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1 | 5118-5121 | Then, the
vector a
+b
r
r
, represented by the third side AC of the triangle ABC, gives us the sum
(or resultant) of the vectors ar and b
r
i e , in triangle ABC (Fig 10 8 (ii)), we have
AB
BC
+
uuur
uuur = AC
uuur
Now again, since AC
CA
= −
uuur
uuur
, from the above equation, we have
AB
BC
CA
+
+
uuur
uuur
uuur
= AA
0
=
uuur
r
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 |
1 | 5119-5122 | e , in triangle ABC (Fig 10 8 (ii)), we have
AB
BC
+
uuur
uuur = AC
uuur
Now again, since AC
CA
= −
uuur
uuur
, from the above equation, we have
AB
BC
CA
+
+
uuur
uuur
uuur
= AA
0
=
uuur
r
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)) |
1 | 5120-5123 | , in triangle ABC (Fig 10 8 (ii)), we have
AB
BC
+
uuur
uuur = AC
uuur
Now again, since AC
CA
= −
uuur
uuur
, from the above equation, we have
AB
BC
CA
+
+
uuur
uuur
uuur
= AA
0
=
uuur
r
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 |
1 | 5121-5124 | 8 (ii)), we have
AB
BC
+
uuur
uuur = AC
uuur
Now again, since AC
CA
= −
uuur
uuur
, from the above equation, we have
AB
BC
CA
+
+
uuur
uuur
uuur
= AA
0
=
uuur
r
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
C’
© NCERT
not to be republished
MATHEMATICS
430
Now, construct a vector BC′
uuuur
so that its magnitude is same as the vector BC
uuur
, but
the direction opposite to that of it (Fig 10 |
1 | 5122-5125 | 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
C’
© NCERT
not to be republished
MATHEMATICS
430
Now, construct a vector BC′
uuuur
so that its magnitude is same as the vector BC
uuur
, but
the direction opposite to that of it (Fig 10 8 (iii)), i |
1 | 5123-5126 | Fig 10 7
a
b
a
b
(i)
(iii)
A
C
a
b
(ii)
a
b
+
A
C
B
B
a
b
–
–b
C’
© NCERT
not to be republished
MATHEMATICS
430
Now, construct a vector BC′
uuuur
so that its magnitude is same as the vector BC
uuur
, but
the direction opposite to that of it (Fig 10 8 (iii)), i e |
1 | 5124-5127 | 7
a
b
a
b
(i)
(iii)
A
C
a
b
(ii)
a
b
+
A
C
B
B
a
b
–
–b
C’
© NCERT
not to be republished
MATHEMATICS
430
Now, construct a vector BC′
uuuur
so that its magnitude is same as the vector BC
uuur
, but
the direction opposite to that of it (Fig 10 8 (iii)), i e ,
BC′
uuuur
=
BC
−
uuur
Then, on applying triangle law from the Fig 10 |
1 | 5125-5128 | 8 (iii)), i e ,
BC′
uuuur
=
BC
−
uuur
Then, on applying triangle law from the Fig 10 8 (iii), we have
AC
AB
BC
′
′
=
+
uuuur
uuur
uuuur = AB
( BC)
+ −
uuur
uuur
a
b
=
−
r
r
The vector AC′
uuuur is said to represent the difference of
aand
rb
r |
1 | 5126-5129 | e ,
BC′
uuuur
=
BC
−
uuur
Then, on applying triangle law from the Fig 10 8 (iii), we have
AC
AB
BC
′
′
=
+
uuuur
uuur
uuuur = AB
( BC)
+ −
uuur
uuur
a
b
=
−
r
r
The vector AC′
uuuur is said to represent the difference of
aand
rb
r 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 |
1 | 5127-5130 | ,
BC′
uuuur
=
BC
−
uuur
Then, on applying triangle law from the Fig 10 8 (iii), we have
AC
AB
BC
′
′
=
+
uuuur
uuur
uuuur = AB
( BC)
+ −
uuur
uuur
a
b
=
−
r
r
The vector AC′
uuuur is said to represent the difference of
aand
rb
r 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 |
1 | 5128-5131 | 8 (iii), we have
AC
AB
BC
′
′
=
+
uuuur
uuur
uuuur = AB
( BC)
+ −
uuur
uuur
a
b
=
−
r
r
The vector AC′
uuuur is said to represent the difference of
aand
rb
r 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 |
1 | 5129-5132 | 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 |
1 | 5130-5133 | 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 |
1 | 5131-5134 | 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 |
1 | 5132-5135 | 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
aand
rb
r
represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10 |
1 | 5133-5136 | e , the resultant velocity) of the boat, we have
the following law of vector addition If we have two vectors
aand
rb
r
represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10 9), then their
sum
a+
rb
r
is represented in magnitude and
direction by the diagonal of the parallelogram
through their common point |
1 | 5134-5137 | , the resultant velocity) of the boat, we have
the following law of vector addition If we have two vectors
aand
rb
r
represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10 9), then their
sum
a+
rb
r
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 |
1 | 5135-5138 | If we have two vectors
aand
rb
r
represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10 9), then their
sum
a+
rb
r
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 |
1 | 5136-5139 | 9), then their
sum
a+
rb
r
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
+
uuur
uuur = OC
uuur
or
OA
OB
+
uuur
uuur = OC
uuur (since AC
uuur=OB
uuur
)
which is parallelogram law |
1 | 5137-5140 | 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
+
uuur
uuur = OC
uuur
or
OA
OB
+
uuur
uuur = OC
uuur (since AC
uuur=OB
uuur
)
which is parallelogram law Thus, we may say that the two laws of vector
addition are equivalent to each other |
1 | 5138-5141 | �Note From Fig 10 9, using the triangle law, one may note that
OA
AC
+
uuur
uuur = OC
uuur
or
OA
OB
+
uuur
uuur = OC
uuur (since AC
uuur=OB
uuur
)
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
aand
rb
r
,
a
b
+
r
r
= b
a
+
r
r
(Commutative property)
Fig 10 |
1 | 5139-5142 | 9, using the triangle law, one may note that
OA
AC
+
uuur
uuur = OC
uuur
