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1 | 5218-5221 | PQ
uuur =
ˆ
ˆ
ˆ
3
5
4 i
j
k
−
−
−
10 5 3 Section formula
Let P and Q be two points represented by the position vectorsOP and OQ
uuur
uuur , respectively,
with respect to the origin O |
1 | 5219-5222 | i
j
k
−
−
−
10 5 3 Section formula
Let P and Q be two points represented by the position vectorsOP and OQ
uuur
uuur , 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 |
1 | 5220-5223 | 5 3 Section formula
Let P and Q be two points represented by the position vectorsOP and OQ
uuur
uuur , 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 |
1 | 5221-5224 | 3 Section formula
Let P and Q be two points represented by the position vectorsOP and OQ
uuur
uuur , 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) |
1 | 5222-5225 | 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 OR
uuur for the point R with respect
to the origin O |
1 | 5223-5226 | 16)
and externally (Fig 10 17) Here, we intend to find
the position vector OR
uuur for the point R with respect
to the origin O We take the two cases one by one |
1 | 5224-5227 | 17) Here, we intend to find
the position vector OR
uuur for the point R with respect
to the origin O We take the two cases one by one Case I When R divides PQ internally (Fig 10 |
1 | 5225-5228 | Here, we intend to find
the position vector OR
uuur for the point R with respect
to the origin O We take the two cases one by one Case I When R divides PQ internally (Fig 10 16) |
1 | 5226-5229 | We take the two cases one by one Case I When R divides PQ internally (Fig 10 16) If R divides PQ
uuur
such that
RQ
m
uuur =
nPR
uuur ,
where m and n are positive scalars, we say that the point R divides PQ
uuur internally in the
ratio of m : n |
1 | 5227-5230 | Case I When R divides PQ internally (Fig 10 16) If R divides PQ
uuur
such that
RQ
m
uuur =
nPR
uuur ,
where m and n are positive scalars, we say that the point R divides PQ
uuur internally in the
ratio of m : n Now from triangles ORQ and OPR, we have
RQ
uuur = OQ
OR
b
r
−
=
−
uuur
uuur
r
r
and
PR
uuur = OR
OP
r
a
−
=
−
uuur
uuur
r
r ,
Therefore, we have
(
)
m b
r
−
r
r =
(
)
n r
r−a
r (Why |
1 | 5228-5231 | 16) If R divides PQ
uuur
such that
RQ
m
uuur =
nPR
uuur ,
where m and n are positive scalars, we say that the point R divides PQ
uuur internally in the
ratio of m : n Now from triangles ORQ and OPR, we have
RQ
uuur = OQ
OR
b
r
−
=
−
uuur
uuur
r
r
and
PR
uuur = OR
OP
r
a
−
=
−
uuur
uuur
r
r ,
Therefore, we have
(
)
m b
r
−
r
r =
(
)
n r
r−a
r (Why )
or
rr = mb
na
m
+n
+
r
r
(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
uuur = mb
na
m
+n
+
r
r
Fig 10 |
1 | 5229-5232 | If R divides PQ
uuur
such that
RQ
m
uuur =
nPR
uuur ,
where m and n are positive scalars, we say that the point R divides PQ
uuur internally in the
ratio of m : n Now from triangles ORQ and OPR, we have
RQ
uuur = OQ
OR
b
r
−
=
−
uuur
uuur
r
r
and
PR
uuur = OR
OP
r
a
−
=
−
uuur
uuur
r
r ,
Therefore, we have
(
)
m b
r
−
r
r =
(
)
n r
r−a
r (Why )
or
rr = mb
na
m
+n
+
r
r
(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
uuur = mb
na
m
+n
+
r
r
Fig 10 16
© NCERT
not to be republished
VECTOR ALGEBRA
439
Case II When R divides PQ externally (Fig 10 |
1 | 5230-5233 | Now from triangles ORQ and OPR, we have
RQ
uuur = OQ
OR
b
r
−
=
−
uuur
uuur
r
r
and
PR
uuur = OR
OP
r
a
−
=
−
uuur
uuur
r
r ,
Therefore, we have
(
)
m b
r
−
r
r =
(
)
n r
r−a
r (Why )
or
rr = mb
na
m
+n
+
r
r
(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
uuur = mb
na
m
+n
+
r
r
Fig 10 16
© NCERT
not to be republished
VECTOR ALGEBRA
439
Case II When R divides PQ externally (Fig 10 17) |
1 | 5231-5234 | )
or
rr = mb
na
m
+n
+
r
r
(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
uuur = mb
na
m
+n
+
r
r
Fig 10 16
© NCERT
not to be republished
VECTOR ALGEBRA
439
Case II When R divides PQ externally (Fig 10 17) We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
m : n
PR
i |
1 | 5232-5235 | 16
© NCERT
not to be republished
VECTOR ALGEBRA
439
Case II When R divides PQ externally (Fig 10 17) We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
