector.

2. Let aGandbG be two nonzero vectors. Then a bG× =G G0 if and only if aG and bG are parallel (or collinear) to each other, i.e.,

a bG×G = 0G⇔a bG&G

In particular, aG G× =a G0 and aG× − =( aG) 0G, since in the first situation, θ = 0 and in the second one, θ = π, making the value of sinθ to be 0.

3. If 2 π

θ = then aG× =bG | ||a bG G|.

4. In view of the Observations 2 and 3, for mutually perpendicular

unit vectors iˆ ˆ, j and kˆ (Fig 10.24), we have

ˆ ˆ

i×i = ˆj× = × =ˆj k kˆ ˆ 0G

ˆ ˆ

i× j_{= k}ˆ_, ˆ_{j k× =}ˆ _iˆ_{, k i}ˆ_{× =}ˆ _jˆ

5. In terms of vector product, the angle between two vectors aG and bG may be given as

sinθ = | | | || | a b a b

×G G

G G

6. It is always true that the vector product is not commutative, as a bG×G = − ×b aG G.

Indeed, a bG× =G | || | sina bG G θnˆ, where a bG, and G nˆ form a right handed system,

i.e., θ is traversed from aG tobG, Fig 10.25 (i). While, bG× =aG | ||a bG G| sinθnˆ₁, where

1

ˆ , and

b aG G n form a right handed system i.e. θ is traversed from bG to aG, Fig 10.25(ii).

Fig 10.25 (i), (ii)

Thus, if we assume aGandbG to lie in the plane of the paper, then nˆ and nˆ₁ both

will be perpendicular to the plane of the paper. But, nˆ being directed above the

paper while nˆ1 directed below the paper. i.e. nˆ₁= −nˆ.

Hence a bG×G = | || | sina bG G θnˆ

= −| || | sina b θnˆ1

G G

b a

= − ×G G 7. In view of the Observations 4 and 6, we have

ˆ ˆ ˆ

ˆ ˆ_{j i× = −k, k× = −}ˆ_{j i}ˆ _{and i}ˆ_{× = −k} ˆ_j.

8. If aandb

G G

represent the adjacent sides of a triangle then its area is given as

1

| |

2 a×b G G

.

By definition of the area of a triangle, we have from Fig 10.26,

Area of triangle ABC = 1AB CD.

2 ⋅

But AB |= bG| (as given), and CD = | |aG sinθ.

Thus, Area of triangle ABC = 1| || | sin

2 b a θ

G G 1

| | . 2 a b = G×G

9. If aG and bG represent the adjacent sides of a parallelogram, then its area is

given by |a bG×G|.

From Fig 10.27, we have

Area of parallelogram ABCD = AB. DE.

But AB=| |bG (as given), and

DE=| | sinaG θ.

Thus,

Area of parallelogram ABCD = | || | sinb aG G θ = ×|a bG G| .

We now state two important properties of vector product.

Property 3 (Distributivity of vector product over addition): If a bG, Gand cG are any three vectors and λ be a scalar, then

(i) aG×(bG+cG) = aG× + ×bG a cG G

(ii) λ ×(a bG G) = (λ × = × λaG) bG aG ( bG)

Fig 10.26

Let aGandbG be two vectors given in component form as a i₁ˆ+a j₂ˆ+a k₃ˆand

1ˆ 2ˆ 3ˆ

b i+b j+b k, respectively. Then their cross product may be given by

a bG×G = 1 2 3

1 2 3

ˆ

ˆ ˆ

i j k

a a a

b b b

Explanation We have

a bG×G = (a i₁ˆ+a j₂ˆ+a k₃ˆ) (× b i₁ˆ+b j₂ˆ+b k₃ˆ)

= a b i1 1(ˆ ˆ× +i) a b i1 2(ˆ× +ˆj) a b i1 3(ˆ× +kˆ) a b j i2 1(ˆ ˆ× )

+ a b2 2(ˆj× +ˆj) a b j k2 3(ˆ×ˆ)