or
OA
OB
+
uuur
uuur = OC
uuur (since AC
uuur=OB
uuur
)
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
aand
rb
r
,
a
b
+
r
r
= b
a
+
r
r
(Commutative property)
Fig 10 9
© NCERT
not to be republished
VECTOR ALGEBRA
431
Proof Consider the parallelogram ABCD
(Fig 10 |
1 | 5140-5143 | 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
aand
rb
r
,
a
b
+
r
r
= b
a
+
r
r
(Commutative property)
Fig 10 9
© NCERT
not to be republished
VECTOR ALGEBRA
431
Proof Consider the parallelogram ABCD
(Fig 10 10) |
1 | 5141-5144 | Properties of vector addition
Property 1 For any two vectors
aand
rb
r
,
a
b
+
r
r
= b
a
+
r
r
(Commutative property)
Fig 10 9
© NCERT
not to be republished
VECTOR ALGEBRA
431
Proof Consider the parallelogram ABCD
(Fig 10 10) Let AB
and BC
,
a
b
uuur
uuur
r
r
then using
the triangle law, from triangle ABC, we have
AC
uuur =
a+
rb
r
Now, since the opposite sides of a
parallelogram are equal and parallel, from
Fig 10 |
1 | 5142-5145 | 9
© NCERT
not to be republished
VECTOR ALGEBRA
431
Proof Consider the parallelogram ABCD
(Fig 10 10) Let AB
and BC
,
a
b
uuur
uuur
r
r
then using
the triangle law, from triangle ABC, we have
AC
uuur =
a+
rb
r
Now, since the opposite sides of a
parallelogram are equal and parallel, from
Fig 10 10, we have, AD = BC = b
uuur
uuur
r
and
DC = AB = a
uuur
uuur
r |
1 | 5143-5146 | 10) Let AB
and BC
,
a
b
uuur
uuur
r
r
then using
the triangle law, from triangle ABC, we have
AC
uuur =
a+
rb
r
Now, since the opposite sides of a
parallelogram are equal and parallel, from
Fig 10 10, we have, AD = BC = b
uuur
uuur
r
and
DC = AB = a
uuur
uuur
r Again using triangle law, from
triangle ADC, we have
AC
uuuur = AD + DC =
b+
a
uuur
uuur
r
r
Hence
a
b
+
r
r
= b
a
+
r
r
Property 2 For any three vectors ,
a band
c
rr
r
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
(Associative property)
Proof Let the vectors ,
a band
c
rr
r be represented by PQ, QR and RS
uuur
uuur
uuur , respectively,
as shown in Fig 10 |
1 | 5144-5147 | Let AB
and BC
,
a
b
uuur
uuur
r
r
then using
the triangle law, from triangle ABC, we have
AC
uuur =
a+
rb
r
Now, since the opposite sides of a
parallelogram are equal and parallel, from
Fig 10 10, we have, AD = BC = b
uuur
uuur
r
and
DC = AB = a
uuur
uuur
r Again using triangle law, from
triangle ADC, we have
AC
uuuur = AD + DC =
b+
a
uuur
uuur
r
r
Hence
a
b
+
r
r
= b
a
+
r
r
Property 2 For any three vectors ,
a band
c
rr
r
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
(Associative property)
Proof Let the vectors ,
a band
c
rr
r be represented by PQ, QR and RS
uuur
uuur
uuur , respectively,
as shown in Fig 10 11(i) and (ii) |
1 | 5145-5148 | 10, we have, AD = BC = b
uuur
uuur
r
and
DC = AB = a
uuur
uuur
r Again using triangle law, from
triangle ADC, we have
AC
uuuur = AD + DC =
b+
a
uuur
uuur
r
r
Hence
a
b
+
r
r
= b
a
+
r
r
Property 2 For any three vectors ,
a band
c
rr
r
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
(Associative property)
Proof Let the vectors ,
a band
c
rr
r be represented by PQ, QR and RS
uuur
uuur
uuur , respectively,
as shown in Fig 10 11(i) and (ii) Fig 10 |
1 | 5146-5149 | Again using triangle law, from
triangle ADC, we have
AC
uuuur = AD + DC =
b+
a
uuur
uuur
r
r
Hence
a
b
+
r
r
= b
a
+
r
r
Property 2 For any three vectors ,
a band
c
rr
r
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
(Associative property)
Proof Let the vectors ,
a band
c
rr
r be represented by PQ, QR and RS
uuur
uuur
uuur , respectively,
as shown in Fig 10 11(i) and (ii) Fig 10 11
Then
a
b
+
r
r
= PQ + QR = PR
uuur
uuur
uuur
and
b
c
+
r
r = QR + RS = QS
uuur
uuur
uuur
So
(
)
a
b
c
+
r+
r
r = PR + RS = PS
uuur
uuur
uur
Fig 10 |
1 | 5147-5150 | 11(i) and (ii) Fig 10 11
Then
a
b
+
r
r
= PQ + QR = PR
uuur
uuur
uuur
and
b
c
+
r
r = QR + RS = QS
uuur
uuur
uuur
So
(
)
a
b
c
+
r+
r
r = PR + RS = PS
uuur
uuur
uur
Fig 10 10
© NCERT
not to be republished
MATHEMATICS
432
a
a
1
2
1
2
a
–2
a
a
2
and
(
)
a
b
c
+
+
r
r
r = PQ + QS = PS
uuur
uuur
uur
Hence
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
Remark The associative property of vector addition enables us to write the sum of
three vectors
,
,
as
a
b
c
a
b
c
+
+
r
r
r
r
r
r without using brackets |
1 | 5148-5151 | Fig 10 11
Then
a
b
+
r
r
= PQ + QR = PR
uuur
uuur
uuur
and
b
c
+
r
r = QR + RS = QS
uuur
uuur
uuur
So
(
)
a
b
c
+
r+
r
r = PR + RS = PS
uuur
uuur
uur
Fig 10 10
© NCERT
not to be republished
MATHEMATICS
432
a
a
1
2
1
2
a
–2
a
a
2
and
(
)
a
b
c
+
+
r
r
r = PQ + QS = PS
uuur
uuur
uur
Hence
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
Remark The associative property of vector addition enables us to write the sum of
three vectors
,
,
as
a
b
c
a
b
c
+
+
r
r
r
r
r
r without using brackets Note that for any vector ar, we have
a +0
r
r
= 0
a
a
+
=
r
r
r
Here, the zero vector 0
r is called the additive identity for the vector addition |
1 | 5149-5152 | 11
Then
a
b
+
r
r
= PQ + QR = PR
uuur
uuur
uuur
and
b
c
+
r
r = QR + RS = QS
uuur
uuur
uuur
So
(
)
a
b
c
+
r+
r
r = PR + RS = PS
uuur
uuur
uur
Fig 10 10
© NCERT
not to be republished
MATHEMATICS
432
a
a
1
2
1
2
a
–2
a
a
2
and
(
)
a
b
c
+
+
r
r
r = PQ + QS = PS
uuur
uuur
uur
Hence
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
Remark The associative property of vector addition enables us to write the sum of
three vectors
,
,
as
a
b
c
a
b
c
+
+
r
r
r
r
r
r without using brackets Note that for any vector ar, we have
