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1 | 5318-5321 | 20
© NCERT
not to be republished
MATHEMATICS
444
is called the projection vector, and its magnitude | pr | is simply called as the projection
of the vector AB
uuur
on the directed line l For example, in each of the following figures (Fig 10 20(i) to (iv)), projection vector
of AB
uuur
along the line l is vector AC
uuur Observations
1 |
1 | 5319-5322 | For example, in each of the following figures (Fig 10 20(i) to (iv)), projection vector
of AB
uuur
along the line l is vector AC
uuur Observations
1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line
l is given by
a pˆ
r |
1 | 5320-5323 | 20(i) to (iv)), projection vector
of AB
uuur
along the line l is vector AC
uuur Observations
1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line
l is given by
a pˆ
r 2 |
1 | 5321-5324 | Observations
1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line
l is given by
a pˆ
r 2 Projection of a vector ar on other vector b
r , is given by
a bˆ,
r⋅
or
, or 1
(
)
|
|
|
|
b
a
a b
b
b
⎛
⎞
⋅
⋅
⎜
⎟
⎝
⎠
r
r
r
r
r
r
3 |
1 | 5322-5325 | If ˆp is the unit vector along a line l, then the projection of a vector ar on the line
l is given by
a pˆ
r 2 Projection of a vector ar on other vector b
r , is given by
a bˆ,
r⋅
or
, or 1
(
)
|
|
|
|
b
a
a b
b
b
⎛
⎞
⋅
⋅
⎜
⎟
⎝
⎠
r
r
r
r
r
r
3 If θ = 0, then the projection vector of AB
uuur
will be AB
uuur
itself and if θ = π, then the
projection vector of AB
uuur
will be BA
uuur |
1 | 5323-5326 | 2 Projection of a vector ar on other vector b
r , is given by
a bˆ,
r⋅
or
, or 1
(
)
|
|
|
|
b
a
a b
b
b
⎛
⎞
⋅
⋅
⎜
⎟
⎝
⎠
r
r
r
r
r
r
3 If θ = 0, then the projection vector of AB
uuur
will be AB
uuur
itself and if θ = π, then the
projection vector of AB
uuur
will be BA
uuur 4 |
1 | 5324-5327 | Projection of a vector ar on other vector b
r , is given by
a bˆ,
r⋅
or
, or 1
(
)
|
|
|
|
b
a
a b
b
b
⎛
⎞
⋅
⋅
⎜
⎟
⎝
⎠
r
r
r
r
r
r
3 If θ = 0, then the projection vector of AB
uuur
will be AB
uuur
itself and if θ = π, then the
projection vector of AB
uuur
will be BA
uuur 4 If
= 2
π
θ
or
3
= 2
π
θ
, then the projection vector of AB
uuur
will be zero vector |
1 | 5325-5328 | If θ = 0, then the projection vector of AB
uuur
will be AB
uuur
itself and if θ = π, then the
projection vector of AB
uuur
will be BA
uuur 4 If
= 2
π
θ
or
3
= 2
π
θ
, then the projection vector of AB
uuur
will be zero vector Remark If α, β and γ are the direction angles of vector
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
, 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
r
r
r
r
r
Also, note that |
a| cos , | |cos and | |cos
a
a
α
β
γ
r
r
r
are respectively the projections of
ar along OX, OY and OZ |
1 | 5326-5329 | 4 If
= 2
π
θ
or
3
= 2
π
θ
, then the projection vector of AB
uuur
will be zero vector Remark If α, β and γ are the direction angles of vector
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
, 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
r
r
r
r
r
Also, note that |
a| cos , | |cos and | |cos
a
a
α
β
γ
r
r
r
are respectively the projections of
ar along OX, OY and OZ i |
1 | 5327-5330 | If
= 2
π
θ
or
3
= 2
π
θ
, then the projection vector of AB
uuur
will be zero vector Remark If α, β and γ are the direction angles of vector
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
, 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
r
r
r
r
r
Also, note that |
a| cos , | |cos and | |cos
a
a
α
β
γ
r
r
r
are respectively the projections of
ar along OX, OY and OZ i e |
1 | 5328-5331 | Remark If α, β and γ are the direction angles of vector
1
2
3 ˆ
ˆ
ˆ
a
a i
a j
a k
=
+
+
r
, 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
r
r
r
r
r
Also, note that |
a| cos , | |cos and | |cos
a
a
α
β
γ
r
r
r
are respectively the projections of
ar along OX, OY and OZ i e , the scalar components a1, a2 and a3 of the vector ar ,
are precisely the projections of ar along x-axis, y-axis and z-axis, respectively |
1 | 5329-5332 | i e , the scalar components a1, a2 and a3 of the vector ar ,
are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further,
if ar is a unit vector, then it may be expressed in terms of its direction cosines as
ˆ
ˆ
ˆ
cos
cos
cos
a
i
j
k
=
α +
β +
γ
r
Example 13 Find the angle between two vectors and
a
rb
r
with magnitudes 1 and 2
respectively and when
a b1
⋅
=
rr |
1 | 5330-5333 | e , the scalar components a1, a2 and a3 of the vector ar ,
are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further,