m : n
PR
i e |
1 | 5233-5236 | 17) We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
m : n
PR
i e QR
nm
is given by
OR
uuur =
mb
na
m
−n
−
r
r
Remark If R is the midpoint of PQ , then m = n |
1 | 5234-5237 | We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
m : n
PR
i e QR
nm
is given by
OR
uuur =
mb
na
m
−n
−
r
r
Remark If R is the midpoint of PQ , then m = n And therefore, from Case I, the
midpoint R of PQ
uuur
, will have its position vector as
OR
uuur =
a2
b
+
r
r
Example 11 Consider two points P and Q with position vectors OP
3
2
a
b
=
−
uuur
r
r
and
OQ
a
b
uuur
r
r |
1 | 5235-5238 | e QR
nm
is given by
OR
uuur =
mb
na
m
−n
−
r
r
Remark If R is the midpoint of PQ , then m = n And therefore, from Case I, the
midpoint R of PQ
uuur
, will have its position vector as
OR
uuur =
a2
b
+
r
r
Example 11 Consider two points P and Q with position vectors OP
3
2
a
b
=
−
uuur
r
r
and
OQ
a
b
uuur
r
r 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 |
1 | 5236-5239 | QR
nm
is given by
OR
uuur =
mb
na
m
−n
−
r
r
Remark If R is the midpoint of PQ , then m = n And therefore, from Case I, the
midpoint R of PQ
uuur
, will have its position vector as
OR
uuur =
a2
b
+
r
r
Example 11 Consider two points P and Q with position vectors OP
3
2
a
b
=
−
uuur
r
r
and
OQ
a
b
uuur
r
r 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
uuur = 2(
)
(3
2 )
5
2
1
3
a
b
a
b
a
+
+
−
=
+
r
r
r
r
r
(ii)
The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
OR
uuur = 2(
)
(3
2 )
4
2
1
a
b
a
b
b
a
+
−
−
=
−
−
r
r
r
r
r
r
Example 12 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A(2
), B(
3
5 ), C(3
4
4 )
i
j
k
i
j
k
i
j
k
are
the vertices of a right angled triangle |
1 | 5237-5240 | And therefore, from Case I, the
midpoint R of PQ
uuur
, will have its position vector as
OR
uuur =
a2
b
+
r
r
Example 11 Consider two points P and Q with position vectors OP
3
2
a
b
=
−
uuur
r
r
and
OQ
a
b
uuur
r
r 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
uuur = 2(
)
(3
2 )
5
2
1
3
a
b
a
b
a
+
+
−
=
+
r
r
r
r
r
(ii)
The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
OR
uuur = 2(
)
(3
2 )
4
2
1
a
b
a
b
b
a
+
−
−
=
−
−
r
r
r
r
r
r
Example 12 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A(2
), B(
3
5 ), C(3
4
4 )
i
j
k
i
j
k
i
j
k
are
the vertices of a right angled triangle Solution We have
AB
uuur =
ˆ
ˆ
ˆ
(1
2)
( 3 1)
( 5 1)
i
j
k
−
+ − +
+ − −
ˆ
ˆ
2ˆ
6
i
j
k
BC
uuur =
ˆ
ˆ
ˆ
(3 1)
( 4
3)
( 4
5)
i
j
k
−
+ − +
+ − +
ˆ
ˆ
ˆ
2i
j
k
=
−
+
and
CA
uuur
=
ˆ
ˆ
ˆ
(2
3)
( 1
4)
(1
4)
i
j
k
−
+ − +
+
+
ˆ
ˆ
3ˆ
5
i
j
k
= − +
+
Fig 10 |
1 | 5238-5241 | 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
uuur = 2(
)
(3
2 )
5
2
1
3
a
b
a
b
a
+
+
−
=
+
r
r
r
r
r
(ii)
The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
OR
uuur = 2(
)
(3
2 )
4
2
1
a
b
a
b
b
a
+
−
−
=
−
−
r
r
r
r
r
r
Example 12 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A(2
), B(
3
5 ), C(3
4
4 )
i
j
k
i
j
k
i
j
k
are
the vertices of a right angled triangle Solution We have
AB
uuur =
ˆ
ˆ
ˆ
(1
2)
( 3 1)
( 5 1)
i
j
k
−
+ − +
+ − −
ˆ
ˆ
2ˆ
6
i
j
k
BC
uuur =
ˆ
ˆ
ˆ
(3 1)
( 4
3)
( 4
5)
i
j
k
−
+ − +
+ − +
ˆ
ˆ
ˆ
2i
j
k
=
−
+
and
CA
uuur
=
ˆ
ˆ
ˆ
(2
3)
( 1
4)
(1
4)
i
j
k
−
+ − +
+
+
ˆ
ˆ
3ˆ
5
i
j
k
= − +
+
Fig 10 17
© NCERT
not to be republished
MATHEMATICS
440
Further, note that
| AB|2
uuur
=
2
2
41
6
35 | BC |
| CA |
=
+
=
+
uuur
uuur
Hence, the triangle is a right angled triangle |
1 | 5239-5242 | Solution
(i)
The position vector of the point R dividing the join of P and Q internally in the
ratio 2:1 is
OR
uuur = 2(
)
(3
2 )
5
2
1
3
a
b
a
b
a
+
+
−
=
+
r
r
r
r
r
(ii)
The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
OR
uuur = 2(
)
(3
2 )
4
2
1
a
b
a
b
b
a
+
−
−
=
−
−
r
r
r
r
r
r
Example 12 Show that the points
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A(2
), B(
3
5 ), C(3
4
4 )
i
j
k
i
j
k
i
j
k
are
the vertices of a right angled triangle Solution We have
AB
uuur =
ˆ
ˆ
ˆ
(1
2)
( 3 1)
( 5 1)
i
j
k
−
+ − +
+ − −
ˆ
ˆ
2ˆ
6
i
j
k
BC
uuur =
ˆ
ˆ
ˆ
(3 1)
( 4
3)
( 4
5)
i
j
k
−
+ − +
+ − +
ˆ