+ _{a b k i3 1(}ˆ_{× +}ˆ_{) a b k3 2(}ˆ_{× +}ˆ_{j) a b k3 3(}ˆ_×kˆ₎ (by Property 1)

= a b i_{1 2}(ˆ× −ˆj) a b k i_{1 3}(ˆ× −ˆ) a b i_{2 1}(ˆ×ˆj)

+ _{a b j k2 3(}ˆ_{× +}ˆ_{) a b k i3 1(}ˆ_{× −}ˆ_{) a b3 2(}ˆ_{j k×} ˆ₎

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

(as i× = × = × =i j j k k 0 and i× = − ×k k i j i, × = − ×i j and k× = − ×j j k)

= a b k_{1 2}ˆ−a b j_{1 3}ˆ−a b k_{2 1}ˆ+a b i_{2 3}ˆ+a b j_{3 1}ˆ−a b i_{3 2}ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

(as i× =j k, j k× =i and k i× =j)

= (a b_{2 3}−a b i_{3 2})ˆ−(a b_{1 3}−a b j_{3 1})ˆ+(a b_{1 2}−a b k_{2 1})ˆ

= _{1 2 3}

1 2 3

ˆ

ˆ ˆ

i j k

a a a

b b b

Example 22Find |a bG×G|, if aG=2iˆ+ +ˆj 3 and kˆ bG=3iˆ+5jˆ−2kˆ

SolutionWe have

a bG×G =

ˆ ˆ ˆ

2 1 3

3 5 2

i j k

−

= iˆ( 2 15)− − − − −( 4 9)ˆj+(10 – 3)kˆ _{= −17i}ˆ₊₁₃ˆ_j+7kˆ

Hence | × | G G

a b = 2 2 2

Example 23 Find a unit vector perpendicular to each of the vectors (a bG+G) and

(a bG−G),where aG= + +iˆ ˆj kˆ, bG= +iˆ 2ˆj+3kˆ.

SolutionWe have _{a b}G_{+ =}G _2iˆ_+3jˆ_{+4 andk}ˆ _aG_{− = − −b}G ˆ_{j 2k}ˆ

A vector which is perpendicular to both a bG+G anda bG−G is given by

(a bG+ ×G) (a bG−G) =

ˆ

ˆ ˆ

ˆ

ˆ ˆ

2 3 4 2 4 2 ( , say)

0 1 2

i j k

i j k c

= − + − =

− −

G

Now | |cG = 4 16+ + =4 24=2 6 Therefore, the required unit vector is

| | c c

G

G = 1ˆ 2 ˆ 1 ˆ

6i 6 j 6k

− _{+ −}

$

6i− 6 j+ 6k But that will

be a consequence of (a bG− ×G) (a bG+G).

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 _ABJJJG_{= +}ˆ_{j 2k}ˆ _{and AC}JJJG_{= +i}ˆ ₂ˆ_j. The area of the given triangle

is 1| AB AC |

2 ×

JJJG JJJG .

Now, AB ACJJJG JJJG× =

ˆ ˆ ˆ

ˆ

ˆ ˆ

0 1 2 4 2

1 2 0

i j k

i j k

= − + −

Therefore | AB AC |× JJJG JJJG

= _{16 4 1}+ + = ₂₁

Example 25 Find the area of a parallelogram whose adjacent sides are given

by the vectors _aG_{= + +3i}ˆ ˆ_{j 4k}ˆ _andbG_{= − +i}ˆ ˆ_{j k}ˆ

SolutionThe area of a parallelogram with aandb

G G

as its adjacent sides is given

by |a bG×G|.

Now a bG×G =

ˆ

ˆ ˆ

ˆ ˆ ˆ

3 1 4 5 4

1 1 1

i j k

i j k

= + − −

Therefore |a bG×G| = 25 1 16+ + = 42

and hence, the required area is 42.