if ar is a unit vector, then it may be expressed in terms of its direction cosines as
ˆ
ˆ
ˆ
cos
cos
cos
a
i
j
k
=
α +
β +
γ
r
Example 13 Find the angle between two vectors and
a
rb
r
with magnitudes 1 and 2
respectively and when
a b1
⋅
=
rr Solution Given
1,|
|
1and | |
2
a b
a
b
r
r
r
r |
1 | 5331-5334 | , the scalar components a1, a2 and a3 of the vector ar ,
are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further,
if ar is a unit vector, then it may be expressed in terms of its direction cosines as
ˆ
ˆ
ˆ
cos
cos
cos
a
i
j
k
=
α +
β +
γ
r
Example 13 Find the angle between two vectors and
a
rb
r
with magnitudes 1 and 2
respectively and when
a b1
⋅
=
rr Solution Given
1,|
|
1and | |
2
a b
a
b
r
r
r
r We have
1
1 1
cos
cos
2
3
|
||
a b|
a b
rr
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
445
Example 14 Find angle ‘θ’ between the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
a
i
j
k
b
i
j
k
= +
−
= −
+
r
r |
1 | 5332-5335 | Further,
if ar is a unit vector, then it may be expressed in terms of its direction cosines as
ˆ
ˆ
ˆ
cos
cos
cos
a
i
j
k
=
α +
β +
γ
r
Example 13 Find the angle between two vectors and
a
rb
r
with magnitudes 1 and 2
respectively and when
a b1
⋅
=
rr Solution Given
1,|
|
1and | |
2
a b
a
b
r
r
r
r We have
1
1 1
cos
cos
2
3
|
||
a b|
a b
rr
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
445
Example 14 Find angle ‘θ’ between the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
a
i
j
k
b
i
j
k
= +
−
= −
+
r
r Solution The angle θ between two vectors
aand
rb
r
is given by
cosθ = |
||
a b|
a b
⋅ r
r
r
r
Now
a b
⋅
rr
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
1 1 1
1
i
j
k
i
j
k
+
−
⋅
−
+
= − − = − |
1 | 5333-5336 | Solution Given
1,|
|
1and | |
2
a b
a
b
r
r
r
r We have
1
1 1
cos
cos
2
3
|
||
a b|
a b
rr
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
445
Example 14 Find angle ‘θ’ between the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
a
i
j
k
b
i
j
k
= +
−
= −
+
r
r Solution The angle θ between two vectors
aand
rb
r
is given by
cosθ = |
||
a b|
a b
⋅ r
r
r
r
Now
a b
⋅
rr
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
1 1 1
1
i
j
k
i
j
k
+
−
⋅
−
+
= − − = − Therefore, we have
cosθ =
31
−
hence the required angle is
θ =
1
1
cos
3
Example 15 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
3 and
3
5
a
i
j
k
b
i
j
k
=
−
−
=
+
−
r
r
, then show that the vectors
and
a
b
a
b
+
−
r
r
r
r
are perpendicular |
1 | 5334-5337 | We have
1
1 1
cos
cos
2
3
|
||
a b|
a b
rr
r
r
© NCERT
not to be republished
VECTOR ALGEBRA
445
Example 14 Find angle ‘θ’ between the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
a
i
j
k
b
i
j
k
= +
−
= −
+
r
r Solution The angle θ between two vectors
aand
rb
r
is given by
cosθ = |
||
a b|
a b
⋅ r
r
r
r
Now
a b
⋅
rr
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
1 1 1
1
i
j
k
i
j
k
+
−
⋅
−
+
= − − = − Therefore, we have
cosθ =
31
−
hence the required angle is
θ =
1
1
cos
3
Example 15 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
3 and
3
5
a
i
j
k
b
i
j
k
=
−
−
=
+
−
r
r
, then show that the vectors
and
a
b
a
b
+
−
r
r
r
r
are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero |
1 | 5335-5338 | Solution The angle θ between two vectors
aand
rb
r
is given by
cosθ = |
||
a b|
a b
⋅ r
r
r
r
Now
a b
⋅
rr
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
1 1 1
1
i
j
k
i
j
k
+
−
⋅
−
+
= − − = − Therefore, we have
cosθ =
31
−
hence the required angle is
θ =
1
1
cos
3
Example 15 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
3 and
3
5
a
i
j
k
b
i
j
k
=
−
−
=
+
−
r
r
, then show that the vectors
and
a
b
a
b
+
−
r
r
r
r
are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero Here
a
b
+
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
6
2
8
i
j
k
i
j
k
i
j
k
−
−
+
+
−
=
+
−
and
a
−b
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
4
4
2
i
j
k
i
j
k
i
j
k
−
−
−
+
−
=
−
+
So
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
(6
2
8 ) (4
4
2 )
24
8 16
0 |
1 | 5336-5339 | Therefore, we have
cosθ =
31
−
hence the required angle is
θ =
1
1
cos
3
Example 15 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
3 and
3
5
a
i
j
k
b
i
j
k
=
−
−
=
+
−
r
r
, then show that the vectors
and
a
b
a
b
+
−
r
r
r
r
are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero Here
a
b
+
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
6
2
8
i
j
k
i
j
k
i
j
k
−
−
+
+
−
=
+
−
and
a
−b
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
4
4
2
i
j
k
i
j
k
i
j
k
−
−
−
+
−
=
−
+
So
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