ˆ
ˆ
2i
j
k
=
−
+
and
CA
uuur
=
ˆ
ˆ
ˆ
(2
3)
( 1
4)
(1
4)
i
j
k
−
+ − +
+
+
ˆ
ˆ
3ˆ
5
i
j
k
= − +
+
Fig 10 17
© NCERT
not to be republished
MATHEMATICS
440
Further, note that
| AB|2
uuur
=
2
2
41
6
35 | BC |
| CA |
=
+
=
+
uuur
uuur
Hence, the triangle is a right angled triangle EXERCISE 10 |
1 | 5240-5243 | Solution We have
AB
uuur =
ˆ
ˆ
ˆ
(1
2)
( 3 1)
( 5 1)
i
j
k
−
+ − +
+ − −
ˆ
ˆ
2ˆ
6
i
j
k
BC
uuur =
ˆ
ˆ
ˆ
(3 1)
( 4
3)
( 4
5)
i
j
k
−
+ − +
+ − +
ˆ
ˆ
ˆ
2i
j
k
=
−
+
and
CA
uuur
=
ˆ
ˆ
ˆ
(2
3)
( 1
4)
(1
4)
i
j
k
−
+ − +
+
+
ˆ
ˆ
3ˆ
5
i
j
k
= − +
+
Fig 10 17
© NCERT
not to be republished
MATHEMATICS
440
Further, note that
| AB|2
uuur
=
2
2
41
6
35 | BC |
| CA |
=
+
=
+
uuur
uuur
Hence, the triangle is a right angled triangle EXERCISE 10 2
1 |
1 | 5241-5244 | 17
© NCERT
not to be republished
MATHEMATICS
440
Further, note that
| AB|2
uuur
=
2
2
41
6
35 | BC |
| CA |
=
+
=
+
uuur
uuur
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
ˆ
ˆ
ˆ
ˆ
;
2
7
3 ;
3
3
3
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
−
=
+
−
r
r
r
2 |
1 | 5242-5245 | EXERCISE 10 2
1 Compute the magnitude of the following vectors:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
2
7
3 ;
3
3
3
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
−
=
+
−
r
r
r
2 Write two different vectors having same magnitude |
1 | 5243-5246 | 2
1 Compute the magnitude of the following vectors:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
2
7
3 ;
3
3
3
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
−
=
+
−
r
r
r
2 Write two different vectors having same magnitude 3 |
1 | 5244-5247 | Compute the magnitude of the following vectors:
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
;
2
7
3 ;
3
3
3
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
−
=
+
−
r
r
r
2 Write two different vectors having same magnitude 3 Write two different vectors having same direction |
1 | 5245-5248 | Write two different vectors having same magnitude 3 Write two different vectors having same direction 4 |
1 | 5246-5249 | 3 Write two different vectors having same direction 4 Find the values of x and y so that the vectors ˆ
ˆ
ˆ
ˆ
2
3 and
i
j
xi
yj
+
+
are equal |
1 | 5247-5250 | Write two different vectors having same direction 4 Find the values of x and y so that the vectors ˆ
ˆ
ˆ
ˆ
2
3 and
i
j
xi
yj
+
+
are equal 5 |
1 | 5248-5251 | 4 Find the values of x and y so that the vectors ˆ
ˆ
ˆ
ˆ
2
3 and
i
j
xi
yj
+
+
are equal 5 Find the scalar and vector components of the vector with initial point (2, 1) and
terminal point (– 5, 7) |
1 | 5249-5252 | Find the values of x and y so that the vectors ˆ
ˆ
ˆ
ˆ
2
3 and
i
j
xi
yj
+
+
are equal 5 Find the scalar and vector components of the vector with initial point (2, 1) and
terminal point (– 5, 7) 6 |
1 | 5250-5253 | 5 Find the scalar and vector components of the vector with initial point (2, 1) and
terminal point (– 5, 7) 6 Find the sum of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
2
4
5
a
i
j
k b
i
j
k
= −
+
= −
+
+
r
r
and
ˆ
ˆ
ˆ
6 – 7
c
i
j
k
=
−
r |
1 | 5251-5254 | Find the scalar and vector components of the vector with initial point (2, 1) and
terminal point (– 5, 7) 6 Find the sum of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
2
4
5
a
i
j
k b
i
j
k
= −
+
= −
+
+
r
r
and
ˆ
ˆ
ˆ
6 – 7
c
i
j
k
=
−
r 7 |
1 | 5252-5255 | 6 Find the sum of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
2
4
5
a
i
j
k b
i
j
k
= −
+
= −
+
+
r
r
and
ˆ
ˆ
ˆ
6 – 7
c
i
j
k
=
−
r 7 Find the unit vector in the direction of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
+
r |
1 | 5253-5256 | Find the sum of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
,
2
4
5
a
i
j
k b
i
j
k
= −
+
= −
+
+
r
r
and
ˆ
ˆ
ˆ
6 – 7
c
i
j
k
=
−
r 7 Find the unit vector in the direction of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
+
r 8 |
1 | 5254-5257 | 7 Find the unit vector in the direction of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
+
r 8 Find the unit vector in the direction of vector PQ,
uuur
where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively |
