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1 | 5518-5521 | (
) c a
b
c c
+
r+
r r
r r
=
2
2
2
|
|
|
|
|
|
a
b
c
+
+
r
r
r
= 9 + 16 + 25 = 50
Therefore
|
|
a
b
c
+
r+
r
r =
50
5 2
=
© NCERT
not to be republished
VECTOR ALGEBRA
457
Example 29 Three vectors , and
a
b
c
r
r
r satisfy the condition
0
a
b
c
+
+
=
r
r
r
r Evaluate
the quantity
, if |
| 1, |
| 4 and |
| 2
a b
b c
c a
a
b
c
μ =
⋅
+
⋅
+ ⋅
=
=
=
r
r
r
r
r
r r
r
r Solution Since
0
a
b
c
+
+
=
r
r
r
r
, we have
(
)
a
a
b
c
r
r
r
r = 0
or
a a
a b
a c
⋅
+
⋅
+
⋅
r
r
r r
r
r
= 0
Therefore
a b
a c
⋅
+
⋅
r
r
r
r
=
2
1
−a
= −
r |
1 | 5519-5522 | c a
b
c c
+
r+
r r
r r
=
2
2
2
|
|
|
|
|
|
a
b
c
+
+
r
r
r
= 9 + 16 + 25 = 50
Therefore
|
|
a
b
c
+
r+
r
r =
50
5 2
=
© NCERT
not to be republished
VECTOR ALGEBRA
457
Example 29 Three vectors , and
a
b
c
r
r
r satisfy the condition
0
a
b
c
+
+
=
r
r
r
r Evaluate
the quantity
, if |
| 1, |
| 4 and |
| 2
a b
b c
c a
a
b
c
μ =
⋅
+
⋅
+ ⋅
=
=
=
r
r
r
r
r
r r
r
r Solution Since
0
a
b
c
+
+
=
r
r
r
r
, we have
(
)
a
a
b
c
r
r
r
r = 0
or
a a
a b
a c
⋅
+
⋅
+
⋅
r
r
r r
r
r
= 0
Therefore
a b
a c
⋅
+
⋅
r
r
r
r
=
2
1
−a
= −
r (1)
Again,
(
)
b
a
b
c
⋅
+
+
r
r
r
r = 0
or
a b
b c
⋅
+
⋅
r
r r
r
=
2
16
−b
= −
r |
1 | 5520-5523 | Evaluate
the quantity
, if |
| 1, |
| 4 and |
| 2
a b
b c
c a
a
b
c
μ =
⋅
+
⋅
+ ⋅
=
=
=
r
r
r
r
r
r r
r
r Solution Since
0
a
b
c
+
+
=
r
r
r
r
, we have
(
)
a
a
b
c
r
r
r
r = 0
or
a a
a b
a c
⋅
+
⋅
+
⋅
r
r
r r
r
r
= 0
Therefore
a b
a c
⋅
+
⋅
r
r
r
r
=
2
1
−a
= −
r (1)
Again,
(
)
b
a
b
c
⋅
+
+
r
r
r
r = 0
or
a b
b c
⋅
+
⋅
r
r r
r
=
2
16
−b
= −
r (2)
Similarly
a c
b c
⋅
+
r⋅
r r
r = – 4 |
1 | 5521-5524 | Solution Since
0
a
b
c
+
+
=
r
r
r
r
, we have
(
)
a
a
b
c
r
r
r
r = 0
or
a a
a b
a c
⋅
+
⋅
+
⋅
r
r
r r
r
r
= 0
Therefore
a b
a c
⋅
+
⋅
r
r
r
r
=
2
1
−a
= −
r (1)
Again,
(
)
b
a
b
c
⋅
+
+
r
r
r
r = 0
or
a b
b c
⋅
+
⋅
r
r r
r
=
2
16
−b
= −
r (2)
Similarly
a c
b c
⋅
+
r⋅
r r
r = – 4 (3)
Adding (1), (2) and (3), we have
2 (
)
a b
b
a
c
c
⋅
+
⋅ +
⋅
r
r
r
r
r
r
= – 21
or
2μ = – 21, i |
1 | 5522-5525 | (1)
Again,
(
)
b
a
b
c
⋅
+
+
r
r
r
r = 0
or
a b
b c
⋅
+
⋅
r
r r
r
=
2
16
−b
= −
r (2)
Similarly
a c
b c
⋅
+
r⋅
r r
r = – 4 (3)
Adding (1), (2) and (3), we have
2 (
)
a b
b
a
c
c
⋅
+
⋅ +
⋅
r
r
r
r
r
r
= – 21
or
2μ = – 21, i e |
1 | 5523-5526 | (2)
Similarly
a c
b c
⋅
+
r⋅
r r
r = – 4 (3)
Adding (1), (2) and (3), we have
2 (
)
a b
b
a
c
c
⋅
+
⋅ +
⋅
r
r
r
r
r
r
= – 21
or
2μ = – 21, i e , μ =
221
−
Example 30 If with reference to the right handed system of mutually perpendicular
unit vectors
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
,
and ,
3
,
2
– 3
i
j
k
i
j
i
j
k
α =
−
β =
+
r
r
, then express β
r in the form
1
2
, where 1
r
r
r
r
is parallel to
and 2
r
r
is perpendicular to αr |
1 | 5524-5527 | (3)
Adding (1), (2) and (3), we have
2 (
)
a b
b
a
c
c
⋅
+
⋅ +
⋅
r
r
r
r
r
r
= – 21
or
2μ = – 21, i e , μ =
221
−
Example 30 If with reference to the right handed system of mutually perpendicular
unit vectors
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
,
and ,
3
,
2
– 3
i
j
k
i
j
i
j
k
α =
−
β =
+
r
r
, then express β
r in the form
1
2
, where 1
r
r
r
r
is parallel to
and 2
r
r
is perpendicular to αr Solution Let
1
,
r
r
is a scalar, i |
1 | 5525-5528 | e , μ =
221
−
Example 30 If with reference to the right handed system of mutually perpendicular
unit vectors
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
,
and ,
3
,
2
– 3
i
j
k
i
j
i
j
k
α =
−
β =
+
r
r
, then express β
r in the form
1
2
, where 1
r
r
r
r
is parallel to
and 2
r
r
is perpendicular to αr Solution Let
1
,
r
r
is a scalar, i e |
1 | 5526-5529 | , μ =
221
−
Example 30 If with reference to the right handed system of mutually perpendicular
unit vectors
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
,
and ,
3
,
2
– 3
i
j
k
i
j
i
j
k
α =
−
β =
+
r
r
, then express β
r in the form
1
2
, where 1
r
r
r
r
is parallel to
and 2
r
r
is perpendicular to αr Solution Let
1
,
r
r
is a scalar, i e ,
1
ˆ
ˆ
3 i
rβ = λ − λj |
1 | 5527-5530 | Solution Let
1
,
r
r
is a scalar, i e ,
1
ˆ
ˆ
3 i
rβ = λ − λj Now
2
rβ = β −β1
r
r =
ˆ
ˆ
ˆ
(2
3 )
(1
)
3
i
j
k
− λ
+
+ λ
− |
1 | 5528-5531 | e ,
1
ˆ
ˆ
3 i
rβ = λ − λj Now
2
rβ = β −β1
r
r =
ˆ
ˆ
ˆ
(2
3 )
(1
)
3
i
j
k
− λ
+
+ λ
− Now, since
β2
r
is to be perpendicular to αr , we should have
2
α⋅β =0
rr |
1 | 5529-5532 | ,
1
ˆ
ˆ
3 i
rβ = λ − λj Now
2
rβ = β −β1
r
r =
ˆ
ˆ
ˆ