a +0
r
r
= 0
a
a
+
=
r
r
r
Here, the zero vector 0
r is called the additive identity for the vector addition 10 |
1 | 5150-5153 | 10
© NCERT
not to be republished
MATHEMATICS
432
a
a
1
2
1
2
a
–2
a
a
2
and
(
)
a
b
c
+
+
r
r
r = PQ + QS = PS
uuur
uuur
uur
Hence
(
)
a
b
c
+
+
r
r
r =
(
)
a
b
c
+
+
r
r
r
Remark The associative property of vector addition enables us to write the sum of
three vectors
,
,
as
a
b
c
a
b
c
+
+
r
r
r
r
r
r without using brackets Note that for any vector ar, we have
a +0
r
r
= 0
a
a
+
=
r
r
r
Here, the zero vector 0
r is called the additive identity for the vector addition 10 5 Multiplication of a Vector by a Scalar
Let ar be a given vector and λ a scalar |
1 | 5151-5154 | Note that for any vector ar, we have
a +0
r
r
= 0
a
a
+
=
r
r
r
Here, the zero vector 0
r is called the additive identity for the vector addition 10 5 Multiplication of a Vector by a Scalar
Let ar be a given vector and λ a scalar Then the product of the vector ar by the scalar
λ, denoted as λ ar , is called the multiplication of vector ar by the scalar λ |
1 | 5152-5155 | 10 5 Multiplication of a Vector by a Scalar
Let ar be a given vector and λ a scalar Then the product of the vector ar by the scalar
λ, denoted as λ ar , is called the multiplication of vector ar by the scalar λ Note that,
λ ar is also a vector, collinear to the vector ar |
1 | 5153-5156 | 5 Multiplication of a Vector by a Scalar
Let ar be a given vector and λ a scalar Then the product of the vector ar by the scalar
λ, denoted as λ ar , is called the multiplication of vector ar by the scalar λ Note that,
λ ar is also a vector, collinear to the vector ar The vector λ ar has the direction same
(or opposite) to that of vector ar according as the value of λ is positive (or negative) |
1 | 5154-5157 | Then the product of the vector ar by the scalar
λ, denoted as λ ar , is called the multiplication of vector ar by the scalar λ Note that,
λ ar is also a vector, collinear to the vector ar The vector λ ar has the direction same
(or opposite) to that of vector ar according as the value of λ is positive (or negative) Also, the magnitude of vector λ ar is |λ| times the magnitude of the vector ar , i |
1 | 5155-5158 | Note that,
λ ar is also a vector, collinear to the vector ar The vector λ ar has the direction same
(or opposite) to that of vector ar according as the value of λ is positive (or negative) Also, the magnitude of vector λ ar is |λ| times the magnitude of the vector ar , i e |
1 | 5156-5159 | The vector λ ar has the direction same
(or opposite) to that of vector ar according as the value of λ is positive (or negative) Also, the magnitude of vector λ ar is |λ| times the magnitude of the vector ar , i e ,
|
a|
λr = |
||
a|
λ
r
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10 |
1 | 5157-5160 | Also, the magnitude of vector λ ar is |λ| times the magnitude of the vector ar , i e ,
|
a|
λr = |
||
a|
λ
r
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10 12 |
1 | 5158-5161 | e ,
|
a|
λr = |
||
a|
λ
r
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10 12 Fig 10 |
1 | 5159-5162 | ,
|
a|
λr = |
||
a|
λ
r
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10 12 Fig 10 12
When λ = –1, then a
λ = −a
r
r, which is a vector having magnitude equal to the
magnitude of ar and direction opposite to that of the direction of ar |
1 | 5160-5163 | 12 Fig 10 12
When λ = –1, then a
λ = −a
r
r, which is a vector having magnitude equal to the
magnitude of ar and direction opposite to that of the direction of ar The vector – ar is
called the negative (or additive inverse) of vector ar and we always have
(– )
a
a
r+
r = (– )
0
a
+a
=
r
r
r
Also, if
= |1
a|
λ
r , provided
0, i |
1 | 5161-5164 | Fig 10 12
When λ = –1, then a
λ = −a
r
r, which is a vector having magnitude equal to the
magnitude of ar and direction opposite to that of the direction of ar The vector – ar is
called the negative (or additive inverse) of vector ar and we always have
(– )
a
a
r+
r = (– )
0
a
+a
=
r
r
r
Also, if
= |1
a|
λ
r , provided
0, i e |
1 | 5162-5165 | 12
When λ = –1, then a
λ = −a
r
r, which is a vector having magnitude equal to the
magnitude of ar and direction opposite to that of the direction of ar The vector – ar is
called the negative (or additive inverse) of vector ar and we always have
(– )
a
a
r+
r = (– )
0
a
+a
=
r
r
r
Also, if
= |1
a|
λ
r , provided
0, i e r
r
a
a is not a null vector, then
|
| |
||
|
a
a
λ
r= λ
r =
1 |
| 1
|
a| a
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
433
So, λ ar represents the unit vector in the direction of ar |
1 | 5163-5166 | The vector – ar is
called the negative (or additive inverse) of vector ar and we always have
(– )
a
a
r+
r = (– )
0
a
+a
=
r
r
r
Also, if
= |1
a|
λ
r , provided
0, i e r
r
a
a is not a null vector, then
|
| |
||
|
a
a
λ
r= λ
r =
1 |
| 1
|
a| a
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
433
So, λ ar represents the unit vector in the direction of ar We write it as
ˆa =
|1
| a
a
r
r
�Note For any scalar k, 0 = 0 |
1 | 5164-5167 | e r
r
a
a is not a null vector, then
|
| |
||
|
a
a
λ
r= λ
r =
1 |
| 1
|
a| a
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
433
So, λ ar represents the unit vector in the direction of ar We write it as
ˆa =
|1
| a
a
r
r
�Note For any scalar k, 0 = 0 k
r
r
10 |
1 | 5165-5168 | r
r
a
a is not a null vector, then
|
| |
||
|
a
a
λ
r= λ
r =
1 |
| 1
|