1 | 5255-5258 | Find the unit vector in the direction of the vector
ˆ
ˆ
ˆ
2
a
i
j
k
= +
+
r 8 Find the unit vector in the direction of vector PQ,
uuur
where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively 9 |
1 | 5256-5259 | 8 Find the unit vector in the direction of vector PQ,
uuur
where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively 9 For given vectors,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2 and
a
i
j
k
b
i
j
k
=
−
+
= − +
−
r
r
, find the unit vector in the
direction of the vector a
b
+
r
r |
1 | 5257-5260 | Find the unit vector in the direction of vector PQ,
uuur
where P and Q are the points
(1, 2, 3) and (4, 5, 6), respectively 9 For given vectors,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2 and
a
i
j
k
b
i
j
k
=
−
+
= − +
−
r
r
, find the unit vector in the
direction of the vector a
b
+
r
r 10 |
1 | 5258-5261 | 9 For given vectors,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2 and
a
i
j
k
b
i
j
k
=
−
+
= − +
−
r
r
, find the unit vector in the
direction of the vector a
b
+
r
r 10 Find a vector in the direction of vector
ˆ
ˆ
ˆ
5
2
i
j
k
−
+
which has magnitude 8 units |
1 | 5259-5262 | For given vectors,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2 and
a
i
j
k
b
i
j
k
=
−
+
= − +
−
r
r
, find the unit vector in the
direction of the vector a
b
+
r
r 10 Find a vector in the direction of vector
ˆ
ˆ
ˆ
5
2
i
j
k
−
+
which has magnitude 8 units 11 |
1 | 5260-5263 | 10 Find a vector in the direction of vector
ˆ
ˆ
ˆ
5
2
i
j
k
−
+
which has magnitude 8 units 11 Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4 and
4
6
8
i
j
k
i
j
k
−
+
−
+
−
are collinear |
1 | 5261-5264 | Find a vector in the direction of vector
ˆ
ˆ
ˆ
5
2
i
j
k
−
+
which has magnitude 8 units 11 Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4 and
4
6
8
i
j
k
i
j
k
−
+
−
+
−
are collinear 12 |
1 | 5262-5265 | 11 Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4 and
4
6
8
i
j
k
i
j
k
−
+
−
+
−
are collinear 12 Find the direction cosines of the vector
ˆ
ˆ
2ˆ
3
i
j
k
+
+ |
1 | 5263-5266 | Show that the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4 and
4
6
8
i
j
k
i
j
k
−
+
−
+
−
are collinear 12 Find the direction cosines of the vector
ˆ
ˆ
2ˆ
3
i
j
k
+
+ 13 |
1 | 5264-5267 | 12 Find the direction cosines of the vector
ˆ
ˆ
2ˆ
3
i
j
k
+
+ 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 |
1 | 5265-5268 | Find the direction cosines of the vector
ˆ
ˆ
2ˆ
3
i
j
k
+
+ 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 |
1 | 5266-5269 | 13 Find the direction cosines of the vector joining the points A (1, 2, –3) and
B(–1, –2, 1), directed from A to B 14 Show that the vector
ˆ
ˆ
ˆ
i
j
k
+
+
is equally inclined to the axes OX, OY and OZ |
1 | 5267-5270 | Find the direction cosines of the vector joining the points A (1, 2, –3) and
B(–1, –2, 1), directed from A to B 14 Show that the vector
ˆ
ˆ
ˆ
i
j
k
+
+
is equally inclined to the axes OX, OY and OZ 15 |
1 | 5268-5271 | 14 Show that the vector
ˆ
ˆ
ˆ
i
j
k
+
+
is equally inclined to the axes OX, OY and OZ 15 Find the position vector of a point R which divides the line joining two points P
and Q whose position vectors are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
and –
i
j
k
i
j
k
+
−
+
+
respectively, in the
ratio 2 : 1
(i) internally
(ii) externally
© NCERT
not to be republished
VECTOR ALGEBRA
441
16 |
1 | 5269-5272 | Show that the vector
ˆ
ˆ
ˆ
i
j
k
+
+
is equally inclined to the axes OX, OY and OZ 15 Find the position vector of a point R which divides the line joining two points P
and Q whose position vectors are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
and –
i
j
k
i
j
k
+
−
+
+
respectively, in the
ratio 2 : 1
(i) internally
(ii) externally
© NCERT
not to be republished
VECTOR ALGEBRA
441
16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2) |
1 | 5270-5273 | 15 Find the position vector of a point R which divides the line joining two points P
and Q whose position vectors are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
and –
i
j
k
i
j
k
+
−
+
+
respectively, in the
ratio 2 : 1