(2
3 )
(1
)
3
i
j
k
− λ
+
+ λ
− Now, since
β2
r
is to be perpendicular to αr , we should have
2
α⋅β =0
rr i |
1 | 5530-5533 | Now
2
rβ = β −β1
r
r =
ˆ
ˆ
ˆ
(2
3 )
(1
)
3
i
j
k
− λ
+
+ λ
− Now, since
β2
r
is to be perpendicular to αr , we should have
2
α⋅β =0
rr i e |
1 | 5531-5534 | Now, since
β2
r
is to be perpendicular to αr , we should have
2
α⋅β =0
rr i e ,
3(2
3 )
(1
)
− λ −
+ λ = 0
or
λ = 1
2
Therefore
β1
r = 3
1
ˆ
ˆ
2
2
i
j
−
and
2
1
3
ˆ
ˆ
ˆ – 3
2
2
i
j
k
β =
+
r
© NCERT
not to be republished
MATHEMATICS
458
Miscellaneous Exercise on Chapter 10
1 |
1 | 5532-5535 | i e ,
3(2
3 )
(1
)
− λ −
+ λ = 0
or
λ = 1
2
Therefore
β1
r = 3
1
ˆ
ˆ
2
2
i
j
−
and
2
1
3
ˆ
ˆ
ˆ – 3
2
2
i
j
k
β =
+
r
© NCERT
not to be republished
MATHEMATICS
458
Miscellaneous Exercise on Chapter 10
1 Write down a unit vector in XY-plane, making an angle of 30° with the positive
direction of x-axis |
1 | 5533-5536 | e ,
3(2
3 )
(1
)
− λ −
+ λ = 0
or
λ = 1
2
Therefore
β1
r = 3
1
ˆ
ˆ
2
2
i
j
−
and
2
1
3
ˆ
ˆ
ˆ – 3
2
2
i
j
k
β =
+
r
© NCERT
not to be republished
MATHEMATICS
458
Miscellaneous Exercise on Chapter 10
1 Write down a unit vector in XY-plane, making an angle of 30° with the positive
direction of x-axis 2 |
1 | 5534-5537 | ,
3(2
3 )
(1
)
− λ −
+ λ = 0
or
λ = 1
2
Therefore
β1
r = 3
1
ˆ
ˆ
2
2
i
j
−
and
2
1
3
ˆ
ˆ
ˆ – 3
2
2
i
j
k
β =
+
r
© NCERT
not to be republished
MATHEMATICS
458
Miscellaneous Exercise on Chapter 10
1 Write down a unit vector in XY-plane, making an angle of 30° with the positive
direction of x-axis 2 Find the scalar components and magnitude of the vector joining the points
P(x1, y1, z1) and Q(x2, y2, z2) |
1 | 5535-5538 | Write down a unit vector in XY-plane, making an angle of 30° with the positive
direction of x-axis 2 Find the scalar components and magnitude of the vector joining the points
P(x1, y1, z1) and Q(x2, y2, z2) 3 |
1 | 5536-5539 | 2 Find the scalar components and magnitude of the vector joining the points
P(x1, y1, z1) and Q(x2, y2, z2) 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of
north and stops |
1 | 5537-5540 | Find the scalar components and magnitude of the vector joining the points
P(x1, y1, z1) and Q(x2, y2, z2) 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of
north and stops Determine the girl’s displacement from her initial point of
departure |
1 | 5538-5541 | 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of
north and stops Determine the girl’s displacement from her initial point of
departure 4 |
1 | 5539-5542 | A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of
north and stops Determine the girl’s displacement from her initial point of
departure 4 If a
b
c
=
r+
r
r , then is it true that |
| |
|
|
|
a
b
c
=
+
r
r
r |
1 | 5540-5543 | Determine the girl’s displacement from her initial point of
departure 4 If a
b
c
=
r+
r
r , then is it true that |
| |
|
|
|
a
b
c
=
+
r
r
r Justify your answer |
1 | 5541-5544 | 4 If a
b
c
=
r+
r
r , then is it true that |
| |
|
|
|
a
b
c
=
+
r
r
r Justify your answer 5 |
1 | 5542-5545 | If a
b
c
=
r+
r
r , then is it true that |
| |
|
|
|
a
b
c
=
+
r
r
r Justify your answer 5 Find the value of x for which
ˆ
ˆ
ˆ
(
)
x i
j
k
+
+
is a unit vector |
1 | 5543-5546 | Justify your answer 5 Find the value of x for which
ˆ
ˆ
ˆ
(
)
x i
j
k
+
+
is a unit vector 6 |
1 | 5544-5547 | 5 Find the value of x for which
ˆ
ˆ
ˆ
(
)
x i
j
k
+
+
is a unit vector 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
and
2
a
i
j
k
b
i
j
k
=
+
−
= −
+
r
r |
1 | 5545-5548 | Find the value of x for which
ˆ
ˆ
ˆ
(
)
x i
j
k
+
+
is a unit vector 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
and
2
a
i
j
k
b
i
j
k
=
+
−
= −
+
r
r 7 |
1 | 5546-5549 | 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
and
2
a
i
j
k
b
i
j
k
=
+
−
= −
+
r
r 7 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
3
and
2
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
+
= −
+
r
r
r
, find a unit vector parallel
to the vector 2
–
3
a
b
r+c
r
r |
1 | 5547-5550 | Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
and
2
a
i
j
k
b
i
j
k
=
+
−
= −
+
r
r 7 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
3
and
2
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
+
= −
+
r
r
r
, find a unit vector parallel
to the vector 2
–
3
a
b
r+c
r
r 8 |
1 | 5548-5551 | 7 If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
3
and
2