a| a
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
433
So, λ ar represents the unit vector in the direction of ar We write it as
ˆa =
|1
| a
a
r
r
�Note For any scalar k, 0 = 0 k
r
r
10 5 |
1 | 5166-5169 | We write it as
ˆa =
|1
| a
a
r
r
�Note For any scalar k, 0 = 0 k
r
r
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 |
1 | 5167-5170 | k
r
r
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
| OA | 1,| OB|
uuur=
uuur = 1 and | OC |
=1
uuur
The vectors OA, OB and OC
uuur
uuur
uuur , each having magnitude 1,
are called unit vectors along the axes OX, OY and OZ,
respectively, and denoted by
ˆ
ˆ ˆ
, and
i j
k , respectively
(Fig 10 |
1 | 5168-5171 | 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
| OA | 1,| OB|
uuur=
uuur = 1 and | OC |
=1
uuur
The vectors OA, OB and OC
uuur
uuur
uuur , each having magnitude 1,
are called unit vectors along the axes OX, OY and OZ,
respectively, and denoted by
ˆ
ˆ ˆ
, and
i j
k , respectively
(Fig 10 13) |
1 | 5169-5172 | 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
| OA | 1,| OB|
uuur=
uuur = 1 and | OC |
=1
uuur
The vectors OA, OB and OC
uuur
uuur
uuur , each having magnitude 1,
are called unit vectors along the axes OX, OY and OZ,
respectively, and denoted by
ˆ
ˆ ˆ
, and
i j
k , respectively
(Fig 10 13) Now, consider the position vector OP
uuur
of a point P(x, y, z) as in Fig 10 |
1 | 5170-5173 | Then, clearly
| OA | 1,| OB|
uuur=
uuur = 1 and | OC |
=1
uuur
The vectors OA, OB and OC
uuur
uuur
uuur , each having magnitude 1,
are called unit vectors along the axes OX, OY and OZ,
respectively, and denoted by
ˆ
ˆ ˆ
, and
i j
k , respectively
(Fig 10 13) Now, consider the position vector OP
uuur
of a point P(x, y, z) as in Fig 10 14 |
1 | 5171-5174 | 13) Now, consider the position vector OP
uuur
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 |
1 | 5172-5175 | Now, consider the position vector OP
uuur
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 |
1 | 5173-5176 | 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
ˆ
ˆ ˆ
, and
i j
k are the unit vectors along the x, y and z-axes,
respectively, and by the definition of the coordinates of P, we have
1
ˆ
P P
OR
zk
=
=
uuur
uuur |
1 | 5174-5177 | 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
ˆ
ˆ ˆ
, and
i j
k are the unit vectors along the x, y and z-axes,
respectively, and by the definition of the coordinates of P, we have
1
ˆ
P P
OR
zk
=
=
uuur
uuur Similarly,
1
ˆ
QP
OS
yj
=
=
uuur
uuur
and
ˆ
OQ
xi
=
uuur |
1 | 5175-5178 | We, thus, see that P1 P is
parallel to z-axis As
ˆ
ˆ ˆ
, and
i j
k are the unit vectors along the x, y and z-axes,
respectively, and by the definition of the coordinates of P, we have
1
ˆ
P P
OR
zk
=
=
uuur
uuur Similarly,
1
ˆ
QP
OS
yj
=
=
uuur
uuur
and
ˆ
OQ
xi
=
uuur Fig 10 |
1 | 5176-5179 | As
ˆ
ˆ ˆ
, and
i j
k are the unit vectors along the x, y and z-axes,
respectively, and by the definition of the coordinates of P, we have
1
ˆ
P P
OR
zk
=
=
uuur
uuur Similarly,
1
ˆ
QP
OS
yj
=
=
uuur
uuur
and
ˆ
OQ
xi
=
uuur Fig 10 13
Fig 10 |
1 | 5177-5180 | Similarly,
1
ˆ
QP
OS
yj
=
=
uuur
uuur
and
ˆ
OQ
xi
=
uuur Fig 10 13
Fig 10 14
© NCERT
not to be republished
MATHEMATICS
434
Therefore, it follows that
OP1
uuur =
1
ˆ
ˆ
OQ + QP
xi
yj
=
+
uuur
uuur
and
OP
uuur =
1
1
ˆ
ˆ
ˆ
OP + P P
xi
yj
zk
=
+
+
uuur
uuuur
Hence, the position vector of P with reference to O is given by
OP (or
)
r
uuur
r =
ˆ
ˆ
ˆ
xi
yj
zk
+
+
This form of any vector is called its component form |
1 | 5178-5181 | Fig 10 13
Fig 10 14
© NCERT
not to be republished
MATHEMATICS
434
Therefore, it follows that
OP1
uuur =
1
ˆ
ˆ
OQ + QP
xi
yj
=
+
uuur
uuur
and
OP
uuur =
1
1
ˆ
ˆ
ˆ
OP + P P
xi
yj
zk
=
+
+
uuur
uuuur
Hence, the position vector of P with reference to O is given by
OP (or
)
r
uuur
r =
ˆ
ˆ
ˆ
xi
yj
zk
+
+
This form of any vector is called its component form Here, x, y and z are called
as the scalar components of rr , and
ˆ
ˆ
ˆ
,
and
xi
yj
zk are called the vector components
of rr along the respective axes |
1 | 5179-5182 | 13
Fig 10 14
© NCERT
not to be republished
MATHEMATICS
434
Therefore, it follows that
OP1
uuur =
1
ˆ
ˆ
OQ + QP
xi
yj
=
+
uuur
uuur
and
OP
uuur =
1
1
ˆ
ˆ
ˆ
OP + P P
xi
yj
zk
=
+
+
uuur
uuuur
Hence, the position vector of P with reference to O is given by
OP (or
)
r
uuur
r =
ˆ
ˆ
ˆ
xi
yj
zk
+
+
This form of any vector is called its component form Here, x, y and z are called
as the scalar components of rr , and
ˆ
ˆ
ˆ
,
and
xi
yj
zk are called the vector components
of rr along the respective axes Sometimes x, y and z are also termed as rectangular
components |
1 | 5180-5183 | 14
© NCERT
not to be republished
MATHEMATICS
434
Therefore, it follows that
OP1
uuur =
1
ˆ
ˆ
OQ + QP
xi
yj
=
+
uuur
uuur
and
OP
uuur =
1
1
ˆ
ˆ
ˆ
OP + P P
xi
yj
zk
=
+
+
uuur
uuuur
Hence, the position vector of P with reference to O is given by
OP (or
)
r
uuur
r =
ˆ
ˆ
ˆ
xi
yj
zk
+
+
This form of any vector is called its component form Here, x, y and z are called
as the scalar components of rr , and
ˆ
ˆ
ˆ
,
and
xi
yj
zk are called the vector components
of rr along the respective axes Sometimes x, y and z are also termed as rectangular
components The length of any vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
, is readily determined by applying the
Pythagoras theorem twice |
1 | 5181-5184 | Here, x, y and z are called
as the scalar components of rr , and
ˆ
ˆ
ˆ
,
and
xi
yj
zk are called the vector components
