Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 5618-5621 | 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 Most of
the above results are obtained in vector form |
1 | 5619-5622 | 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 Most of
the above results are obtained in vector form Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation |
1 | 5620-5623 | 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 Most of
the above results are obtained in vector form Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation 11 |
1 | 5621-5624 | Most of
the above results are obtained in vector form Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation 11 2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos α, cos β and cos γ are called direction cosines of the
directed line L |
1 | 5622-5625 | Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation 11 2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos α, cos β and cos γ are called direction cosines of the
directed line L If we reverse the direction of L, then the direction angles are replaced by their supplements,
i |
1 | 5623-5626 | 11 2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos α, cos β and cos γ are called direction cosines of the
directed line L If we reverse the direction of L, then the direction angles are replaced by their supplements,
i e |
1 | 5624-5627 | 2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos α, cos β and cos γ are called direction cosines of the
directed line L If we reverse the direction of L, then the direction angles are replaced by their supplements,
i e , π α
− , π β
− and π γ
− |
1 | 5625-5628 | If we reverse the direction of L, then the direction angles are replaced by their supplements,
i e , π α
− , π β
− and π γ
− Thus, the signs of the direction cosines are reversed |
1 | 5626-5629 | e , π α
− , π β
− and π γ
− Thus, the signs of the direction cosines are reversed Chapter 11
THREE DIMENSIONAL GEOMETRY
* Fo r vari ou s acti vi ti es i n th ree di mens io nal geomet ry, on e may refer to t he Boo k
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
Leonhard Euler
(1707-1783)
© NCERT
not to be republished
MATHEMATI CS
464
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines |
1 | 5627-5630 | , π α
− , π β
− and π γ
− Thus, the signs of the direction cosines are reversed Chapter 11
THREE DIMENSIONAL GEOMETRY
* Fo r vari ou s acti vi ti es i n th ree di mens io nal geomet ry, on e may refer to t he Boo k
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
Leonhard Euler
(1707-1783)
© NCERT
not to be republished
MATHEMATI CS
464
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines In order to have a unique set of direction cosines for
a given line in space, we must take the given line as a directed line |
1 | 5628-5631 | Thus, the signs of the direction cosines are reversed Chapter 11
THREE DIMENSIONAL GEOMETRY
* Fo r vari ou s acti vi ti es i n th ree di mens io nal geomet ry, on e may refer to t he Boo k
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
Leonhard Euler
(1707-1783)
© NCERT
not to be republished
MATHEMATI CS
464
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines In order to have a unique set of direction cosines for
a given line in space, we must take the given line as a directed line These unique
direction cosines are denoted by l, m and n |
1 | 5629-5632 | Chapter 11
THREE DIMENSIONAL GEOMETRY
* Fo r vari ou s acti vi ti es i n th ree di mens io nal geomet ry, on e may refer to t he Boo k
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
Leonhard Euler
(1707-1783)
© NCERT
not to be republished
MATHEMATI CS
464
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines In order to have a unique set of direction cosines for
a given line in space, we must take the given line as a directed line These unique
direction cosines are denoted by l, m and n Remark If the given line in space does not pass through the origin, then, in order to find
its direction cosines, we draw a line through the origin and parallel to the given line |
1 | 5630-5633 | In order to have a unique set of direction cosines for
a given line in space, we must take the given line as a directed line These unique
direction cosines are denoted by l, m and n Remark If the given line in space does not pass through the origin, then, in order to find
its direction cosines, we draw a line through the origin and parallel to the given line Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines |
1 | 5631-5634 | These unique
direction cosines are denoted by l, m and n Remark If the given line in space does not pass through the origin, then, in order to find
its direction cosines, we draw a line through the origin and parallel to the given line Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines Any three numbers which are proportional to the direction cosines of a line are
called the direction ratios of the line |
1 | 5632-5635 | Remark If the given line in space does not pass through the origin, then, in order to find
its direction cosines, we draw a line through the origin and parallel to the given line Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines Any three numbers which are proportional to the direction cosines of a line are
called the direction ratios of the line If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R |
1 | 5633-5636 | Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines Any three numbers which are proportional to the direction cosines of a line are
called the direction ratios of the line If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R �Note Some authors also call direction ratios as direction numbers |
1 | 5634-5637 | Any three numbers which are proportional to the direction cosines of a line are
called the direction ratios of the line If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R �Note Some authors also call direction ratios as direction numbers Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d |
1 | 5635-5638 | If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R �Note Some authors also call direction ratios as direction numbers Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d c’s) of the line |