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
+
= −
+
r
r
r
, find a unit vector parallel
to the vector 2
–
3
a
b
r+c
r
r 8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and
find the ratio in which B divides AC |
1 | 5549-5552 | If
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
3
and
2
a
i
j
k
b
i
j
k
c
i
j
k
= +
+
=
−
+
= −
+
r
r
r
, find a unit vector parallel
to the vector 2
–
3
a
b
r+c
r
r 8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and
find the ratio in which B divides AC 9 |
1 | 5550-5553 | 8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and
find the ratio in which B divides AC 9 Find the position vector of a point R which divides the line joining two points
P and Q whose position vectors are (2
) and ( – 3 )
a
b
a
b
+
r
r
r
r
externally in the ratio
1 : 2 |
1 | 5551-5554 | Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and
find the ratio in which B divides AC 9 Find the position vector of a point R which divides the line joining two points
P and Q whose position vectors are (2
) and ( – 3 )
a
b
a
b
+
r
r
r
r
externally in the ratio
1 : 2 Also, show that P is the mid point of the line segment RQ |
1 | 5552-5555 | 9 Find the position vector of a point R which divides the line joining two points
P and Q whose position vectors are (2
) and ( – 3 )
a
b
a
b
+
r
r
r
r
externally in the ratio
1 : 2 Also, show that P is the mid point of the line segment RQ 10 |
1 | 5553-5556 | Find the position vector of a point R which divides the line joining two points
P and Q whose position vectors are (2
) and ( – 3 )
a
b
a
b
+
r
r
r
r
externally in the ratio
1 : 2 Also, show that P is the mid point of the line segment RQ 10 The two adjacent sides of a parallelogram are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
5 and
2
3
i
j
k
i
j
k
−
+
−
− |
1 | 5554-5557 | Also, show that P is the mid point of the line segment RQ 10 The two adjacent sides of a parallelogram are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
5 and
2
3
i
j
k
i
j
k
−
+
−
− Find the unit vector parallel to its diagonal |
1 | 5555-5558 | 10 The two adjacent sides of a parallelogram are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
5 and
2
3
i
j
k
i
j
k
−
+
−
− Find the unit vector parallel to its diagonal Also, find its area |
1 | 5556-5559 | The two adjacent sides of a parallelogram are
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
5 and
2
3
i
j
k
i
j
k
−
+
−
− Find the unit vector parallel to its diagonal Also, find its area 11 |
1 | 5557-5560 | Find the unit vector parallel to its diagonal Also, find its area 11 Show that the direction cosines of a vector equally inclined to the axes OX, OY
and OZ are 1
1
1
,
, |
1 | 5558-5561 | Also, find its area 11 Show that the direction cosines of a vector equally inclined to the axes OX, OY
and OZ are 1
1
1
,
, 3
3
3
12 |
1 | 5559-5562 | 11 Show that the direction cosines of a vector equally inclined to the axes OX, OY
and OZ are 1
1
1
,
, 3
3
3
12 Let
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
2 ,
3
2
7 and
2
4
a
i
j
k b
i
j
k
c
i
j
k
= +
+
=
−
+
=
−
+
r
r
r |
1 | 5560-5563 | Show that the direction cosines of a vector equally inclined to the axes OX, OY
and OZ are 1
1
1
,
, 3
3
3
12 Let
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
2 ,
3
2
7 and
2
4
a
i
j
k b
i
j
k
c
i
j
k
= +
+
=
−
+
=
−
+
r
r
r Find a vector d
r
which is perpendicular to both
a and
rb
r
, and
15
c d
⋅
rr= |
1 | 5561-5564 | 3
3
3
12 Let
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
2 ,
3
2
7 and
2
4
a
i
j
k b
i
j
k
c
i
j
k
= +
+
=
−
+
=
−
+
r
r
r Find a vector d
r
which is perpendicular to both
a and
rb
r
, and
15
c d
⋅
rr= 13 |
1 | 5562-5565 | Let
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
2 ,
3
2
7 and
2
4
a
i
j
k b
i
j
k
c
i
j
k
= +
+
=
−
+
=
−
+
r
r
r Find a vector d
r
which is perpendicular to both
a and
rb
r
, and
15
c d
⋅
rr= 13 The scalar product of the vector
ˆ
ˆ
ˆ
i
j
k
+
+
with a unit vector along the sum of
vectors
ˆ
ˆ
ˆ
2
4
5
i
j
k
+
−
and
ˆ
ˆ
2ˆ
3
i
j
k
λ +
+
is equal to one |
1 | 5563-5566 | Find a vector d
r
which is perpendicular to both
a and
rb
r
, and
15
c d
⋅
rr= 13 The scalar product of the vector
ˆ
ˆ
ˆ
i
j
k
+
+
with a unit vector along the sum of
vectors
ˆ
ˆ
ˆ
2
4
5
i
j
k
+
−
and
ˆ
ˆ
2ˆ
3
i
j
k
λ +
+
is equal to one Find the value of λ |
1 | 5564-5567 | 13 The scalar product of the vector
ˆ
ˆ