of rr along the respective axes Sometimes x, y and z are also termed as rectangular
components The length of any vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
, is readily determined by applying the
Pythagoras theorem twice We note that in the right angle triangle OQP1 (Fig 10 |
1 | 5182-5185 | Sometimes x, y and z are also termed as rectangular
components The length of any vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
, is readily determined by applying the
Pythagoras theorem twice We note that in the right angle triangle OQP1 (Fig 10 14)
| OP |1
uuuur =
2
2
2
2
|OQ| +|QP |1
x
y
=
+
uuuur
uuur
,
and in the right angle triangle OP1P, we have
OP
uuur =
2
2
2
2
2
1
1
| OP |
| P P |
(
)
x
y
z
uuur
uuur
Hence, the length of any vector
ˆ
ˆ
ˆ+
r
xi
yj
zk
=
+
r
is given by
|
|
rr =
2
2
2
ˆ
ˆ
ˆ
|
| =
xi
yj
zk
x
y
z
+
+
+
+
If and
a
rb
r
are any two vectors given in the component form
1
2
3 ˆ
ˆ
ˆ+
a i
a j
a k
+
and
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
, respectively, then
(i)
the sum (or resultant) of the vectors
aand
rb
r
is given by
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
+
+
+
+
+
(ii)
the difference of the vector
aand
rb
r
is given by
a
−b
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
−
+
−
+
−
(iii)
the vectors
aand
rb
r
are equal if and only if
a1 = b1, a2 = b2 and a3 = b3
(iv)
the multiplication of vector ar by any scalar λ is given by
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
© NCERT
not to be republished
VECTOR ALGEBRA
435
The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let
a and
rb
r
be any two vectors, and k and m be any scalars |
1 | 5183-5186 | The length of any vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
, is readily determined by applying the
Pythagoras theorem twice We note that in the right angle triangle OQP1 (Fig 10 14)
| OP |1
uuuur =
2
2
2
2
|OQ| +|QP |1
x
y
=
+
uuuur
uuur
,
and in the right angle triangle OP1P, we have
OP
uuur =
2
2
2
2
2
1
1
| OP |
| P P |
(
)
x
y
z
uuur
uuur
Hence, the length of any vector
ˆ
ˆ
ˆ+
r
xi
yj
zk
=
+
r
is given by
|
|
rr =
2
2
2
ˆ
ˆ
ˆ
|
| =
xi
yj
zk
x
y
z
+
+
+
+
If and
a
rb
r
are any two vectors given in the component form
1
2
3 ˆ
ˆ
ˆ+
a i
a j
a k
+
and
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
, respectively, then
(i)
the sum (or resultant) of the vectors
aand
rb
r
is given by
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
+
+
+
+
+
(ii)
the difference of the vector
aand
rb
r
is given by
a
−b
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
−
+
−
+
−
(iii)
the vectors
aand
rb
r
are equal if and only if
a1 = b1, a2 = b2 and a3 = b3
(iv)
the multiplication of vector ar by any scalar λ is given by
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
© NCERT
not to be republished
VECTOR ALGEBRA
435
The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let
a and
rb
r
be any two vectors, and k and m be any scalars Then
(i)
(
)
ka
ma
k
m a
+
=
+
r
r
r
(ii)
(
)
(
)
k ma
r=km a
r
(iii)
(
)
k a
b
ka
kb
r
r
r
r
Remarks
(i)
One may observe that whatever be the value of λ, the vector a
λr is always
collinear to the vector ar |
1 | 5184-5187 | We note that in the right angle triangle OQP1 (Fig 10 14)
| OP |1
uuuur =
2
2
2
2
|OQ| +|QP |1
x
y
=
+
uuuur
uuur
,
and in the right angle triangle OP1P, we have
OP
uuur =
2
2
2
2
2
1
1
| OP |
| P P |
(
)
x
y
z
uuur
uuur
Hence, the length of any vector
ˆ
ˆ
ˆ+
r
xi
yj
zk
=
+
r
is given by
|
|
rr =
2
2
2
ˆ
ˆ
ˆ
|
| =
xi
yj
zk
x
y
z
+
+
+
+
If and
a
rb
r
are any two vectors given in the component form
1
2
3 ˆ
ˆ
ˆ+
a i
a j
a k
+
and
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
, respectively, then
(i)
the sum (or resultant) of the vectors
aand
rb
r
is given by
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
+
+
+
+
+
(ii)
the difference of the vector
aand
rb
r
is given by
a
−b
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
−
+
−
+
−
(iii)
the vectors
aand
rb
r
are equal if and only if
a1 = b1, a2 = b2 and a3 = b3
(iv)
the multiplication of vector ar by any scalar λ is given by
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
© NCERT
not to be republished
VECTOR ALGEBRA
435
The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let
a and
rb
r
be any two vectors, and k and m be any scalars Then
(i)
(
)
ka
ma
k
m a
+
=
+
r
r
r
(ii)
(
)
(
)
k ma
r=km a
r
(iii)
(
)
k a
b
ka
kb
r
r
r
r
Remarks
(i)
One may observe that whatever be the value of λ, the vector a
λr is always
collinear to the vector ar In fact, two vectors and
a
rb
r
are collinear if and only
if there exists a nonzero scalar λ such that b
r= λa
r |
1 | 5185-5188 | 14)
| OP |1
uuuur =
2
2
2
2
|OQ| +|QP |1
x
y
=
+
uuuur
uuur
,
and in the right angle triangle OP1P, we have
OP
uuur =
2
2
2
2
2
1
1
| OP |
| P P |
(
)
x
y
z
uuur
uuur
Hence, the length of any vector
ˆ
ˆ
ˆ+
r
xi
yj
zk
=
+
r
is given by
|
|
rr =
2
2
2
ˆ
ˆ
ˆ
|
| =
xi
yj
zk
x
y
z
+
+
+
+
If and
a
rb
r
are any two vectors given in the component form
1
2
3 ˆ
ˆ
ˆ+
a i
a j
a k
+
and
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
, respectively, then
(i)
the sum (or resultant) of the vectors
aand
rb
r
is given by
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
+
+
+
+
+
(ii)
the difference of the vector
aand
rb
r
is given by
a
−b
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b k
−
+
−
+
−
(iii)
the vectors
aand
rb
r
are equal if and only if