(i) internally
(ii) externally
© NCERT
not to be republished
VECTOR ALGEBRA
441
16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2) 17 |
1 | 5271-5274 | Find the position vector of a point R which divides the line joining two points P
and Q whose position vectors are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
and –
i
j
k
i
j
k
+
−
+
+
respectively, in the
ratio 2 : 1
(i) internally
(ii) externally
© NCERT
not to be republished
VECTOR ALGEBRA
441
16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2) 17 Show that the points A, B and C with position vectors,
ˆ
ˆ
ˆ
3
4
4 ,
a
i
j
k
=
−
−
r
ˆ
ˆ
ˆ
2
b
i
j
k
=
−
+
r
and
ˆ
ˆ
3ˆ
5
c
i
j
k
= −
−
r
, respectively form the vertices of a right angled
triangle |
1 | 5272-5275 | Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2) 17 Show that the points A, B and C with position vectors,
ˆ
ˆ
ˆ
3
4
4 ,
a
i
j
k
=
−
−
r
ˆ
ˆ
ˆ
2
b
i
j
k
=
−
+
r
and
ˆ
ˆ
3ˆ
5
c
i
j
k
= −
−
r
, respectively form the vertices of a right angled
triangle 18 |
1 | 5273-5276 | 17 Show that the points A, B and C with position vectors,
ˆ
ˆ
ˆ
3
4
4 ,
a
i
j
k
=
−
−
r
ˆ
ˆ
ˆ
2
b
i
j
k
=
−
+
r
and
ˆ
ˆ
3ˆ
5
c
i
j
k
= −
−
r
, respectively form the vertices of a right angled
triangle 18 In triangle ABC (Fig 10 |
1 | 5274-5277 | Show that the points A, B and C with position vectors,
ˆ
ˆ
ˆ
3
4
4 ,
a
i
j
k
=
−
−
r
ˆ
ˆ
ˆ
2
b
i
j
k
=
−
+
r
and
ˆ
ˆ
3ˆ
5
c
i
j
k
= −
−
r
, 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 = 0
uuur
uuuur
uuur
r
(B) AB
BC
AC
0
+
−
=
uuur
uuur
uuur
r
(C) AB
BC
CA
0
+
−
=
uuur
uuur
uuur
r
(D) AB
CB
CA
0
−
+
=
uuur
uuur
uuur
r
19 |
1 | 5275-5278 | 18 In triangle ABC (Fig 10 18), which of the following is not true:
(A) AB + BC + CA = 0
uuur
uuuur
uuur
r
(B) AB
BC
AC
0
+
−
=
uuur
uuur
uuur
r
(C) AB
BC
CA
0
+
−
=
uuur
uuur
uuur
r
(D) AB
CB
CA
0
−
+
=
uuur
uuur
uuur
r
19 If
aand
rb
r
are two collinear vectors, then which of the following are incorrect:
(A)
, for some scalar
b
= λa
λ
r
r
(B) a
= ±b
r
r
(C) the respective components of
aand
rb
r
are not proportional
(D) both the vectors
aand
rb
r
have same direction, but different magnitudes |
1 | 5276-5279 | In triangle ABC (Fig 10 18), which of the following is not true:
(A) AB + BC + CA = 0
uuur
uuuur
uuur
r
(B) AB
BC
AC
0
+
−
=
uuur
uuur
uuur
r
(C) AB
BC
CA
0
+
−
=
uuur
uuur
uuur
r
(D) AB
CB
CA
0
−
+
=
uuur
uuur
uuur
r
19 If
aand
rb
r
are two collinear vectors, then which of the following are incorrect:
(A)
, for some scalar
b
= λa
λ
r
r
(B) a
= ±b
r
r
(C) the respective components of
aand
rb
r
are not proportional
(D) both the vectors
aand
rb
r
have same direction, but different magnitudes 10 |
1 | 5277-5280 | 18), which of the following is not true:
(A) AB + BC + CA = 0
uuur
uuuur
uuur
r
(B) AB
BC
AC
0
+
−
=
uuur
uuur
uuur
r
(C) AB
BC
CA
0
+
−
=
uuur
uuur
uuur
r
(D) AB
CB
CA
0
−
+
=
uuur
uuur
uuur
r
19 If
aand
rb
r
are two collinear vectors, then which of the following are incorrect:
(A)
, for some scalar
b
= λa
λ
r
r
(B) a
= ±b
r
r
(C) the respective components of
aand
rb
r
are not proportional
(D) both the vectors
aand
rb
r
have same direction, but different magnitudes 10 6 Product of Two Vectors
So far we have studied about addition and subtraction of vectors |
1 | 5278-5281 | If
aand
rb
r
are two collinear vectors, then which of the following are incorrect:
(A)
, for some scalar
b
= λa
λ
r
r
(B) a
= ±b
r
r
(C) the respective components of
aand
rb
r
are not proportional
(D) both the vectors
aand
rb
r
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 |
1 | 5279-5282 | 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 |
1 | 5280-5283 | 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 |
1 | 5281-5284 | 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 |
1 | 5282-5285 | 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 |
1 | 5283-5286 | 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 |
1 | 5284-5287 | 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 |
1 | 5285-5288 | 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 | 5286-5289 | In this section,