ˆ
i
j
k
+
+
with a unit vector along the sum of
vectors
ˆ
ˆ
ˆ
2
4
5
i
j
k
+
−
and
ˆ
ˆ
2ˆ
3
i
j
k
λ +
+
is equal to one Find the value of λ 14 |
1 | 5565-5568 | The scalar product of the vector
ˆ
ˆ
ˆ
i
j
k
+
+
with a unit vector along the sum of
vectors
ˆ
ˆ
ˆ
2
4
5
i
j
k
+
−
and
ˆ
ˆ
2ˆ
3
i
j
k
λ +
+
is equal to one Find the value of λ 14 If ,
a b, c
r r
r
are mutually perpendicular vectors of equal magnitudes, show that
the vector a
b
c
+
r+
r
r is equally inclined to ,
and
a
b
c
r
r
r |
1 | 5566-5569 | Find the value of λ 14 If ,
a b, c
r r
r
are mutually perpendicular vectors of equal magnitudes, show that
the vector a
b
c
+
r+
r
r is equally inclined to ,
and
a
b
c
r
r
r © NCERT
not to be republished
VECTOR ALGEBRA
459
15 |
1 | 5567-5570 | 14 If ,
a b, c
r r
r
are mutually perpendicular vectors of equal magnitudes, show that
the vector a
b
c
+
r+
r
r is equally inclined to ,
and
a
b
c
r
r
r © NCERT
not to be republished
VECTOR ALGEBRA
459
15 Prove that
2
2
(
) (
) |
|
| |
a
b
a
b
a
b
+
⋅
+
=
+
r
r
r
r
r
r
, if and only if ,a b
rr
are perpendicular,
given
0,
0
a
b
≠
rr≠
r
r |
1 | 5568-5571 | If ,
a b, c
r r
r
are mutually perpendicular vectors of equal magnitudes, show that
the vector a
b
c
+
r+
r
r is equally inclined to ,
and
a
b
c
r
r
r © NCERT
not to be republished
VECTOR ALGEBRA
459
15 Prove that
2
2
(
) (
) |
|
| |
a
b
a
b
a
b
+
⋅
+
=
+
r
r
r
r
r
r
, if and only if ,a b
rr
are perpendicular,
given
0,
0
a
b
≠
rr≠
r
r Choose the correct answer in Exercises 16 to 19 |
1 | 5569-5572 | © NCERT
not to be republished
VECTOR ALGEBRA
459
15 Prove that
2
2
(
) (
) |
|
| |
a
b
a
b
a
b
+
⋅
+
=
+
r
r
r
r
r
r
, if and only if ,a b
rr
are perpendicular,
given
0,
0
a
b
≠
rr≠
r
r Choose the correct answer in Exercises 16 to 19 16 |
1 | 5570-5573 | Prove that
2
2
(
) (
) |
|
| |
a
b
a
b
a
b
+
⋅
+
=
+
r
r
r
r
r
r
, if and only if ,a b
rr
are perpendicular,
given
0,
0
a
b
≠
rr≠
r
r Choose the correct answer in Exercises 16 to 19 16 If θ is the angle between two vectors
a and
rb
r
, then
0
a b
⋅
rr≥
only when
(A) 0
< θ <π2
(B) 0
π2
≤ θ ≤
(C) 0 < θ < π
(D) 0 ≤ θ ≤ π
17 |
1 | 5571-5574 | Choose the correct answer in Exercises 16 to 19 16 If θ is the angle between two vectors
a and
rb
r
, then
0
a b
⋅
rr≥
only when
(A) 0
< θ <π2
(B) 0
π2
≤ θ ≤
(C) 0 < θ < π
(D) 0 ≤ θ ≤ π
17 Let and
a
rb
r
be two unit vectors and θ is the angle between them |
1 | 5572-5575 | 16 If θ is the angle between two vectors
a and
rb
r
, then
0
a b
⋅
rr≥
only when
(A) 0
< θ <π2
(B) 0
π2
≤ θ ≤
(C) 0 < θ < π
(D) 0 ≤ θ ≤ π
17 Let and
a
rb
r
be two unit vectors and θ is the angle between them Then a
+b
r
r
is a unit vector if
(A)
θ =π4
(B)
θ =π3
(C)
θ =π2
(D)
π32
θ =
18 |
1 | 5573-5576 | If θ is the angle between two vectors
a and
rb
r
, then
0
a b
⋅
rr≥
only when
(A) 0
< θ <π2
(B) 0
π2
≤ θ ≤
(C) 0 < θ < π
(D) 0 ≤ θ ≤ π
17 Let and
a
rb
r
be two unit vectors and θ is the angle between them Then a
+b
r
r
is a unit vector if
(A)
θ =π4
(B)
θ =π3
(C)
θ =π2
(D)
π32
θ =
18 The value of
ˆ
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ |
1 | 5574-5577 | Let and
a
rb
r
be two unit vectors and θ is the angle between them Then a
+b
r
r
is a unit vector if
(A)
θ =π4
(B)
θ =π3
(C)
θ =π2
(D)
π32
θ =
18 The value of
ˆ
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ (
)
(
)
(
)
i
j
k
j
i
k
k
i
j
is
(A) 0
(B) –1
(C) 1
(D) 3
19 |
1 | 5575-5578 | Then a
+b
r
r
is a unit vector if
(A)
θ =π4
(B)
θ =π3
(C)
θ =π2
(D)
π32
θ =
18 The value of
ˆ
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ (
)
(
)
(
)
i
j
k
j
i
k
k
i
j
is
(A) 0
(B) –1
(C) 1
(D) 3
19 If θ is the angle between any two vectors and
a
rb
r
, then |
| |
|
a b
a b
⋅
=
×
r
r
r
r
when
θ is equal to
(A) 0
(B)
π4
(C)
π2
(D) π
Summary
� Position vector of a point P(x, y, z) is given as
ˆ
ˆ
ˆ
OP(
r)
xi
yj
zk
=
=
+
+
uuur
r
, and its
magnitude by
2
2
2
x
y
z
+
+ |
1 | 5576-5579 | The value of
ˆ
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
ˆ
ˆ (
)
(
)
(
)
i
j
k
j
i
k
k
i
j
is
(A) 0
(B) –1
(C) 1
(D) 3
19 If θ is the angle between any two vectors and
a
rb
r
, then |
| |
|
a b
a b
⋅
=
×
r
r
r
r
when
θ is equal to
(A) 0
(B)
π4
(C)
π2
(D) π
Summary
� Position vector of a point P(x, y, z) is given as
ˆ
ˆ
ˆ
OP(
r)
xi
yj
zk
=
=
+
+
uuur
r
, and its
magnitude by
2
2
2
x
y
z
+
+ � The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes |
1 | 5577-5580 | (
)
(
)
(
)
i
j
k