a1 = b1, a2 = b2 and a3 = b3
(iv)
the multiplication of vector ar by any scalar λ is given by
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
© NCERT
not to be republished
VECTOR ALGEBRA
435
The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let
a and
rb
r
be any two vectors, and k and m be any scalars Then
(i)
(
)
ka
ma
k
m a
+
=
+
r
r
r
(ii)
(
)
(
)
k ma
r=km a
r
(iii)
(
)
k a
b
ka
kb
r
r
r
r
Remarks
(i)
One may observe that whatever be the value of λ, the vector a
λr is always
collinear to the vector ar In fact, two vectors and
a
rb
r
are collinear if and only
if there exists a nonzero scalar λ such that b
r= λa
r If the vectors
a and
rb
r
are
given in the component form, i |
1 | 5186-5189 | Then
(i)
(
)
ka
ma
k
m a
+
=
+
r
r
r
(ii)
(
)
(
)
k ma
r=km a
r
(iii)
(
)
k a
b
ka
kb
r
r
r
r
Remarks
(i)
One may observe that whatever be the value of λ, the vector a
λr is always
collinear to the vector ar In fact, two vectors and
a
rb
r
are collinear if and only
if there exists a nonzero scalar λ such that b
r= λa
r If the vectors
a and
rb
r
are
given in the component form, i e |
1 | 5187-5190 | In fact, two vectors and
a
rb
r
are collinear if and only
if there exists a nonzero scalar λ such that b
r= λa
r If the vectors
a and
rb
r
are
given in the component form, i e 1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
and
1
2
3 ˆ
ˆ
ˆ
b
b i
b j
b k
=
+
+
r
,
then the two vectors are collinear if and only if
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
a i
a j
a k
λ
+
+
⇔
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
⇔
1
1
b
= λa
,
2
2
3
3
,
b
a
b
a
= λ
= λ
⇔
1
1
b
a =
3
2
2
3
b
ab
=a
= λ
(ii)
If
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
r
, then a1, a2, a3 are also called direction ratios of ar |
1 | 5188-5191 | If the vectors
a and
rb
r
are
given in the component form, i e 1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
and
1
2
3 ˆ
ˆ
ˆ
b
b i
b j
b k
=
+
+
r
,
then the two vectors are collinear if and only if
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
a i
a j
a k
λ
+
+
⇔
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
⇔
1
1
b
= λa
,
2
2
3
3
,
b
a
b
a
= λ
= λ
⇔
1
1
b
a =
3
2
2
3
b
ab
=a
= λ
(ii)
If
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
r
, then a1, a2, a3 are also called direction ratios of ar (iii)
In case if it is given that l, m, n are direction cosines of a vector, then
ˆ
ˆ
ˆ
li
mj
nk
+
+
=
ˆ
ˆ
ˆ
(cos )
(cos )
(cos )
i
j
k
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 |
1 | 5189-5192 | e 1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
and
1
2
3 ˆ
ˆ
ˆ
b
b i
b j
b k
=
+
+
r
,
then the two vectors are collinear if and only if
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
a i
a j
a k
λ
+
+
⇔
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
⇔
1
1
b
= λa
,
2
2
3
3
,
b
a
b
a
= λ
= λ
⇔
1
1
b
a =
3
2
2
3
b
ab
=a
= λ
(ii)
If
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
r
, then a1, a2, a3 are also called direction ratios of ar (iii)
In case if it is given that l, m, n are direction cosines of a vector, then
ˆ
ˆ
ˆ
li
mj
nk
+
+
=
ˆ
ˆ
ˆ
(cos )
(cos )
(cos )
i
j
k
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
ˆ
ˆ
2ˆ
a
xi
j
zk
=
+
+
r
and
ˆ
ˆ
ˆ
2
b
i
yj
k
=
+
+
r
are equal |
1 | 5190-5193 | 1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
and
1
2
3 ˆ
ˆ
ˆ
b
b i
b j
b k
=
+
+
r
,
then the two vectors are collinear if and only if
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
a i
a j
a k
λ
+
+
⇔
1
2
3 ˆ
ˆ
ˆ
b i
b j
b k
+
+
=
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
⇔
1
1
b
= λa
,
2
2
3
3
,
b
a
b
a
= λ
= λ
⇔
1
1
b
a =
3
2
2
3
b
ab
=a
= λ
(ii)
If
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
r
, then a1, a2, a3 are also called direction ratios of ar (iii)
In case if it is given that l, m, n are direction cosines of a vector, then
ˆ
ˆ
ˆ
li
mj
nk
+
+
=
ˆ
ˆ
ˆ
(cos )
(cos )
(cos )
i
j
k
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
ˆ
ˆ
2ˆ
a
xi
j
zk
=
+
+
r
and
ˆ
ˆ
ˆ
2
b
i
yj
k
=
+
+
r
are equal Solution Note that two vectors are equal if and only if their corresponding components
are equal |
1 | 5191-5194 | (iii)
In case if it is given that l, m, n are direction cosines of a vector, then
ˆ
ˆ
ˆ
li
mj
nk
+
+
=
ˆ
ˆ
ˆ
(cos )
(cos )
(cos )
i
j
k
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
ˆ
ˆ
2ˆ
a
xi
j
zk
=
+
+
r
and
ˆ
ˆ
ˆ
2
b
i
yj
k
=
+
+
r
are equal Solution Note that two vectors are equal if and only if their corresponding components
are equal Thus, the given vectors
aand
rb
r
will be equal if and only if
x = 2, y = 2, z = 1
© NCERT
not to be republished
MATHEMATICS
436
Example 5 Let
ˆ
2ˆ
a
i
j
r= +
and
ˆ
ˆ
2
b
i
j
=
+
r |
1 | 5192-5195 | Example 4 Find the values of x, y and z so that the vectors
ˆ
ˆ
2ˆ
a
xi
j
zk
=
+
+
r
and
ˆ
ˆ
ˆ
2
b
i
yj
k
=
+
+
r
are equal Solution Note that two vectors are equal if and only if their corresponding components
are equal Thus, the given vectors
aand
rb
r
will be equal if and only if
x = 2, y = 2, z = 1
© NCERT
not to be republished
MATHEMATICS
436
Example 5 Let
ˆ
2ˆ
a
i
j
r= +
and
ˆ
ˆ
2
b
i
j
=
+
r Is |
|
|
|
a
b
=
r
r |
1 | 5193-5196 | Solution Note that two vectors are equal if and only if their corresponding components
are equal Thus, the given vectors
aand
rb
r
will be equal if and only if
x = 2, y = 2, z = 1
© NCERT
not to be republished
MATHEMATICS
436
Example 5 Let
ˆ
2ˆ
a
i
j
r= +
and
ˆ
ˆ
2
b
i