(6
2
8 ) (4
4
2 )
24
8 16
0 a
b
a
b
i
j
k
i
j
k
+
⋅
−
=
+
−
⋅
−
+
=
−
−
=
r
r
r
r
Hence
and
a
b
a
b
+
−
r
r
r
r
are perpendicular vectors |
1 | 5337-5340 | Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero Here
a
b
+
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
6
2
8
i
j
k
i
j
k
i
j
k
−
−
+
+
−
=
+
−
and
a
−b
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
4
4
2
i
j
k
i
j
k
i
j
k
−
−
−
+
−
=
−
+
So
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
(6
2
8 ) (4
4
2 )
24
8 16
0 a
b
a
b
i
j
k
i
j
k
+
⋅
−
=
+
−
⋅
−
+
=
−
−
=
r
r
r
r
Hence
and
a
b
a
b
+
−
r
r
r
r
are perpendicular vectors Example 16 Find the projection of the vector
ˆ
ˆ
ˆ
2
3
2
a
i
j
k
=
+
+
r
on the vector
ˆ
ˆ
2ˆ
b
i
j
k
=
+
+
r |
1 | 5338-5341 | Here
a
b
+
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
6
2
8
i
j
k
i
j
k
i
j
k
−
−
+
+
−
=
+
−
and
a
−b
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(5
3 )
(
3
5 )
4
4
2
i
j
k
i
j
k
i
j
k
−
−
−
+
−
=
−
+
So
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
(6
2
8 ) (4
4
2 )
24
8 16
0 a
b
a
b
i
j
k
i
j
k
+
⋅
−
=
+
−
⋅
−
+
=
−
−
=
r
r
r
r
Hence
and
a
b
a
b
+
−
r
r
r
r
are perpendicular vectors Example 16 Find the projection of the vector
ˆ
ˆ
ˆ
2
3
2
a
i
j
k
=
+
+
r
on the vector
ˆ
ˆ
2ˆ
b
i
j
k
=
+
+
r Solution The projection of vector ar on the vector b
r
is given by
1 (
)
|
|
a b
b
⋅
rr
r
=
2
2
2
(2 1
3 2
2 1)
10
5
6
3
6
(1)
(2)
(1)
× + ×
+
×
=
=
+
+
Example 17 Find |
|
a
b
−
r
r
, if two vectors
aand
rb
r
are such that |
|
2, |
|
3
a
b
r
r
and
4
a b
⋅
rr= |
1 | 5339-5342 | a
b
a
b
i
j
k
i
j
k
+
⋅
−
=
+
−
⋅
−
+
=
−
−
=
r
r
r
r
Hence
and
a
b
a
b
+
−
r
r
r
r
are perpendicular vectors Example 16 Find the projection of the vector
ˆ
ˆ
ˆ
2
3
2
a
i
j
k
=
+
+
r
on the vector
ˆ
ˆ
2ˆ
b
i
j
k
=
+
+
r Solution The projection of vector ar on the vector b
r
is given by
1 (
)
|
|
a b
b
⋅
rr
r
=
2
2
2
(2 1
3 2
2 1)
10
5
6
3
6
(1)
(2)
(1)
× + ×
+
×
=
=
+
+
Example 17 Find |
|
a
b
−
r
r
, if two vectors
aand
rb
r
are such that |
|
2, |
|
3
a
b
r
r
and
4
a b
⋅
rr= Solution We have
2
|
|
a
b
r
r
= (
) (
)
a
b
a
b
−
⋅
−
r
r
r
r
= |
1 | 5340-5343 | Example 16 Find the projection of the vector
ˆ
ˆ
ˆ
2
3
2
a
i
j
k
=
+
+
r
on the vector
ˆ
ˆ
2ˆ
b
i
j
k
=
+
+
r Solution The projection of vector ar on the vector b
r
is given by
1 (
)
|
|
a b
b
⋅
rr
r
=
2
2
2
(2 1
3 2
2 1)
10
5
6
3
6
(1)
(2)
(1)
× + ×
+
×
=
=
+
+
Example 17 Find |
|
a
b
−
r
r
, if two vectors
aand
rb
r
are such that |
|
2, |
|
3
a
b
r
r
and
4
a b
⋅
rr= Solution We have
2
|
|
a
b
r
r
= (
) (
)
a
b
a
b
−
⋅
−
r
r
r
r
= a a
a b
b a
b b
−
⋅
−
⋅
+
⋅
r
r
r r
r r
r
r
© NCERT
not to be republished
MATHEMATICS
446
B
C
A
a
b
+
a
b
=
2
2
|
|
2(
) |
|
a
a b
b
−
⋅
+
r
r
r
r
=
2
2
(2)
2(4)
(3)
−
+
Therefore
|
|
a
−b
r
r
=
5
Example 18 If ar is a unit vector and (
) (
)
8
x
a
x
a
−
⋅
+
=
r
r
r
r
, then find |
|
xr |
1 | 5341-5344 | Solution The projection of vector ar on the vector b
r
is given by
1 (
)
|
|
a b
b
⋅
rr
r
=
2
2
2
(2 1
3 2
2 1)
10
5
6
3
6
(1)
(2)
(1)
× + ×
+
×
=
=
+
+
Example 17 Find |
|
a
b
−
r
r
, if two vectors
aand
rb
r
are such that |
|
2, |
|
3
a
b
r
r
and
4
a b
⋅
rr= Solution We have
2
|
|
a
b
r
r
= (
) (
)
a
b
a
b
−
⋅
−
r
r
r
r
= a a
a b
b a
b b
−
⋅
−
⋅
+
⋅
r
r
r r
r r
r
r
© NCERT
not to be republished
MATHEMATICS
446
B
C
A
a
b
+
a
b
=
2
2
|
|
2(
) |
|
a
a b
b
−
⋅
+
r
r
r
r
=
2
2
(2)
2(4)
(3)
−
+
Therefore
|
|
a
−b
r
r
=
5
Example 18 If ar is a unit vector and (
) (
)
8
x
a
x
a
−
⋅
+
=
r
r
r
r
, then find |
|
xr Solution Since ar is a unit vector, |
ra =| 1 |
1 | 5342-5345 | Solution We have
2
|
|
a
b
r
r
= (
) (
)
a
b
a
b
−
⋅
−
r
r
r
r
= a a
a b
b a
b b
−
⋅
−
⋅
+
⋅
r
r
r r
r r
r
r
© NCERT
not to be republished
MATHEMATICS
446
B
C
A
a
b
+
a
b
=
2
2
|
|
2(
) |
|
a
a b
b
−
⋅
+
r
r
r
r
=
2
2
(2)
2(4)
(3)
−
+
Therefore
|
|
a
−b
r
r
=
5
Example 18 If ar is a unit vector and (
) (
)
8
x
a
x
a
−
⋅
+
=
r
r
r
r
, then find |
|
xr Solution Since ar is a unit vector, |
ra =| 1 Also,
(
) (
)
x
a
x
a
−
⋅
+
r
r
r
r = 8
or
x x
x a
a x
a a
⋅
+
⋅
−
⋅
−
⋅
r r
r r
r r
r r = 8
or
2
|
|
1
rx
= 8 i |
1 | 5343-5346 | a a
a b
b a
b b
−
⋅
−
⋅
+
⋅
r
r
r r
r r
r
r
© NCERT
not to be republished
MATHEMATICS
446
B
C
A
a
b
+
a
b
=
2
2
|
|
2(
) |
|
a
a b
b
−
⋅
+
r
r
r
r
=
2
2
(2)
2(4)
(3)
−
+
Therefore
|
|
a
−b
r
r
=
5
Example 18 If ar is a unit vector and (
) (
)
8
x
a
x
a
−
⋅
+
=
r
r
r
r
, then find |
|
xr Solution Since ar is a unit vector, |
ra =| 1 Also,
(
) (
)
x
a
x
a
−
⋅
+
r
r
r
r = 8
or
x x
x a
a x
a a
⋅
+
⋅
−
⋅
−
⋅
r r
r r
r r
r r = 8