j
i
k
k
i
j
is
(A) 0
(B) –1
(C) 1
(D) 3
19 If θ is the angle between any two vectors and
a
rb
r
, then |
| |
|
a b
a b
⋅
=
×
r
r
r
r
when
θ is equal to
(A) 0
(B)
π4
(C)
π2
(D) π
Summary
� Position vector of a point P(x, y, z) is given as
ˆ
ˆ
ˆ
OP(
r)
xi
yj
zk
=
=
+
+
uuur
r
, and its
magnitude by
2
2
2
x
y
z
+
+ � The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes � The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of
any vector are related as:
,
,
a
b
c
l
m
n
r
r
r
=
=
=
� The vector sum of the three sides of a triangle taken in order is 0
r |
1 | 5578-5581 | If θ is the angle between any two vectors and
a
rb
r
, then |
| |
|
a b
a b
⋅
=
×
r
r
r
r
when
θ is equal to
(A) 0
(B)
π4
(C)
π2
(D) π
Summary
� Position vector of a point P(x, y, z) is given as
ˆ
ˆ
ˆ
OP(
r)
xi
yj
zk
=
=
+
+
uuur
r
, and its
magnitude by
2
2
2
x
y
z
+
+ � The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes � The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of
any vector are related as:
,
,
a
b
c
l
m
n
r
r
r
=
=
=
� The vector sum of the three sides of a triangle taken in order is 0
r © NCERT
not to be republished
MATHEMATICS
460
� The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors |
1 | 5579-5582 | � The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes � The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of
any vector are related as:
,
,
a
b
c
l
m
n
r
r
r
=
=
=
� The vector sum of the three sides of a triangle taken in order is 0
r © NCERT
not to be republished
MATHEMATICS
460
� The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors � The multiplication of a given vector by a scalar λ, changes the magnitude of
the vector by the multiple |λ|, and keeps the direction same (or makes it
opposite) according as the value of λ is positive (or negative) |
1 | 5580-5583 | � The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of
any vector are related as:
,
,
a
b
c
l
m
n
r
r
r
=
=
=
� The vector sum of the three sides of a triangle taken in order is 0
r © NCERT
not to be republished
MATHEMATICS
460
� The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors � The multiplication of a given vector by a scalar λ, changes the magnitude of
the vector by the multiple |λ|, and keeps the direction same (or makes it
opposite) according as the value of λ is positive (or negative) � For a given vector ar , the vector ˆ
|
a|
a
=a
r
r gives the unit vector in the direction
of ar |
1 | 5581-5584 | © NCERT
not to be republished
MATHEMATICS
460
� The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors � The multiplication of a given vector by a scalar λ, changes the magnitude of
the vector by the multiple |λ|, and keeps the direction same (or makes it
opposite) according as the value of λ is positive (or negative) � For a given vector ar , the vector ˆ
|
a|
a
=a
r
r gives the unit vector in the direction
of ar � The position vector of a point R dividing a line segment joining the points
P and Q whose position vectors are
aand
rb
r
respectively, in the ratio m : n
(i)
internally, is given by na
mmb
+n
+
r
r |
1 | 5582-5585 | � The multiplication of a given vector by a scalar λ, changes the magnitude of
the vector by the multiple |λ|, and keeps the direction same (or makes it
opposite) according as the value of λ is positive (or negative) � For a given vector ar , the vector ˆ
|
a|
a
=a
r
r gives the unit vector in the direction
of ar � The position vector of a point R dividing a line segment joining the points
P and Q whose position vectors are
aand
rb
r
respectively, in the ratio m : n
(i)
internally, is given by na
mmb
+n
+
r
r (ii)
externally, is given by mb
na
m
−n
−
r
r |
1 | 5583-5586 | � For a given vector ar , the vector ˆ
|
a|
a
=a
r
r gives the unit vector in the direction
of ar � The position vector of a point R dividing a line segment joining the points
P and Q whose position vectors are
aand
rb
r
respectively, in the ratio m : n
(i)
internally, is given by na
mmb
+n
+
r
r (ii)
externally, is given by mb
na
m
−n
−
r
r � The scalar product of two given vectors
aand
rb
r
having angle θ between
them is defined as
|
||
| cos
a b
a b
⋅
=
θ
r
r
r
r |
1 | 5584-5587 | � The position vector of a point R dividing a line segment joining the points
P and Q whose position vectors are
aand
rb
r
respectively, in the ratio m : n
(i)
internally, is given by na
mmb
+n
+
r