we will discuss these two types of products 10 6 1 Scalar (or dot) product of two vectors
Definition 2 The scalar product of two nonzero vectors
aand
rb
r
, denoted by a b
⋅
r
r
, is
Fig 10 |
1 | 5287-5290 | 10 6 1 Scalar (or dot) product of two vectors
Definition 2 The scalar product of two nonzero vectors
aand
rb
r
, denoted by a b
⋅
r
r
, is
Fig 10 18
© NCERT
not to be republished
MATHEMATICS
442
defined as
a b
⋅
r
r
= |
| |
| cos ,
a
b
θ
r
r
where, θ is the angle between
and
, 0
a
b
r
r
(Fig 10 |
1 | 5288-5291 | 6 1 Scalar (or dot) product of two vectors
Definition 2 The scalar product of two nonzero vectors
aand
rb
r
, denoted by a b
⋅
r
r
, is
Fig 10 18
© NCERT
not to be republished
MATHEMATICS
442
defined as
a b
⋅
r
r
= |
| |
| cos ,
a
b
θ
r
r
where, θ is the angle between
and
, 0
a
b
r
r
(Fig 10 19) |
1 | 5289-5292 | 1 Scalar (or dot) product of two vectors
Definition 2 The scalar product of two nonzero vectors
aand
rb
r
, denoted by a b
⋅
r
r
, is
Fig 10 18
© NCERT
not to be republished
MATHEMATICS
442
defined as
a b
⋅
r
r
= |
| |
| cos ,
a
b
θ
r
r
where, θ is the angle between
and
, 0
a
b
r
r
(Fig 10 19) 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
Observations
1 |
1 | 5290-5293 | 18
© NCERT
not to be republished
MATHEMATICS
442
defined as
a b
⋅
r
r
= |
| |
| cos ,
a
b
θ
r
r
where, θ is the angle between
and
, 0
a
b
r
r
(Fig 10 19) 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
Observations
1 a b
⋅
r
r
is a real number |
1 | 5291-5294 | 19) 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
Observations
1 a b
⋅
r
r
is a real number 2 |
1 | 5292-5295 | 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
Observations
1 a b
⋅
r
r
is a real number 2 Let
aand
rb
r
be two nonzero vectors, then
0
a b
⋅
r=
r
if and only if
aand
rb
r
are
perpendicular to each other |
1 | 5293-5296 | a b
⋅
r
r
is a real number 2 Let
aand
rb
r
be two nonzero vectors, then
0
a b
⋅
r=
r
if and only if
aand
rb
r
are
perpendicular to each other i |
1 | 5294-5297 | 2 Let
aand
rb
r
be two nonzero vectors, then
0
a b
⋅
r=
r
if and only if
aand
rb
r
are
perpendicular to each other i e |
1 | 5295-5298 | Let
aand
rb
r
be two nonzero vectors, then
0
a b
⋅
r=
r
if and only if
aand
rb
r
are
perpendicular to each other i e 0
a b
a
b
⋅
=
⇔
⊥
r
r
r
r
3 |
1 | 5296-5299 | i e 0
a b
a
b
⋅
=
⇔
⊥
r
r
r
r
3 If θ = 0, then
|
| |
|
a b
a
b
⋅
r=
r
r
r
In particular,
2
|
| ,
a a
a
⋅
r r=
r
as θ in this case is 0 |
1 | 5297-5300 | e 0
a b
a
b
⋅
=
⇔
⊥
r
r
r
r
3 If θ = 0, then
|
| |
|
a b
a
b
⋅
r=
r
r
r
In particular,
2
|
| ,
a a
a
⋅
r r=
r
as θ in this case is 0 4 |
1 | 5298-5301 | 0
a b
a
b
⋅
=
⇔
⊥
r
r
r
r
3 If θ = 0, then
|
| |
|
a b
a
b
⋅
r=
r
r
r
In particular,
2
|
| ,
a a
a
⋅
r r=
r
as θ in this case is 0 4 If θ = π, then
|
| |
|
a b
a
b
⋅
r= −
r
r
r
In particular,
2
(
)
|
|
a
a
a
r
r
r
, as θ in this case is π |
1 | 5299-5302 | If θ = 0, then
|
| |
|
a b
a
b
⋅
r=
r
r
r
In particular,
2
|
| ,
a a
a
⋅
r r=
r
as θ in this case is 0 4 If θ = π, then
|
| |
|
a b
a
b
⋅
r= −
r
r
r
In particular,
2
(
)
|
|
a
a
a
r
r
r
, as θ in this case is π 5 |
1 | 5300-5303 | 4 If θ = π, then
|
| |
|
a b
a
b
⋅
r= −
r
r
r
In particular,
2
(
)
|
|
a
a
a
r
r
r
, 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
⋅
= ˆ ˆ
1,
k k
⋅
=
ˆ
ˆ ˆ
ˆ
i
j
j k
⋅
=
⋅
= ˆ ˆ
0
k i
6 |
1 | 5301-5304 | If θ = π, then
|
| |
|
a b
a
b
⋅
r= −
r
r
r
In particular,
2
(
)
|
|
a
a
a
r
r
r
, 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
⋅
= ˆ ˆ
1,
k k
⋅
=
ˆ
ˆ ˆ
ˆ
i
j
j k
⋅
=
⋅
= ˆ ˆ
0
k i
6 The angle between two nonzero vectors ra and
r
b is given by
cos
,
|
||
a b|
a b
r
r
r
r
or
–1 |
1 | 5302-5305 | 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
⋅
= ˆ ˆ
1,
k k
⋅
=
ˆ
ˆ ˆ
ˆ
i
j
j k
⋅
=
⋅
= ˆ ˆ
0
k i
6 The angle between two nonzero vectors ra and
r
b is given by
cos
,
|
||
a b|
a b
r
r
r
r
or
–1 cos
|
||
a b|
a b
⎛
⎞
θ =
⎜
⎟
⎝
⎠
rr
r
r
7 |
1 | 5303-5306 | 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
⋅
= ˆ ˆ
1,
k k
⋅
=
ˆ
ˆ ˆ
ˆ
i
j
j k
⋅
=
⋅
= ˆ ˆ
0
k i
6 The angle between two nonzero vectors ra and
r
b is given by
cos
,
|
||
a b|
a b
r
r
r
r