or
2
|
|
1
rx
= 8 i e |
1 | 5344-5347 | Solution Since ar is a unit vector, |
ra =| 1 Also,
(
) (
)
x
a
x
a
−
⋅
+
r
r
r
r = 8
or
x x
x a
a x
a a
⋅
+
⋅
−
⋅
−
⋅
r r
r r
r r
r r = 8
or
2
|
|
1
rx
= 8 i e | rx |2 = 9
Therefore
|
|
xr = 3 (as magnitude of a vector is non negative) |
1 | 5345-5348 | Also,
(
) (
)
x
a
x
a
−
⋅
+
r
r
r
r = 8
or
x x
x a
a x
a a
⋅
+
⋅
−
⋅
−
⋅
r r
r r
r r
r r = 8
or
2
|
|
1
rx
= 8 i e | rx |2 = 9
Therefore
|
|
xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors
aand
rb
r
, we always have |
|
|
||
|
a b
a
b
⋅
≤
r
r
r
r
(Cauchy-
Schwartz inequality) |
1 | 5346-5349 | e | rx |2 = 9
Therefore
|
|
xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors
aand
rb
r
, we always have |
|
|
||
|
a b
a
b
⋅
≤
r
r
r
r
(Cauchy-
Schwartz inequality) Solution The inequality holds trivially when either
0 or
0
a
b
=
r=
r
r
r |
1 | 5347-5350 | | rx |2 = 9
Therefore
|
|
xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors
aand
rb
r
, we always have |
|
|
||
|
a b
a
b
⋅
≤
r
r
r
r
(Cauchy-
Schwartz inequality) Solution The inequality holds trivially when either
0 or
0
a
b
=
r=
r
r
r Actually, in such a
situation we have |
|
0
|
||
|
a b
a
b
⋅
=
=
r
r
r
r |
1 | 5348-5351 | Example 19 For any two vectors
aand
rb
r
, we always have |
|
|
||
|
a b
a
b
⋅
≤
r
r
r
r
(Cauchy-
Schwartz inequality) Solution The inequality holds trivially when either
0 or
0
a
b
=
r=
r
r
r Actually, in such a
situation we have |
|
0
|
||
|
a b
a
b
⋅
=
=
r
r
r
r So, let us assume that |
|
0
|
|
a
b
≠
≠
r
r |
1 | 5349-5352 | Solution The inequality holds trivially when either
0 or
0
a
b
=
r=
r
r
r Actually, in such a
situation we have |
|
0
|
||
|
a b
a
b
⋅
=
=
r
r
r
r So, let us assume that |
|
0
|
|
a
b
≠
≠
r
r Then, we have
|
|
|
||
a b|
a b
⋅
rr
r
r
= | cos |
1
θ ≤
Therefore
|
a b|
⋅
rr
≤ |
||
|
a
rb
r
Example 20 For any two vectors
aand
rb
r
, we always
have |
|
|
|
|
|
a
b
a
b
+
≤
+
r
r
r
r
(triangle inequality) |
1 | 5350-5353 | Actually, in such a
situation we have |
|
0
|
||
|
a b
a
b
⋅
=
=
r
r
r
r So, let us assume that |
|
0
|
|
a
b
≠
≠
r
r Then, we have
|
|
|
||
a b|
a b
⋅
rr
r
r
= | cos |
1
θ ≤
Therefore
|
a b|
⋅
rr
≤ |
||
|
a
rb
r
Example 20 For any two vectors
aand
rb
r
, we always
have |
|
|
|
|
|
a
b
a
b
+
≤
+
r
r
r
r
(triangle inequality) Solution The inequality holds trivially in case either
0 or
0
a
b
=
r=
r
r
r
(How |
1 | 5351-5354 | So, let us assume that |
|
0
|
|
a
b
≠
≠
r
r Then, we have
|
|
|
||
a b|
a b
⋅
rr
r
r
= | cos |
1
θ ≤
Therefore
|
a b|
⋅
rr
≤ |
||
|
a
rb
r
Example 20 For any two vectors
aand
rb
r
, we always
have |
|
|
|
|
|
a
b
a
b
+
≤
+
r
r
r
r
(triangle inequality) Solution The inequality holds trivially in case either
0 or
0
a
b
=
r=
r
r
r
(How ) |
1 | 5352-5355 | Then, we have
|
|
|
||
a b|
a b
⋅
rr
r
r
= | cos |
1
θ ≤
Therefore
|
a b|
⋅
rr
≤ |
||
|
a
rb
r
Example 20 For any two vectors
aand
rb
r
, we always
have |
|
|
|
|
|
a
b
a
b
+
≤
+
r
r
r
r
(triangle inequality) Solution The inequality holds trivially in case either
0 or
0
a
b
=
r=
r
r
r
(How ) So, let |
|
0
| |
a
b
r
r
r |
1 | 5353-5356 | Solution The inequality holds trivially in case either
0 or
0
a
b
=
r=
r
r
r
(How ) So, let |
|
0
| |
a
b
r
r
r Then,
2
|
|
a
+b
r
r
=
2
(
)
(
) (
)
a
b
a
b
a
b
+
=
+
⋅
+
r
r
r
r
r
r
= a a
a b
b a
b b
⋅
+
⋅
+
⋅
+
⋅
r
r
r r
r r
r
r
=
2
2
|
|
2
|
|
a
a b
b
+
⋅ +
r
r
r
r
(scalar product is commutative)
≤
2
2
|
|
2|
|
|
|
a
a b
b
+
⋅
r+
r
r
r
(since
|
|
x
x
x
≤
∀ ∈ R )
≤
2
2
|
|
2|
||
|
|
|
a
a b
b
+
+
r
r
r
r
(from Example 19)
=
2
(|
|
|
|)
a
b
r
r
Fig 10 |
1 | 5354-5357 | ) So, let |
|
0
| |
a
b
r
r
r Then,
2
|
|
a
+b
r
r
=
2
(
)
(
) (
)
a
b
a
b
a
b
+
=
+
⋅
+
r
r
r
r
r
r
= a a
a b
b a
b b
⋅
+
⋅
+
⋅
+
⋅
r
r
r r
r r
r
r
=
2
2
|
|
2
|
|
a
a b
b
+
⋅ +
r
r
r
r
(scalar product is commutative)
≤
2
2
|
|
2|
|
|
|
a
a b
b
+
⋅
r+
r
r
r
(since
|
|
x
x
x
≤
∀ ∈ R )
≤
2
2
|
|
2|
||
|
|
|
a
a b
b
+
+
r
r
r
r
(from Example 19)
=
2
(|
|
|
|)
a
b
r
r
Fig 10 21
© NCERT
not to be republished
VECTOR ALGEBRA
447
Hence
|
|
a
b
r
r
≤ |
|
|
|
a
b
r
r
Remark If the equality holds in triangle inequality (in the above Example 20), i |
1 | 5355-5358 | So, let |
|
0
| |
a
b
r
r
r Then,
2
|
|
a
+b
r
r
=
2
(
)
(
) (
)
a
b
a
b
a
b
+
=
+
⋅
+
r
r
r
r
r
r
= a a
a b
b a
b b
⋅
+
⋅
+
⋅
+
⋅
r
r
r r
r r
r
r
=
2
2
|
|
2
|
|
a
a b
b
+
⋅ +
r
r
r
r
(scalar product is commutative)
≤
2
2
|
|
2|
|
|
|
a
a b
b
+
⋅
r+
r
r
r
(since
|
|
x
x
x
≤
∀ ∈ R )
≤
2
2
|
|
2|
||
|
|
|
a
a b
b
+
+
r
r
r
r
(from Example 19)
=
2
(|
|
|
|)
a
b
r
r
Fig 10 21
© NCERT