r (ii)
externally, is given by mb
na
m
−n
−
r
r � The scalar product of two given vectors
aand
rb
r
having angle θ between
them is defined as
|
||
| cos
a b
a b
⋅
=
θ
r
r
r
r Also, when a b
⋅
rr
is given, the angle ‘θ’ between the vectors
aand
rb
r
may be
determined by
cosθ = |
||
a b|
a b
⋅
rr
r
r
� If θ is the angle between two vectors
aand
rb
r
, then their cross product is
given as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
where ˆn is a unit vector perpendicular to the plane containing
aand
rb
r |
1 | 5585-5588 | (ii)
externally, is given by mb
na
m
−n
−
r
r � The scalar product of two given vectors
aand
rb
r
having angle θ between
them is defined as
|
||
| cos
a b
a b
⋅
=
θ
r
r
r
r Also, when a b
⋅
rr
is given, the angle ‘θ’ between the vectors
aand
rb
r
may be
determined by
cosθ = |
||
a b|
a b
⋅
rr
r
r
� If θ is the angle between two vectors
aand
rb
r
, then their cross product is
given as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
where ˆn is a unit vector perpendicular to the plane containing
aand
rb
r Such
that
ˆ
, ,
r
ra b n form right handed system of coordinate axes |
1 | 5586-5589 | � The scalar product of two given vectors
aand
rb
r
having angle θ between
them is defined as
|
||
| cos
a b
a b
⋅
=
θ
r
r
r
r Also, when a b
⋅
rr
is given, the angle ‘θ’ between the vectors
aand
rb
r
may be
determined by
cosθ = |
||
a b|
a b
⋅
rr
r
r
� If θ is the angle between two vectors
aand
rb
r
, then their cross product is
given as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
where ˆn is a unit vector perpendicular to the plane containing
aand
rb
r Such
that
ˆ
, ,
r
ra b n form right handed system of coordinate axes � If we have two vectors
aand
rb
r
, given in component form as
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
and λ any scalar,
© NCERT
not to be republished
VECTOR ALGEBRA
461
then
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b
k
+
+
+
+
+
;
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
;
rr |
1 | 5587-5590 | Also, when a b
⋅
rr
is given, the angle ‘θ’ between the vectors
aand
rb
r
may be
determined by
cosθ = |
||
a b|
a b
⋅
rr
r
r
� If θ is the angle between two vectors
aand
rb
r
, then their cross product is
given as
a
b
×
r
r
=
ˆ
|
||
a b|sin
n
θ
r
r
where ˆn is a unit vector perpendicular to the plane containing
aand
rb
r Such
that
ˆ
, ,
r
ra b n form right handed system of coordinate axes � If we have two vectors
aand
rb
r
, given in component form as
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
and λ any scalar,
© NCERT
not to be republished
VECTOR ALGEBRA
461
then
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b
k
+
+
+
+
+
;
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
;
rr a b
=
1 1
2 2
3 3
a b
a b
a b
+
+
;
and
a
b
×
r
r
=
1
1
1
2
2
2
ˆ
ˆ
ˆ |
1 | 5588-5591 | Such
that
ˆ
, ,
r
ra b n form right handed system of coordinate axes � If we have two vectors
aand
rb
r
, given in component form as
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
and λ any scalar,
© NCERT
not to be republished
VECTOR ALGEBRA
461
then
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b
k
+
+
+
+
+
;
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
;
rr a b
=
1 1
2 2
3 3
a b
a b
a b
+
+
;
and
a
b
×
r
r
=
1
1
1
2
2
2
ˆ
ˆ
ˆ i
j
k
a
b
c
a
b
c
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry” |
1 | 5589-5592 | � If we have two vectors
aand
rb
r
, given in component form as
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
and λ any scalar,
© NCERT
not to be republished
VECTOR ALGEBRA
461
then
a
b
+
r
r
=
1
1
2
2
3
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a
b i
a
b
j
a
b
k
+
+
+
+
+
;
a
λr =
1
2
3 ˆ
ˆ
ˆ
(
)
(
)
(
)
a i
a
j
a k
λ
+ λ
+ λ
;
rr a b
=
1 1
2 2
3 3
a b
a b
a b
+
+
;
and
a
b
×
r
r
=
1
1
1
2
2
2
ˆ
ˆ
ˆ i
j
k
a
b
c
a
b
c
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry” The germinal ideas of modern vector theory date from around 1800
when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described
that how a complex number a + ib could be given a geometric interpretation with
the help of a directed line segment in a coordinate plane |
1 | 5590-5593 | a b
=
1 1
2 2
3 3
a b
a b
a b
+
+
;
and
a
b
×
r
r
=
1
1
1
2
2
2
ˆ
ˆ
ˆ i
j
k
a
b
c
a
b
c
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry” The germinal ideas of modern vector theory date from around 1800