EXERCISE 10.4

1. Find |a bG×G|, if aG= −iˆ 7ˆj+7 andkˆ bG=3iˆ−2ˆj+2kˆ.

2. Find a unit vector perpendicular to each of the vector a bG+G anda bG−G, where

ˆ ˆ

ˆ ˆ ˆ ˆ

3 2 2 and 2 2

aG= i+ j+ k bG= +i j− k .

3. If a unit vector aG makes angles with ,ˆ with ˆ

3 i 4 j

π π

and an acute angle θ with

ˆ

k, then find θ and hence, the components of aG. 4. Show that

(a bG− ×G) (a bG+G) = 2(a bG×G)

5. Find λ and μ if (2iˆ+6ˆj+27 ) (kˆ × + λ + μ =iˆ ˆj kˆ) 0G.

6. Given that a bG⋅ =G 0 and a bG× =G G0. What can you conclude about the vectors

and aG bG?

7. Let the vectors a b cG, ,G G be given as a i₁ˆ+a j₂ˆ+a k b i₃ˆ, ₁ˆ+b j₂ˆ+b k₃ˆ,

1ˆ 2ˆ 3ˆ

c i+c j+c k. Then show that aG×(bG+cG)= × + ×a bG G a cG G.

8. If either aG=0 or G bG=0,G then a bG× =G G0. Is the converse true? Justify your answer with an example.

10. Find the area of the parallelogram whose adjacent sides are determined by the vectors _aG_{= − +i}ˆ ˆ_{j 3k}ˆ and _bG_=2iˆ₋₇ˆ_j+kˆ.

11. Let the vectors aG and bG be such that | | 3 and | | 2 3

aG = bG = , then a bG×G is a

unit vector, if the angle between aG and bG 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

1 _ˆ 1 _ˆ

ˆ ˆ ˆ ˆ

– 4 , 4

2 2

i+ j+ k i+ j+ k, ˆ 1 ˆ 4ˆ 2

i− j+ k and –ˆ 1 ˆ 4ˆ 2

i− j+ k, respectively is

(A) 1

2 (B) 1

(C) 2 (D) 4

Miscellaneous Examples

Example 26Write all the unit vectors in XY-plane.

SolutionLet rG= +x i∧ y j∧ be a unit vector in XY-plane (Fig 10.28). Then, from the figure, we have x = cos θ and y = sin θ (since |rG| = 1). So, we may write the vector rG as

(

)

rG =JJJG = cosθ +iˆ sinθ ˆj ... (1)

Clearly, | |rG = 2 2

cos θ +sin θ =1

Fig 10.28

Example 27 If iˆ+ +ˆj kˆ, 2iˆ+5 , 3ˆj iˆ+2ˆj−3 and kˆ iˆ−6ˆj−kˆ are the position

vectors of points A, B, C and D respectively, then find the angle between ABJJJG and

CD JJJG

. Deduce that ABJJJG and CDJJJG are collinear.

SolutionNote that if θ is the angle between AB and CD, then θ is also the angle

between

AB and CD

JJJG

JJJG

Now ABJJJG = Position vector of B – Position vector of A

= (2iˆ+5 )ˆj − + +(iˆ ˆj kˆ)= +iˆ 4ˆj−kˆ

Therefore | AB |JJJG = ₍₁₎2₊₍₄₎2 _{+ −( 1)}2 _{=3 2}

Similarly _CDJJJG = _{− −2i}ˆ ₈ˆ_{j+2 and |CD | 6 2k}ˆ JJJG ₌

Thus cosθ = AB CD

|AB||CD| ⋅ JJJG JJJG JJJG JJJG

= 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 AB JJJG

and CD JJJG

are collinear.

Alternatively, AB 1CD 2 = − JJJG JJJG

which implies that AB and CDJJJG JJJG are collinear vectors.

Example 28Let a bG,G and cG be three vectors such that | | 3, | | 4, | | 5aG = bG = cG = and

each one of them being perpendicular to the sum of the other two, find

|

a b c

G

+ +

G

G

|

SolutionGiven a bG⋅(G+cG) = 0, bG⋅ +(cG Ga)=0, cG G⋅ +(a bG)=0.