not to be republished
VECTOR ALGEBRA
447
Hence
|
|
a
b
r
r
≤ |
|
|
|
a
b
r
r
Remark If the equality holds in triangle inequality (in the above Example 20), i e |
1 | 5356-5359 | Then,
2
|
|
a
+b
r
r
=
2
(
)
(
) (
)
a
b
a
b
a
b
+
=
+
⋅
+
r
r
r
r
r
r
= a a
a b
b a
b b
⋅
+
⋅
+
⋅
+
⋅
r
r
r r
r r
r
r
=
2
2
|
|
2
|
|
a
a b
b
+
⋅ +
r
r
r
r
(scalar product is commutative)
≤
2
2
|
|
2|
|
|
|
a
a b
b
+
⋅
r+
r
r
r
(since
|
|
x
x
x
≤
∀ ∈ R )
≤
2
2
|
|
2|
||
|
|
|
a
a b
b
+
+
r
r
r
r
(from Example 19)
=
2
(|
|
|
|)
a
b
r
r
Fig 10 21
© NCERT
not to be republished
VECTOR ALGEBRA
447
Hence
|
|
a
b
r
r
≤ |
|
|
|
a
b
r
r
Remark If the equality holds in triangle inequality (in the above Example 20), i e |
|
a
b
+
r
r
= |
|
|
|
a
+b
r
r
,
then
| AC|
uuur
= | AB|
uuur+| BC |
uuur
showing that the points A, B and C are collinear |
1 | 5357-5360 | 21
© NCERT
not to be republished
VECTOR ALGEBRA
447
Hence
|
|
a
b
r
r
≤ |
|
|
|
a
b
r
r
Remark If the equality holds in triangle inequality (in the above Example 20), i e |
|
a
b
+
r
r
= |
|
|
|
a
+b
r
r
,
then
| AC|
uuur
= | AB|
uuur+| BC |
uuur
showing that the points A, B and C are collinear Example 21 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A( 2
3
5 ), B(
2
3 )
i
j
k
i
j
k
−
+
+
+
+
and
ˆ
C(7ˆ
)
i
k
−
are collinear |
1 | 5358-5361 | e |
|
a
b
+
r
r
= |
|
|
|
a
+b
r
r
,
then
| AC|
uuur
= | AB|
uuur+| BC |
uuur
showing that the points A, B and C are collinear Example 21 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A( 2
3
5 ), B(
2
3 )
i
j
k
i
j
k
−
+
+
+
+
and
ˆ
C(7ˆ
)
i
k
−
are collinear Solution We have
AB
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(1
2)
(2
3)
(3
5)
3
2
i
j
k
i
j
k
,
BC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
1)
(0
2)
( 1
3)
6
2
4
i
j
k
i
j
k
,
AC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
2)
(0
3)
( 1
5)
9
3
6
i
j
k
i
j
k
| AB|
uuur
=
14, | BC|
2 14 and | AC| 3 14
uuur
uuur
Therefore
AC
uuur
= | AB|
uuur+| BC |
uuur
Hence the points A, B and C are collinear |
1 | 5359-5362 | |
|
a
b
+
r
r
= |
|
|
|
a
+b
r
r
,
then
| AC|
uuur
= | AB|
uuur+| BC |
uuur
showing that the points A, B and C are collinear Example 21 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A( 2
3
5 ), B(
2
3 )
i
j
k
i
j
k
−
+
+
+
+
and
ˆ
C(7ˆ
)
i
k
−
are collinear Solution We have
AB
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(1
2)
(2
3)
(3
5)
3
2
i
j
k
i
j
k
,
BC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
1)
(0
2)
( 1
3)
6
2
4
i
j
k
i
j
k
,
AC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
2)
(0
3)
( 1
5)
9
3
6
i
j
k
i
j
k
| AB|
uuur
=
14, | BC|
2 14 and | AC| 3 14
uuur
uuur
Therefore
AC
uuur
= | AB|
uuur+| BC |
uuur
Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB
BC
CA
0
+
+
=
uuur
uuur
uuur
r but the
points A, B and C do not form the vertices of a triangle |
1 | 5360-5363 | Example 21 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A( 2
3
5 ), B(
2
3 )
i
j
k
i
j
k
−
+
+
+
+
and
ˆ
C(7ˆ
)
i
k
−
are collinear Solution We have
AB
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(1
2)
(2
3)
(3
5)
3
2
i
j
k
i
j
k
,
BC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
1)
(0
2)
( 1
3)
6
2
4
i
j
k
i
j
k
,
AC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
2)
(0
3)
( 1
5)
9
3
6
i
j
k
i
j
k
| AB|
uuur
=
14, | BC|
2 14 and | AC| 3 14
uuur
uuur
Therefore
AC
uuur
= | AB|
uuur+| BC |
uuur
Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB
BC
CA
0
+
+
=
uuur
uuur
uuur
r but the
points A, B and C do not form the vertices of a triangle EXERCISE 10 |
1 | 5361-5364 | Solution We have
AB
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(1
2)
(2
3)
(3
5)
3
2
i
j
k
i
j
k
,
BC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
1)
(0
2)
( 1
3)
6
2
4
i
j
k
i
j
k
,
AC
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(7
2)
(0
3)
( 1
5)
9
3
6
i
j
k
i
j
k
| AB|
uuur
=
14, | BC|
2 14 and | AC| 3 14
uuur
uuur
Therefore
AC
uuur
= | AB|
uuur+| BC |
uuur
Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB
BC
CA
0
+
+
=
uuur
uuur
uuur
r but the
points A, B and C do not form the vertices of a triangle EXERCISE 10 3
1 |
1 | 5362-5365 | �Note In Example 21, one may note that although AB
BC
CA
0
+
+
=
uuur
uuur
uuur
r but the
points A, B and C do not form the vertices of a triangle EXERCISE 10 3
1 Find the angle between two vectors
aand
rb
r
with magnitudes
3 and 2 ,
respectively having
6
a b
⋅
=
rr |
1 | 5363-5366 | EXERCISE 10 3
1 Find the angle between two vectors
aand
rb
r
with magnitudes
3 and 2 ,
respectively having