when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described
that how a complex number a + ib could be given a geometric interpretation with
the help of a directed line segment in a coordinate plane William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853) |
1 | 5591-5594 | i
j
k
a
b
c
a
b
c
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry” The germinal ideas of modern vector theory date from around 1800
when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described
that how a complex number a + ib could be given a geometric interpretation with
the help of a directed line segment in a coordinate plane William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853) Hamilton’s
method of quaternions (an ordered set of four real numbers given as:
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
, , ,
a
bi
cj
dk i
j k
+
+
+
following certain algebraic rules) was a solution to the
problem of multiplying vectors in three dimensional space |
1 | 5592-5595 | The germinal ideas of modern vector theory date from around 1800
when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described
that how a complex number a + ib could be given a geometric interpretation with
the help of a directed line segment in a coordinate plane William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853) Hamilton’s
method of quaternions (an ordered set of four real numbers given as:
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
, , ,
a
bi
cj
dk i
j k
+
+
+
following certain algebraic rules) was a solution to the
problem of multiplying vectors in three dimensional space Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B |
1 | 5593-5596 | William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853) Hamilton’s
method of quaternions (an ordered set of four real numbers given as:
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
, , ,
a
bi
cj
dk i
j k
+
+
+
following certain algebraic rules) was a solution to the
problem of multiplying vectors in three dimensional space Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B C |
1 | 5594-5597 | Hamilton’s
method of quaternions (an ordered set of four real numbers given as:
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
, , ,
a
bi
cj
dk i
j k
+
+
+
following certain algebraic rules) was a solution to the
problem of multiplying vectors in three dimensional space Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B C ), a Greek
philosopher, and pupil of Plato (427-348 B |
1 | 5595-5598 | Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B C ), a Greek
philosopher, and pupil of Plato (427-348 B C |
1 | 5596-5599 | C ), a Greek
philosopher, and pupil of Plato (427-348 B C ) |
1 | 5597-5600 | ), a Greek
philosopher, and pupil of Plato (427-348 B C ) That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law |
1 | 5598-5601 | C ) That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law The correct law for the composition of
forces, that forces add vectorially, had been discovered in the case of perpendicular
forces by Stevin-Simon (1548-1620) |
1 | 5599-5602 | ) That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law The correct law for the composition of
forces, that forces add vectorially, had been discovered in the case of perpendicular
forces by Stevin-Simon (1548-1620) In 1586 A |
1 | 5600-5603 | That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law The correct law for the composition of
forces, that forces add vectorially, had been discovered in the case of perpendicular
forces by Stevin-Simon (1548-1620) In 1586 A D |
1 | 5601-5604 | The correct law for the composition of
forces, that forces add vectorially, had been discovered in the case of perpendicular
forces by Stevin-Simon (1548-1620) In 1586 A D , he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics |
1 | 5602-5605 | In 1586 A D , he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics But it took another 200 years for the general concept
of vectors to form |
1 | 5603-5606 | D , he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics But it took another 200 years for the general concept
of vectors to form In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist
and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
© NCERT
not to be republished
MATHEMATICS
462
part of quaternion from its imaginary (vector) part |