Now |aG+ +bG cG|2 = (aG+ +bG cG)2 =(aG+ + ⋅ + +bG cG) (aG bG cG)

= a aG G G⋅ + ⋅ +a b(G cG)+ ⋅ + ⋅b bG G Gb (aG G+c)

+ c aG G.( +bG)+c cG G.

= _{| |a}G 2 _{+| |b}G 2_{+| |c}G 2

= 9 + 16 + 25 = 50

Example 29Three vectors a bG, and G cG satisfy the condition aG+ + =bG cG 0G. Evaluate

the quantity μ = ⋅ + ⋅ + ⋅a bG G Gb cG G Gc a, if | | 1, | | 4 and | | 2aG = bG = cG = .

SolutionSince aG+ + =bG cG 0G, we have

( )

a aG G⋅ + +bG cG _{= 0}

or a aG G⋅ + ⋅ + ⋅a bG G a cG G = 0

Therefore a b⋅ + ⋅a c

G _G

G G

= − aG 2 = −1 ... (1)

Again, bG⋅

(

)

or a bG⋅ + ⋅G G Gb c =

2

16 b

− G = − ... (2)

Similarly a cG G⋅ + ⋅b cG G = – 4. ... (3)

Adding (1), (2) and (3), we have

2 (a b⋅ + ⋅ + ⋅b c a c) G G _G G _G G

= – 21

or 2μ = – 21, i.e., μ = 21 2 −

Example 30If with reference to the right handed system of mutually perpendicular

unit vectors i jˆ ˆ, and ,kˆ α =G 3iˆ−ˆj, β =G 2iˆ+ ˆj– 3kˆ, then express βG in the form

1 2, where 1

β = β + βG G G βG is parallel to αG and βG₂ is perpendicular to

α

G

SolutionLet β = λα λ₁ ,

G _G

is a scalar, i.e., β = λ − λG₁ 3 iˆ ˆj.

Now β = β − βG₂ G G₁ = (2 3 )− λ + + λ −iˆ (1 )ˆj 3kˆ.

Now, since βG₂ is to be perpendicular to

α

G

3(2− λ − + λ3 ) (1 ) = 0

or λ = 1

2

Therefore βG₁ = 3ˆ 1 ˆ

2i−2 j and 2

1_ˆ 3 _ˆ

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 (x₁, y₁, z₁) and Q (x₂, y₂, z₂).

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 aG= +bG cG, then is it true that | | | | | |aG =bG + cG ? Justify your answer.

5. Find the value of x for which _{x i(}ˆ_{+ +}ˆ_{j k}ˆ₎ is a unit vector.

6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

ˆ ˆ

ˆ ˆ ˆ ˆ

2 3 and 2

aG= i+ j−k bG= −i j+k.

7. If _aG_{= + +i}ˆ ˆ_{j k}ˆ_{, b}G_=2iˆ_{− +j}ˆ _3kˆ _{and c}G_{= −i}ˆ _2jˆ_+kˆ, find a unit vector parallel

to the vector 2 – 3aG bG+ cG.

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(2aG+bG) and ( – 3 )aG bG 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 _2iˆ₋₄ˆ_{j+5 and k}ˆ _iˆ₋₂ˆ_j−3kˆ. 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 , 1 , 1 .

3 3 3

12. Let aG= +iˆ 4ˆj+2 ,k bˆ G= −3iˆ 2jˆ+7 kˆ and cG=2iˆ− +ˆj 4kˆ. Find a vector dG

which is perpendicular to both aG and bG, and c dG⋅ =G 15.

13. The scalar product of the vector _iˆ_{+ +}ˆ_{j k}ˆ with a unit vector along the sum of

vectors _2iˆ₊₄ˆ_j−5kˆ and λ +iˆ 2ˆj+3kˆ is equal to one. Find the value of λ.

14. If a bG, G G, c are mutually perpendicular vectors of equal magnitudes, show that

15. Prove that ₍