6
a b
⋅
=
rr 2 |
1 | 5364-5367 | 3
1 Find the angle between two vectors
aand
rb
r
with magnitudes
3 and 2 ,
respectively having
6
a b
⋅
=
rr 2 Find the angle between the vectors
ˆ
ˆ
2ˆ
3
i
j
k
−
+
and
ˆ
ˆ
ˆ
3
2
i
j
k
−
+
3 |
1 | 5365-5368 | Find the angle between two vectors
aand
rb
r
with magnitudes
3 and 2 ,
respectively having
6
a b
⋅
=
rr 2 Find the angle between the vectors
ˆ
ˆ
2ˆ
3
i
j
k
−
+
and
ˆ
ˆ
ˆ
3
2
i
j
k
−
+
3 Find the projection of the vector ˆ
ˆ
i
−j
on the vector ˆ
ˆ
i
+j |
1 | 5366-5369 | 2 Find the angle between the vectors
ˆ
ˆ
2ˆ
3
i
j
k
−
+
and
ˆ
ˆ
ˆ
3
2
i
j
k
−
+
3 Find the projection of the vector ˆ
ˆ
i
−j
on the vector ˆ
ˆ
i
+j 4 |
1 | 5367-5370 | Find the angle between the vectors
ˆ
ˆ
2ˆ
3
i
j
k
−
+
and
ˆ
ˆ
ˆ
3
2
i
j
k
−
+
3 Find the projection of the vector ˆ
ˆ
i
−j
on the vector ˆ
ˆ
i
+j 4 Find the projection of the vector
ˆ
ˆ
3ˆ
7
i
j
k
+
+
on the vector
ˆ
ˆ
ˆ
7
8
i
j
k
−
+ |
1 | 5368-5371 | Find the projection of the vector ˆ
ˆ
i
−j
on the vector ˆ
ˆ
i
+j 4 Find the projection of the vector
ˆ
ˆ
3ˆ
7
i
j
k
+
+
on the vector
ˆ
ˆ
ˆ
7
8
i
j
k
−
+ 5 |
1 | 5369-5372 | 4 Find the projection of the vector
ˆ
ˆ
3ˆ
7
i
j
k
+
+
on the vector
ˆ
ˆ
ˆ
7
8
i
j
k
−
+ 5 Show that each of the given three vectors is a unit vector:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6 ),
(3
6
2 ),
(6
2
3 )
7
7
7
i
j
k
i
j
k
i
j
k
+
+
−
+
+
−
Also, show that they are mutually perpendicular to each other |
1 | 5370-5373 | Find the projection of the vector
ˆ
ˆ
3ˆ
7
i
j
k
+
+
on the vector
ˆ
ˆ
ˆ
7
8
i
j
k
−
+ 5 Show that each of the given three vectors is a unit vector:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6 ),
(3
6
2 ),
(6
2
3 )
7
7
7
i
j
k
i
j
k
i
j
k
+
+
−
+
+
−
Also, show that they are mutually perpendicular to each other © NCERT
not to be republished
MATHEMATICS
448
6 |
1 | 5371-5374 | 5 Show that each of the given three vectors is a unit vector:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6 ),
(3
6
2 ),
(6
2
3 )
7
7
7
i
j
k
i
j
k
i
j
k
+
+
−
+
+
−
Also, show that they are mutually perpendicular to each other © NCERT
not to be republished
MATHEMATICS
448
6 Find |
| and |
|
a
rb
r
, if (
) (
)
8 and | | 8|
|
a
b
a
b
a
b
+
⋅
−
=
=
r
r
r
r
r
r |
1 | 5372-5375 | Show that each of the given three vectors is a unit vector:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6 ),
(3
6
2 ),
(6
2
3 )
7
7
7
i
j
k
i
j
k
i
j
k
+
+
−
+
+
−
Also, show that they are mutually perpendicular to each other © NCERT
not to be republished
MATHEMATICS
448
6 Find |
| and |
|
a
rb
r
, if (
) (
)
8 and | | 8|
|
a
b
a
b
a
b
+
⋅
−
=
=
r
r
r
r
r
r 7 |
1 | 5373-5376 | © NCERT
not to be republished
MATHEMATICS
448
6 Find |
| and |
|
a
rb
r
, if (
) (
)
8 and | | 8|
|
a
b
a
b
a
b
+
⋅
−
=
=
r
r
r
r
r
r 7 Evaluate the product (3
5 ) (2
7 )
a
b
a
b
−
⋅
+
r
r
r
r |
1 | 5374-5377 | Find |
| and |
|
a
rb
r
, if (
) (
)
8 and | | 8|
|
a
b
a
b
a
b
+
⋅
−
=
=
r
r
r
r
r
r 7 Evaluate the product (3
5 ) (2
7 )
a
b
a
b
−
⋅
+
r
r
r
r 8 |
1 | 5375-5378 | 7 Evaluate the product (3
5 ) (2
7 )
a
b
a
b
−
⋅
+
r
r
r
r 8 Find the magnitude of two vectors
aand
rb
r
, having the same magnitude and
such that the angle between them is 60o and their scalar product is 1
2 |
1 | 5376-5379 | Evaluate the product (3
5 ) (2
7 )
a
b
a
b
−
⋅
+
r
r
r
r 8 Find the magnitude of two vectors
aand
rb
r
, having the same magnitude and
such that the angle between them is 60o and their scalar product is 1
2 9 |
1 | 5377-5380 | 8 Find the magnitude of two vectors
aand
rb
r
, having the same magnitude and
such that the angle between them is 60o and their scalar product is 1
2 9 Find |
|
xr , if for a unit vector ar , (
) (
)
12
x
a
x
a
−
⋅
+
=
r
r
r
r |
1 | 5378-5381 | Find the magnitude of two vectors
aand
rb
r
, having the same magnitude and
such that the angle between them is 60o and their scalar product is 1
2 9 Find |
|
xr , if for a unit vector ar , (
) (
)
12
x
a
x
a
−
⋅
+
=
r
r
r
r 10 |
1 | 5379-5382 | 9 Find |
|
xr , if for a unit vector ar , (
) (
)
12
x
a
x
a
−
⋅
+
=
r
r
r
r 10 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
3 ,
2
and
3
a
i
j
k
b
i
j
k
c
i
j
=
+
+
= − +
+
=
+
r
r
r
are such that a
+ λb
r
r
is
perpendicular to cr , then find the value of λ |
1 | 5380-5383 | Find |
|
xr , if for a unit vector ar , (
) (
)
12
x
a
x
a
−
⋅
+
=
r
r
r
r 10 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
3 ,
2
and
3
a
i
j
k
b
i
j
k
c
i
j
=
+
+
= − +
+
=
+
r
r
r
are such that a
+ λb
r
r
is
perpendicular to cr , then find the value of λ 11 |