j
=
+
r Is |
|
|
|
a
b
=
r
r Are the vectors
a and
rb
r
equal |
1 | 5194-5197 | Thus, the given vectors
aand
rb
r
will be equal if and only if
x = 2, y = 2, z = 1
© NCERT
not to be republished
MATHEMATICS
436
Example 5 Let
ˆ
2ˆ
a
i
j
r= +
and
ˆ
ˆ
2
b
i
j
=
+
r Is |
|
|
|
a
b
=
r
r Are the vectors
a and
rb
r
equal Solution We have
2
2
|
|
1
2
5
a =
+
=
r
and
2
2
|
|
2
1
5
b
r
So, |
| |
|
a
=b
r
r |
1 | 5195-5198 | Is |
|
|
|
a
b
=
r
r Are the vectors
a and
rb
r
equal Solution We have
2
2
|
|
1
2
5
a =
+
=
r
and
2
2
|
|
2
1
5
b
r
So, |
| |
|
a
=b
r
r But, the two vectors are not equal since their corresponding components
are distinct |
1 | 5196-5199 | Are the vectors
a and
rb
r
equal Solution We have
2
2
|
|
1
2
5
a =
+
=
r
and
2
2
|
|
2
1
5
b
r
So, |
| |
|
a
=b
r
r But, the two vectors are not equal since their corresponding components
are distinct Example 6 Find unit vector in the direction of vector
ˆ
ˆ
ˆ
2
3
a
i
j
k
=
+
+
r
Solution The unit vector in the direction of a vector ar is given by
1
ˆ
|
|
a
=aa
rr |
1 | 5197-5200 | Solution We have
2
2
|
|
1
2
5
a =
+
=
r
and
2
2
|
|
2
1
5
b
r
So, |
| |
|
a
=b
r
r But, the two vectors are not equal since their corresponding components
are distinct Example 6 Find unit vector in the direction of vector
ˆ
ˆ
ˆ
2
3
a
i
j
k
=
+
+
r
Solution The unit vector in the direction of a vector ar is given by
1
ˆ
|
|
a
=aa
rr Now
|
|
ar =
2
2
2
2
3
1
14
+
+
=
Therefore
1
ˆ
ˆ
ˆ
ˆ
(2
3
)
14
a
i
j
k
=
+
+
=
2
3
1
ˆ
ˆ
ˆ
14
14
14
i
j
k
+
+
Example 7 Find a vector in the direction of vector
ˆ
2ˆ
a
i
j
r= −
that has magnitude
7 units |
1 | 5198-5201 | But, the two vectors are not equal since their corresponding components
are distinct Example 6 Find unit vector in the direction of vector
ˆ
ˆ
ˆ
2
3
a
i
j
k
=
+
+
r
Solution The unit vector in the direction of a vector ar is given by
1
ˆ
|
|
a
=aa
rr Now
|
|
ar =
2
2
2
2
3
1
14
+
+
=
Therefore
1
ˆ
ˆ
ˆ
ˆ
(2
3
)
14
a
i
j
k
=
+
+
=
2
3
1
ˆ
ˆ
ˆ
14
14
14
i
j
k
+
+
Example 7 Find a vector in the direction of vector
ˆ
2ˆ
a
i
j
r= −
that has magnitude
7 units Solution The unit vector in the direction of the given vector ar is
1
ˆ
|
|
a
=aa
rr
=
1
1
2
ˆ
ˆ
ˆ
ˆ
(
2 )
5
5
5
i
j
i
j
−
=
−
Therefore, the vector having magnitude equal to 7 and in the direction of ar is
7a
∧ =
1
2
7
5
5
i
j
∧
∧
⎛
⎞
−
⎜
⎟
⎝
⎠
=
7
14
ˆ
ˆ
5
5
i
j
−
Example 8 Find the unit vector in the direction of the sum of the vectors,
ˆ
ˆ
ˆ
2
2 – 5
a
i
j
k
=
+
r
and
ˆ
ˆ
ˆ
2
3
b
i
j
k
=
+
+
r |
1 | 5199-5202 | Example 6 Find unit vector in the direction of vector
ˆ
ˆ
ˆ
2
3
a
i
j
k
=
+
+
r
Solution The unit vector in the direction of a vector ar is given by
1
ˆ
|
|
a
=aa
rr Now
|
|
ar =
2
2
2
2
3
1
14
+
+
=
Therefore
1
ˆ
ˆ
ˆ
ˆ
(2
3
)
14
a
i
j
k
=
+
+
=
2
3
1
ˆ
ˆ
ˆ
14
14
14
i
j
k
+
+
Example 7 Find a vector in the direction of vector
ˆ
2ˆ
a
i
j
r= −
that has magnitude
7 units Solution The unit vector in the direction of the given vector ar is
1
ˆ
|
|
a
=aa
rr
=
1
1
2
ˆ
ˆ
ˆ
ˆ
(
2 )
5
5
5
i
j
i
j
−
=
−
Therefore, the vector having magnitude equal to 7 and in the direction of ar is
7a
∧ =
1
2
7
5
5
i
j
∧
∧
⎛
⎞
−
⎜
⎟
⎝
⎠
=
7
14
ˆ
ˆ
5
5
i
j
−
Example 8 Find the unit vector in the direction of the sum of the vectors,
ˆ
ˆ
ˆ
2
2 – 5
a
i
j
k
=
+
r
and
ˆ
ˆ
ˆ
2
3
b
i
j
k
=
+
+
r Solution The sum of the given vectors is
ˆ
ˆ
ˆ
(
, say) = 4
3
2
r
r
r
a
b
c
i
j
k
and
|
|
cr =
2
2
2
4
3
( 2)
29
+
+ −
=
© NCERT
not to be republished
VECTOR ALGEBRA
437
Thus, the required unit vector is
1
1
4
3
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(4
3
2 )
|
|
29
29
29
29
c
c
i
j
k
i
j
k
=c
=
+
−
=
+
−
r
r
Example 9 Write the direction ratio’s of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
−
r
and hence calculate
its direction cosines |
1 | 5200-5203 | Now
|
|
ar =
2
2
2
2
3
1
14
+
+
=
Therefore
1
ˆ
ˆ
ˆ
ˆ
(2
3
)
14
a
i
j
k
=
+
+
=
2
3
1
ˆ
ˆ
ˆ
14
14
14
i
j
k
+
+
Example 7 Find a vector in the direction of vector
ˆ
2ˆ
a
i
j
r= −
that has magnitude
7 units Solution The unit vector in the direction of the given vector ar is
1
ˆ
|
|
a
=aa
rr
=
1
1
2
ˆ
ˆ
ˆ
ˆ
(
2 )
5
5
5
i
j
i
j
−
=
−
Therefore, the vector having magnitude equal to 7 and in the direction of ar is
7a
∧ =
1
2
7
5
5
i
j
∧
∧
⎛
⎞
−
⎜
⎟
⎝
⎠
=
7
14
ˆ
ˆ
5
5
i
j
−
Example 8 Find the unit vector in the direction of the sum of the vectors,
ˆ
ˆ
ˆ
2
2 – 5
a
i
j
k
=
+
r
and
ˆ
ˆ
ˆ
2
3
b
i
j
k
=
+
+
r Solution The sum of the given vectors is
ˆ
ˆ
ˆ
(
, say) = 4
3
2
r
r
r
a
b
c
i
j
k
and
|
|
cr =
2
2
2
4
3
( 2)
29
+
+ −
=
© NCERT
not to be republished
VECTOR ALGEBRA
437
Thus, the required unit vector is
1
1
4
3
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(4
3
2 )
|
|
29
29
29
29
c
c
i
j
k
i
j
k
=c
=
+
−
=
+
−
r
r
Example 9 Write the direction ratio’s of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
−
r
and hence calculate
its direction cosines Solution Note that the direction ratio’s a, b, c of a vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
are just
the respective components x, y and z of the vector |
1 | 5201-5204 | Solution The unit vector in the direction of the given vector ar is
1
ˆ
|
|
a
=aa
rr
=
1
1
2
ˆ
ˆ
ˆ
ˆ
(
2 )
5
5
5
i
j
i
j
−
=
−
Therefore, the vector having magnitude equal to 7 and in the direction of ar is
7a
∧ =
1
2
7
5
5
i
j
∧
∧
⎛
⎞
−
⎜
⎟
⎝
⎠
=
7
14
ˆ
ˆ
5
5
i
j
−
Example 8 Find the unit vector in the direction of the sum of the vectors,
ˆ
ˆ
ˆ
2
2 – 5
a
i
j
k
=
+
r
and
ˆ
ˆ
ˆ
2
3
b
i
j
k
=
+
+