1 | 5604-5607 | , he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics But it took another 200 years for the general concept
of vectors to form In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist
and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
© NCERT
not to be republished
MATHEMATICS
462
part of quaternion from its imaginary (vector) part In 1881 and 1884, Gibbs
printed a treatise entitled Element of Vector Analysis |
1 | 5605-5608 | But it took another 200 years for the general concept
of vectors to form In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist
and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
© NCERT
not to be republished
MATHEMATICS
462
part of quaternion from its imaginary (vector) part In 1881 and 1884, Gibbs
printed a treatise entitled Element of Vector Analysis This book gave a systematic
and concise account of vectors |
1 | 5606-5609 | In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist
and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
© NCERT
not to be republished
MATHEMATICS
462
part of quaternion from its imaginary (vector) part In 1881 and 1884, Gibbs
printed a treatise entitled Element of Vector Analysis This book gave a systematic
and concise account of vectors However, much of the credit for demonstrating
the applications of vectors is due to the D |
1 | 5607-5610 | In 1881 and 1884, Gibbs
printed a treatise entitled Element of Vector Analysis This book gave a systematic
and concise account of vectors However, much of the credit for demonstrating
the applications of vectors is due to the D Heaviside and P |
1 | 5608-5611 | This book gave a systematic
and concise account of vectors However, much of the credit for demonstrating
the applications of vectors is due to the D Heaviside and P G |
1 | 5609-5612 | However, much of the credit for demonstrating
the applications of vectors is due to the D Heaviside and P G Tait (1831-1901)
who contributed significantly to this subject |
1 | 5610-5613 | Heaviside and P G Tait (1831-1901)
who contributed significantly to this subject —�
�
�
�
�—
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
463
� The moving power of mathematical invention is not
reasoning but imagination |
1 | 5611-5614 | G Tait (1831-1901)
who contributed significantly to this subject —�
�
�
�
�—
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
463
� The moving power of mathematical invention is not
reasoning but imagination – A |
1 | 5612-5615 | Tait (1831-1901)
who contributed significantly to this subject —�
�
�
�
�—
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
463
� The moving power of mathematical invention is not
reasoning but imagination – A DEMORGAN �
11 |
1 | 5613-5616 | —�
�
�
�
�—
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
463
� The moving power of mathematical invention is not
reasoning but imagination – A DEMORGAN �
11 1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only |
1 | 5614-5617 | – A DEMORGAN �
11 1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only In
the previous chapter of this book, we have studied some
basic concepts of vectors |
1 | 5615-5618 | DEMORGAN �
11 1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only In
the previous chapter of this book, we have studied some
basic concepts of vectors We will now use vector algebra
to three dimensional geometry |
1 | 5616-5619 | 1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only In
the previous chapter of this book, we have studied some
basic concepts of vectors We will now use vector algebra
to three dimensional geometry The purpose of this
approach to 3-dimensional geometry is that it makes the
study simple and elegant* |
1 | 5617-5620 | In
the previous chapter of this book, we have studied some
basic concepts of vectors We will now use vector algebra
to three dimensional geometry The purpose of this
approach to 3-dimensional geometry is that it makes the
study simple and elegant* In this chapter, we shall study the direction cosines
and direction ratios of a line joining two points and also
discuss about the equations of lines and planes in space
under different conditions, angle between two lines, two
planes, a line and a plane, shortest distance between two
skew lines and distance of a point from a plane |
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