1 | 5381-5384 | 10 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
3 ,
2
and
3
a
i
j
k
b
i
j
k
c
i
j
=
+
+
= − +
+
=
+
r
r
r
are such that a
+ λb
r
r
is
perpendicular to cr , then find the value of λ 11 Show that |
|
|
|
a b
r+b a
r
r
r is perpendicular to |
|
|
|
a b
r−b a
r
r
r , for any two nonzero
vectors
aand
rb
r |
1 | 5382-5385 | If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
3 ,
2
and
3
a
i
j
k
b
i
j
k
c
i
j
=
+
+
= − +
+
=
+
r
r
r
are such that a
+ λb
r
r
is
perpendicular to cr , then find the value of λ 11 Show that |
|
|
|
a b
r+b a
r
r
r is perpendicular to |
|
|
|
a b
r−b a
r
r
r , for any two nonzero
vectors
aand
rb
r 12 |
1 | 5383-5386 | 11 Show that |
|
|
|
a b
r+b a
r
r
r is perpendicular to |
|
|
|
a b
r−b a
r
r
r , for any two nonzero
vectors
aand
rb
r 12 If
0 and
0
a a
a b
⋅
=
⋅
=
r
r r
r
, then what can be concluded about the vector b
r |
1 | 5384-5387 | Show that |
|
|
|
a b
r+b a
r
r
r is perpendicular to |
|
|
|
a b
r−b a
r
r
r , for any two nonzero
vectors
aand
rb
r 12 If
0 and
0
a a
a b
⋅
=
⋅
=
r
r r
r
, then what can be concluded about the vector b
r 13 |
1 | 5385-5388 | 12 If
0 and
0
a a
a b
⋅
=
⋅
=
r
r r
r
, then what can be concluded about the vector b
r 13 If
, ,
a b c
rr
r are unit vectors such that
0
a
b
c
+
+
=
r
r
r
r
, find the value of
a b
b c
c a
⋅
+
⋅
+
⋅
r
r
r
r
r r |
1 | 5386-5389 | If
0 and
0
a a
a b
⋅
=
⋅
=
r
r r
r
, then what can be concluded about the vector b
r 13 If
, ,
a b c
rr
r are unit vectors such that
0
a
b
c
+
+
=
r
r
r
r
, find the value of
a b
b c
c a
⋅
+
⋅
+
⋅
r
r
r
r
r r 14 |
1 | 5387-5390 | 13 If
, ,
a b c
rr
r are unit vectors such that
0
a
b
c
+
+
=
r
r
r
r
, find the value of
a b
b c
c a
⋅
+
⋅
+
⋅
r
r
r
r
r r 14 If either vector
0 or
0, then
0
a
b
a b
=
=
⋅
=
r
r
r
r
r
r |
1 | 5388-5391 | If
, ,
a b c
rr
r are unit vectors such that
0
a
b
c
+
+
=
r
r
r
r
, find the value of
a b
b c
c a
⋅
+
⋅
+
⋅
r
r
r
r
r r 14 If either vector
0 or
0, then
0
a
b
a b
=
=
⋅
=
r
r
r
r
r
r But the converse need not be
true |
1 | 5389-5392 | 14 If either vector
0 or
0, then
0
a
b
a b
=
=
⋅
=
r
r
r
r
r
r But the converse need not be
true Justify your answer with an example |
1 | 5390-5393 | If either vector
0 or
0, then
0
a
b
a b
=
=
⋅
=
r
r
r
r
r
r But the converse need not be
true Justify your answer with an example 15 |
1 | 5391-5394 | 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 |
1 | 5392-5395 | 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 BA
uuur
and BC
uuur ] |
1 | 5393-5396 | 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 BA
uuur
and BC
uuur ] 16 |
1 | 5394-5397 | 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 BA
uuur
and BC
uuur ] 16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear |
1 | 5395-5398 | [∠ABC is the angle between the vectors BA
uuur
and BC
uuur ] 16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear 17 |
1 | 5396-5399 | 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
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
3
5
and 3
4
4
i
j
k i
j
k
i
j
k
−
+
−
−
−
−
form the vertices
of a right angled triangle |
1 | 5397-5400 | Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear 17 Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
3
5
and 3
4
4
i
j
k i
j
k
i
j
k
−
+
−
−
−
−
form the vertices
of a right angled triangle 18 |
1 | 5398-5401 | 17 Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
3
5
and 3
4
4
i
j
k i
j
k
i
j
k
−
+
−
−
−
−
form the vertices
of a right angled triangle 18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit
vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = 1/|λ|
10 |
1 | 5399-5402 | Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
3
5
and 3
4
4
i
j
k i
j
k
i
j
k
−
+
−
−
−
−
form the vertices
of a right angled triangle 18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit
vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = 1/|λ|
10 6 |
1 | 5400-5403 | 18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit
vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = 1/|λ|
10 6 3 Vector (or cross) product of two vectors
In Section 10 |
1 | 5401-5404 | If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit
vector if
(A) λ = 1
(B) λ = – 1
(C) a = |λ|
(D) a = 1/|λ|
10 6 3 Vector (or cross) product of two vectors
In Section 10 2, we have discussed on the three dimensional right handed rectangular
coordinate system |
1 | 5402-5405 | 6 3 Vector (or cross) product of two vectors
In Section 10 2, we have discussed on the three dimensional right handed rectangular