or
–1 cos
|
||
a b|
a b
⎛
⎞
θ =
⎜
⎟
⎝
⎠
rr
r
r
7 The scalar product is commutative |
1 | 5304-5307 | The angle between two nonzero vectors ra and
r
b is given by
cos
,
|
||
a b|
a b
r
r
r
r
or
–1 cos
|
||
a b|
a b
⎛
⎞
θ =
⎜
⎟
⎝
⎠
rr
r
r
7 The scalar product is commutative i |
1 | 5305-5308 | cos
|
||
a b|
a b
⎛
⎞
θ =
⎜
⎟
⎝
⎠
rr
r
r
7 The scalar product is commutative i e |
1 | 5306-5309 | The scalar product is commutative i e a b
⋅
rr
= b a
⋅
r r (Why |
1 | 5307-5310 | i e a b
⋅
rr
= b a
⋅
r r (Why )
Two important properties of scalar product
Property 1 (Distributivity of scalar product over addition) Let , and
a b
c
rr
r be
any three vectors, then
(
)
a
b
c
⋅
r+
r
r =
a b
a c
⋅
+
⋅
rr
r r
Fig 10 |
1 | 5308-5311 | e a b
⋅
rr
= b a
⋅
r r (Why )
Two important properties of scalar product
Property 1 (Distributivity of scalar product over addition) Let , and
a b
c
rr
r be
any three vectors, then
(
)
a
b
c
⋅
r+
r
r =
a b
a c
⋅
+
⋅
rr
r r
Fig 10 19
© NCERT
not to be republished
VECTOR ALGEBRA
443
(i)
B
C
A
l
B
l
A
C
(ii)
A
B
C
l
(iv)
l
C
B
A
(iii)
θ
θ
θ
θ
p
p
p
p
a
a
a
a
(90 <
< 180 )
0
0
θ
(0 <
< 90 )
0
0
θ
(270 <
< 360 )
0
0
θ
(180 <
< 270 )
0
0
θ
Property 2 Let
aand
rb
r
be any two vectors, and λ be any scalar |
1 | 5309-5312 | a b
⋅
rr
= b a
⋅
r r (Why )
Two important properties of scalar product
Property 1 (Distributivity of scalar product over addition) Let , and
a b
c
rr
r be
any three vectors, then
(
)
a
b
c
⋅
r+
r
r =
a b
a c
⋅
+
⋅
rr
r r
Fig 10 19
© NCERT
not to be republished
VECTOR ALGEBRA
443
(i)
B
C
A
l
B
l
A
C
(ii)
A
B
C
l
(iv)
l
C
B
A
(iii)
θ
θ
θ
θ
p
p
p
p
a
a
a
a
(90 <
< 180 )
0
0
θ
(0 <
< 90 )
0
0
θ
(270 <
< 360 )
0
0
θ
(180 <
< 270 )
0
0
θ
Property 2 Let
aand
rb
r
be any two vectors, and λ be any scalar Then
(
a)
b
λ
⋅
r
r
= (
)
(
)
(
)
a
b
a b
a
b
r
r
r
r
r
r
If two vectors
aand
rb
r
are given in component form as
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
+
+
, then their scalar product is given as
a b
⋅
rr
=
1
2
3
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
a i
a j
a k
b i
b j
b k
+
+
⋅
+
+
=
1
1
2
3
2
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
a i
b i
b j
b k
a j
b i
b j
b k
⋅
+
+
+
⋅
+
+
+
3
1
2
3
ˆ
ˆ
ˆ
ˆ
(
)
a k
b i
b j
b k
⋅
+
+
=
1 1
1 2
1 3
2 1
2 2
2 3
ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
(
)
(
)
(
)
(
)
(
)
(
)
a b i i
a b i
j
a b i k
a b
j i
a b
j j
a b
j k
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
3 1
3 2
3 3
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
(
)
(
)
(
)
a b k i
a b k j
a b k k
⋅
+
⋅
+
⋅
(Using the above Properties 1 and 2)
= a1b1 + a2b2 + a3b3
(Using Observation 5)
Thus
a b
⋅
rr
=
1 1
2 2
3 3
a b
a b
a b
+
+
10 |
1 | 5310-5313 | )
Two important properties of scalar product
Property 1 (Distributivity of scalar product over addition) Let , and
a b
c
rr
r be
any three vectors, then
(
)
a
b
c
⋅
r+
r
r =
a b
a c
⋅
+
⋅
rr
r r
Fig 10 19
© NCERT
not to be republished
VECTOR ALGEBRA
443
(i)
B
C
A
l
B
l
A
C
(ii)
A
B
C
l
(iv)
l
C
B
A
(iii)
θ
θ
θ
θ
p
p
p
p
a
a
a
a
(90 <
< 180 )
0
0
θ
(0 <
< 90 )
0
0
θ
(270 <
< 360 )
0
0
θ
(180 <
< 270 )
0
0
θ
Property 2 Let
aand
rb
r
be any two vectors, and λ be any scalar Then
(
a)
b
λ
⋅
r
r
= (
)
(
)
(
)
a
b
a b
a
b
r
r
r
r
r
r
If two vectors
aand
rb
r
are given in component form as
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
+
+
, then their scalar product is given as
a b
⋅
rr
=
1
2
3
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
a i
a j
a k
b i
b j
b k
+
+
⋅
+
+
=
1
1
2
3
2
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
a i
b i
b j
b k
a j
b i
b j
b k
⋅
+
+
+
⋅
+
+
+
3
1
2
3
ˆ
ˆ
ˆ
ˆ
(
)
a k
b i
b j
b k
⋅
+
+
=
1 1
1 2
1 3
2 1
2 2
2 3
ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
(
)
(
)
(
)
(
)
(
)
(
)
a b i i
a b i
j
a b i k
a b
j i
a b
j j
a b
j k
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
3 1
3 2
3 3
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
(
)
(
)
(
)
a b k i
a b k j
a b k k
⋅
+
⋅
+
⋅
(Using the above Properties 1 and 2)
= a1b1 + a2b2 + a3b3
(Using Observation 5)
Thus
a b
⋅
rr
=
1 1
2 2
3 3
a b