r Solution The sum of the given vectors is
ˆ
ˆ
ˆ
(
, say) = 4
3
2
r
r
r
a
b
c
i
j
k
and
|
|
cr =
2
2
2
4
3
( 2)
29
+
+ −
=
© NCERT
not to be republished
VECTOR ALGEBRA
437
Thus, the required unit vector is
1
1
4
3
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(4
3
2 )
|
|
29
29
29
29
c
c
i
j
k
i
j
k
=c
=
+
−
=
+
−
r
r
Example 9 Write the direction ratio’s of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
−
r
and hence calculate
its direction cosines Solution Note that the direction ratio’s a, b, c of a vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
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 |
1 | 5202-5205 | Solution The sum of the given vectors is
ˆ
ˆ
ˆ
(
, say) = 4
3
2
r
r
r
a
b
c
i
j
k
and
|
|
cr =
2
2
2
4
3
( 2)
29
+
+ −
=
© NCERT
not to be republished
VECTOR ALGEBRA
437
Thus, the required unit vector is
1
1
4
3
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(4
3
2 )
|
|
29
29
29
29
c
c
i
j
k
i
j
k
=c
=
+
−
=
+
−
r
r
Example 9 Write the direction ratio’s of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
−
r
and hence calculate
its direction cosines Solution Note that the direction ratio’s a, b, c of a vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
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
−
=
=
=
=
=
=
r=
r
r
r
Thus, the direction cosines are
1
1
2
,
,–
6
6
6
⎛
⎞
⎜
⎟
⎝
⎠ |
1 | 5203-5206 | Solution Note that the direction ratio’s a, b, c of a vector
ˆ
ˆ
ˆ
r
xi
yj
zk
=
+
+
r
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
−
=
=
=
=
=
=
r=
r
r
r
Thus, the direction cosines are
1
1
2
,
,–
6
6
6
⎛
⎞
⎜
⎟
⎝
⎠ 10 |
1 | 5204-5207 | 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
−
=
=
=
=
=
=
r=
r
r
r
Thus, the direction cosines are
1
1
2
,
,–
6
6
6
⎛
⎞
⎜
⎟
⎝
⎠ 10 5 |
1 | 5205-5208 | 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
−
=
=
=
=
=
=
r=
r
r
r
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
1 2
P P
uuuur (Fig 10 |
1 | 5206-5209 | 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
1 2
P P
uuuur (Fig 10 15) |
1 | 5207-5210 | 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
1 2
P P
uuuur (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
uuur+P P
uuuur =
OP |
1 | 5208-5211 | 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
1 2
P P
uuuur (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
uuur+P P
uuuur =
OP 2
uuuur
Using the properties of vector addition, the
above equation becomes
1 2
P P
uuuur =
2
1
OP
OP
−
uuuur
uuur
i |
1 | 5209-5212 | 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
uuur+P P
uuuur =
OP 2
uuuur
Using the properties of vector addition, the
above equation becomes
1 2
P P
uuuur =
2
1
OP
OP
−
uuuur
uuur
i e |
1 | 5210-5213 | Joining the points P1 and P2 with the origin
O, and applying triangle law, from the triangle
OP1P2, we have
1
1 2
OP
uuur+P P
uuuur =
OP 2
uuuur
Using the properties of vector addition, the
above equation becomes
1 2
P P
uuuur =
2
1
OP
OP
−
uuuur
uuur
i e 1 2
P P
uuuur =
2
2
2
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
x i
y j
z k
x i
y j
z k
+
+
−
+
+
=
2
1
2
1
2
1 ˆ
ˆ
ˆ
(
)
(
)
(
)
x
x i
y
y
j
z
z k
−
+
−
+
−
The magnitude of vector
1 2
P P
uuuur
is given by
1 2
P P
uuuur =
2
2
2
2
1
2
1
2
1
(
)
(
)
(
)
x
x
y
y
z
z
−
+
−
+
−
Fig 10 |
1 | 5211-5214 | 2
uuuur
Using the properties of vector addition, the
above equation becomes
1 2
P P
uuuur =
2
1
OP
OP
−
uuuur
uuur
i e 1 2
P P
uuuur =
2
2
2
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
x i
y j
z k
x i
y j
z k
+
+
−
+
+
=
2
1
2
1
2
1 ˆ
ˆ
ˆ
(
)
(
)
(
)
x
x i
y
y
j
z
z k
−
+
−
+
−
The magnitude of vector
1 2
P P
uuuur
is given by
1 2
P P
uuuur =
2
2
2
2
1
2
1
2
1
(
)
(
)
(
)
x
x
y
y
z
z
−
+
−
+
−
Fig 10 15
© NCERT
not to be republished
MATHEMATICS
438
Example 10 Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed
from P to Q |
1 | 5212-5215 | e 1 2
P P
uuuur =
2
2
2
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
x i
y j
z k
x i
y j
z k
+
+
−
+
+
=
2
1
2
1
2
1 ˆ
ˆ
ˆ
(
)
(
)
(
)
x
x i
y
y
j
z
z k
−
+
−
+
−
The magnitude of vector
1 2
P P
uuuur
is given by
1 2
P P
uuuur =
2
2
2
2
1
2
1
2
1
(
)
(
)
(
)
x
x
y
y
z
z
−
+
−
+
−
Fig 10 15
© NCERT
not to be republished
MATHEMATICS
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 |
1 | 5213-5216 | 1 2
P P
uuuur =
2
2
2
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
x i
y j
z k
x i
y j
z k
+
+
−
+
+
=
2
1
2
1
2
1 ˆ
ˆ
ˆ
(
)
(
)
(
)
x
x i
y
y
j
z
z k
−
+
−
+
−
The magnitude of vector
1 2
P P
uuuur
is given by
1 2
P P
uuuur =
2
2
2
2
1
2
1
2
1
(
)
(
)
(
)
x
x
y
y
z
z
−
+
−
+
−
Fig 10 15
© NCERT
not to be republished
MATHEMATICS
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
uuur ,
given by
PQ
uuur =
ˆ
ˆ
ˆ
( 1
2)
( 2
3)
( 4
0)
i
j
k
− −
+ − −
+ − −
i |
1 | 5214-5217 | 15
© NCERT
not to be republished
MATHEMATICS
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
uuur ,
given by
PQ
uuur =
ˆ
ˆ
ˆ
( 1
2)
( 2
3)
( 4
0)
i
j
k
− −
+ − −
+ − −
i e |
1 | 5215-5218 | 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
uuur ,
given by
PQ
uuur =
ˆ
ˆ
ˆ
( 1
2)
( 2
3)
( 4
0)
i
j
k
− −
+ − −
+ − −
i e PQ
uuur =
ˆ
ˆ
ˆ
3
5
4 |
1 | 5216-5219 | So, the required vector joining P and Q is the vector PQ
uuur ,
given by
PQ
uuur =
ˆ
ˆ
ˆ
( 1
2)
( 2
3)
( 4
0)
i
j
k
− −
+ − −
+ − −
i e PQ
uuur =
ˆ
ˆ
ˆ
3
5
4 i
j
k
−
−
−
10 |
1 | 5217-5220 | e PQ
uuur =
ˆ
ˆ
ˆ
3
5
4 i
j
k
−
−
−
10 5 |
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