coordinate system In this system, when the positive x-axis is rotated counterclockwise
© NCERT
not to be republished
VECTOR ALGEBRA
449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10 |
1 | 5403-5406 | 3 Vector (or cross) product of two vectors
In Section 10 2, we have discussed on the three dimensional right handed rectangular
coordinate system In this system, when the positive x-axis is rotated counterclockwise
© NCERT
not to be republished
VECTOR ALGEBRA
449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10 22(i)) |
1 | 5404-5407 | 2, we have discussed on the three dimensional right handed rectangular
coordinate system In this system, when the positive x-axis is rotated counterclockwise
© NCERT
not to be republished
VECTOR ALGEBRA
449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10 22(i)) In a right handed coordinate system, the thumb of the right hand points in the
direction of the positive z-axis when the fingers are curled in the direction away from
the positive x-axis toward the positive y-axis (Fig 10 |
1 | 5405-5408 | In this system, when the positive x-axis is rotated counterclockwise
© NCERT
not to be republished
VECTOR ALGEBRA
449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10 22(i)) In a right handed coordinate system, the thumb of the right hand points in the
direction of the positive z-axis when the fingers are curled in the direction away from
the positive x-axis toward the positive y-axis (Fig 10 22(ii)) |
1 | 5406-5409 | 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 |
1 | 5407-5410 | In a right handed coordinate system, the thumb of the right hand points in the
direction of the positive z-axis when the fingers are curled in the direction away from
the positive x-axis toward the positive y-axis (Fig 10 22(ii)) Fig 10 22 (i), (ii)
Definition 3 The vector product of two nonzero vectors
aand
rb
r
, is denoted by a
b
r
r
and defined as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
,
where, θ is the angle between
aand
rb
r
, 0 ≤ θ ≤ π and ˆn is
a unit vector perpendicular to both
a and
rb
r
, such that
ˆ
,
a b and
n
rr
form a right handed system (Fig 10 |
1 | 5408-5411 | 22(ii)) Fig 10 22 (i), (ii)
Definition 3 The vector product of two nonzero vectors
aand
rb
r
, is denoted by a
b
r
r
and defined as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
,
where, θ is the angle between
aand
rb
r
, 0 ≤ θ ≤ π and ˆn is
a unit vector perpendicular to both
a and
rb
r
, such that
ˆ
,
a b and
n
rr
form a right handed system (Fig 10 23) |
1 | 5409-5412 | Fig 10 22 (i), (ii)
Definition 3 The vector product of two nonzero vectors
aand
rb
r
, is denoted by a
b
r
r
and defined as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
,
where, θ is the angle between
aand
rb
r
, 0 ≤ θ ≤ π and ˆn is
a unit vector perpendicular to both
a and
rb
r
, such that
ˆ
,
a b and
n
rr
form a right handed system (Fig 10 23) i |
1 | 5410-5413 | 22 (i), (ii)
Definition 3 The vector product of two nonzero vectors
aand
rb
r
, is denoted by a
b
r
r
and defined as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
,
where, θ is the angle between
aand
rb
r
, 0 ≤ θ ≤ π and ˆn is
a unit vector perpendicular to both
a and
rb
r
, such that
ˆ
,
a b and
n
rr
form a right handed system (Fig 10 23) i e |
1 | 5411-5414 | 23) i e , the
right handed system rotated from
ato
rb
r
moves in the
direction of ˆn |
1 | 5412-5415 | i e , the
right handed system rotated from
ato
rb
r
moves in the
direction of ˆn If either
0 or
0
a
b
=
r=
r
r
r
, then θ is not defined and in this case, we define
0
a
×b
r=
r
r |
1 | 5413-5416 | e , the
right handed system rotated from
ato
rb
r
moves in the
direction of ˆn If either
0 or
0
a
b
=
r=
r
r
r
, then θ is not defined and in this case, we define
0
a
×b
r=
r
r Observations
1 |
1 | 5414-5417 | , the
right handed system rotated from
ato
rb
r
moves in the
direction of ˆn If either
0 or
0
a
b
=
r=
r
r
r
, then θ is not defined and in this case, we define
0
a
×b
r=
r
r Observations
1 a
×b
r
r
is a vector |
1 | 5415-5418 | If either
0 or
0
a
b
=
r=
r
r
r
, then θ is not defined and in this case, we define
0
a
×b
r=
r
r Observations
1 a
×b
r
r
is a vector 2 |
1 | 5416-5419 | Observations
1 a
×b
r
r
is a vector 2 Let
aand
rb
r
be two nonzero vectors |
1 | 5417-5420 | a
×b
r
r
is a vector 2 Let
aand
rb
r
be two nonzero vectors Then
0
a
×b
r=
r
r
if and only if
a and
rb
r
are parallel (or collinear) to each other, i |
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