a b
a b
+
+
10 6 |
1 | 5311-5314 | 19
© NCERT
not to be republished
VECTOR ALGEBRA
443
(i)
B
C
A
l
B
l
A
C
(ii)
A
B
C
l
(iv)
l
C
B
A
(iii)
θ
θ
θ
θ
p
p
p
p
a
a
a
a
(90 <
< 180 )
0
0
θ
(0 <
< 90 )
0
0
θ
(270 <
< 360 )
0
0
θ
(180 <
< 270 )
0
0
θ
Property 2 Let
aand
rb
r
be any two vectors, and λ be any scalar Then
(
a)
b
λ
⋅
r
r
= (
)
(
)
(
)
a
b
a b
a
b
r
r
r
r
r
r
If two vectors
aand
rb
r
are given in component form as
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
+
+
, then their scalar product is given as
a b
⋅
rr
=
1
2
3
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
a i
a j
a k
b i
b j
b k
+
+
⋅
+
+
=
1
1
2
3
2
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
a i
b i
b j
b k
a j
b i
b j
b k
⋅
+
+
+
⋅
+
+
+
3
1
2
3
ˆ
ˆ
ˆ
ˆ
(
)
a k
b i
b j
b k
⋅
+
+
=
1 1
1 2
1 3
2 1
2 2
2 3
ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
(
)
(
)
(
)
(
)
(
)
(
)
a b i i
a b i
j
a b i k
a b
j i
a b
j j
a b
j k
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
3 1
3 2
3 3
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
(
)
(
)
(
)
a b k i
a b k j
a b k k
⋅
+
⋅
+
⋅
(Using the above Properties 1 and 2)
= a1b1 + a2b2 + a3b3
(Using Observation 5)
Thus
a b
⋅
rr
=
1 1
2 2
3 3
a b
a b
a b
+
+
10 6 2 Projection of a vector on a line
Suppose a vector AB
uuur
makes an angle θ with a given directed line l (say), in the
anticlockwise direction (Fig 10 |
1 | 5312-5315 | Then
(
a)
b
λ
⋅
r
r
= (
)
(
)
(
)
a
b
a b
a
b
r
r
r
r
r
r
If two vectors
aand
rb
r
are given in component form as
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
+
+
, then their scalar product is given as
a b
⋅
rr
=
1
2
3
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
a i
a j
a k
b i
b j
b k
+
+
⋅
+
+
=
1
1
2
3
2
1
2
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
a i
b i
b j
b k
a j
b i
b j
b k
⋅
+
+
+
⋅
+
+
+
3
1
2
3
ˆ
ˆ
ˆ
ˆ
(
)
a k
b i
b j
b k
⋅
+
+
=
1 1
1 2
1 3
2 1
2 2
2 3
ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
(
)
(
)
(
)
(
)
(
)
(
)
a b i i
a b i
j
a b i k
a b
j i
a b
j j
a b
j k
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
3 1
3 2
3 3
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
(
)
(
)
(
)
a b k i
a b k j
a b k k
⋅
+
⋅
+
⋅
(Using the above Properties 1 and 2)
= a1b1 + a2b2 + a3b3
(Using Observation 5)
Thus
a b
⋅
rr
=
1 1
2 2
3 3
a b
a b
a b
+
+
10 6 2 Projection of a vector on a line
Suppose a vector AB
uuur
makes an angle θ with a given directed line l (say), in the
anticlockwise direction (Fig 10 20) |
1 | 5313-5316 | 6 2 Projection of a vector on a line
Suppose a vector AB
uuur
makes an angle θ with a given directed line l (say), in the
anticlockwise direction (Fig 10 20) Then the projection of AB
uuur
on l is a vector pr
(say) with magnitude | AB | cosθ
uuur
, and the direction of pr being the same (or opposite)
to that of the line l, depending upon whether cosθ is positive or negative |
1 | 5314-5317 | 2 Projection of a vector on a line
Suppose a vector AB
uuur
makes an angle θ with a given directed line l (say), in the
anticlockwise direction (Fig 10 20) Then the projection of AB
uuur
on l is a vector pr
(say) with magnitude | AB | cosθ
uuur
, and the direction of pr being the same (or opposite)
to that of the line l, depending upon whether cosθ is positive or negative The vector pr
Fig 10 |
1 | 5315-5318 | 20) Then the projection of AB
uuur
on l is a vector pr
(say) with magnitude | AB | cosθ
uuur
, and the direction of pr being the same (or opposite)
to that of the line l, depending upon whether cosθ is positive or negative The vector pr
Fig 10 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 |
1 | 5316-5319 | Then the projection of AB
uuur
on l is a vector pr
(say) with magnitude | AB | cosθ
uuur
, and the direction of pr being the same (or opposite)
to that of the line l, depending upon whether cosθ is positive or negative The vector pr
Fig 10 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 |
1 | 5317-5320 | The vector pr
Fig 10 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 |
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