1 | 5636-5639 | �Note Some authors also call direction ratios as direction numbers Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d c’s) of the line Then
l
a = m
b = n
k
c
=
(say), k being a constant |
1 | 5637-5640 | Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d c’s) of the line Then
l
a = m
b = n
k
c
=
(say), k being a constant Therefore
l = ak, m = bk, n = ck |
1 | 5638-5641 | c’s) of the line Then
l
a = m
b = n
k
c
=
(say), k being a constant Therefore
l = ak, m = bk, n = ck (1)
But
l2 + m 2 + n2 = 1
Therefore
k2 (a2 + b2 + c2) = 1
or
k =
2
2
2
1
a
b
c
±
+
+
Fig 11 |
1 | 5639-5642 | Then
l
a = m
b = n
k
c
=
(say), k being a constant Therefore
l = ak, m = bk, n = ck (1)
But
l2 + m 2 + n2 = 1
Therefore
k2 (a2 + b2 + c2) = 1
or
k =
2
2
2
1
a
b
c
±
+
+
Fig 11 1
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
465
Hence, from (1), the d |
1 | 5640-5643 | Therefore
l = ak, m = bk, n = ck (1)
But
l2 + m 2 + n2 = 1
Therefore
k2 (a2 + b2 + c2) = 1
or
k =
2
2
2
1
a
b
c
±
+
+
Fig 11 1
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
465
Hence, from (1), the d c |
1 | 5641-5644 | (1)
But
l2 + m 2 + n2 = 1
Therefore
k2 (a2 + b2 + c2) = 1
or
k =
2
2
2
1
a
b
c
±
+
+
Fig 11 1
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
465
Hence, from (1), the d c ’s of the line are
2
2
2
2
2
2
2
2
2
,
,
a
b
c
l
m
n
a
b
c
a
b
c
a
b
c
=±
= ±
= ±
+
+
+
+
+
+
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n |
1 | 5642-5645 | 1
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
465
Hence, from (1), the d c ’s of the line are
2
2
2
2
2
2
2
2
2
,
,
a
b
c
l
m
n
a
b
c
a
b
c
a
b
c
=±
= ±
= ±
+
+
+
+
+
+
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a
set of direction ratios |
1 | 5643-5646 | c ’s of the line are
2
2
2
2
2
2
2
2
2
,
,
a
b
c
l
m
n
a
b
c
a
b
c
a
b
c
=±
= ±
= ±
+
+
+
+
+
+
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a
set of direction ratios So, any two sets of direction ratios of a line are also proportional |
1 | 5644-5647 | ’s of the line are
2
2
2
2
2
2
2
2
2
,
,
a
b
c
l
m
n
a
b
c
a
b
c
a
b
c
=±
= ±
= ±
+
+
+
+
+
+
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a
set of direction ratios So, any two sets of direction ratios of a line are also proportional Also, for any line there are infinitely many sets of direction ratios |
1 | 5645-5648 | For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a
set of direction ratios So, any two sets of direction ratios of a line are also proportional Also, for any line there are infinitely many sets of direction ratios 11 |
1 | 5646-5649 | So, any two sets of direction ratios of a line are also proportional Also, for any line there are infinitely many sets of direction ratios 11 2 |
1 | 5647-5650 | Also, for any line there are infinitely many sets of direction ratios 11 2 1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n |
1 | 5648-5651 | 11 2 1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n Through
the origin draw a line parallel to the given line and take a
point P(x, y, z) on this line |
1 | 5649-5652 | 2 1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n Through
the origin draw a line parallel to the given line and take a
point P(x, y, z) on this line From P draw a perpendicular
PA on the x-axis (Fig |
1 | 5650-5653 | 1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n Through
the origin draw a line parallel to the given line and take a
point P(x, y, z) on this line From P draw a perpendicular
PA on the x-axis (Fig 11 |
1 | 5651-5654 | Through
the origin draw a line parallel to the given line and take a
point P(x, y, z) on this line From P draw a perpendicular
PA on the x-axis (Fig 11 2) |
1 | 5652-5655 | From P draw a perpendicular
PA on the x-axis (Fig 11 2) Let OP = r |
1 | 5653-5656 | 11 2) Let OP = r Then
OA
cos
α=OP
x
r= |
1 | 5654-5657 | 2) Let OP = r Then
OA
cos
α=OP
x
r= This gives x = lr |
1 | 5655-5658 | Let OP = r Then
OA
cos
α=OP
x
r= This gives x = lr Similarly,
y = mr and z = nr
Thus
x2 + y2 + z2 = r2 (l2 + m 2 + n2)
But
x2 + y2 + z2 = r2
Hence
l2 + m2 + n2 = 1
11 |
1 | 5656-5659 | Then
OA
cos
α=OP
x
r= This gives x = lr Similarly,
y = mr and z = nr
Thus
x2 + y2 + z2 = r2 (l2 + m 2 + n2)
But
x2 + y2 + z2 = r2
Hence
l2 + m2 + n2 = 1
11 2 |
1 | 5657-5660 | This gives x = lr Similarly,
y = mr and z = nr
Thus
x2 + y2 + z2 = r2 (l2 + m 2 + n2)
But
x2 + y2 + z2 = r2
Hence
l2 + m2 + n2 = 1
11 2 2 Direction cosines of a line passing through two points
Since one and only one line passes through two given points, we can determine the
direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2)
as follows (Fig 11 |
1 | 5658-5661 | Similarly,
y = mr and z = nr
Thus
x2 + y2 + z2 = r2 (l2 + m 2 + n2)
But
x2 + y2 + z2 = r2
Hence
l2 + m2 + n2 = 1
11 2 2 Direction cosines of a line passing through two points
Since one and only one line passes through two given points, we can determine the
direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2)
as follows (Fig 11 3 (a)) |
1 | 5659-5662 | 2 2 Direction cosines of a line passing through two points
Since one and only one line passes through two given points, we can determine the
direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2)
as follows (Fig 11 3 (a)) Fig 11 |
1 | 5660-5663 | 2 Direction cosines of a line passing through two points
Since one and only one line passes through two given points, we can determine the
direction cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2)
as follows (Fig 11 3 (a)) Fig 11 3
r
Z
X
Y
R
S
P ( , , )
x y z
A
Oα
A
O
P
x
α
α
Fig 11 |
1 | 5661-5664 | 3 (a)) Fig 11 3
r
Z
X
Y
R
S
P ( , , )
x y z
A
Oα
A
O
P
x
α
α
Fig 11 2
© NCERT
not to be republished
MATHEMATI CS
466
Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ
with the x, y and z-axis, respectively |
1 | 5662-5665 | Fig 11 3
r
Z
X
Y
R
S
P ( , , )
x y z
A
Oα
A
O
P
x
α
α
Fig 11 2
© NCERT
not to be republished
MATHEMATI CS
466
Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ
with the x, y and z-axis, respectively Draw perpendiculars from P and Q to XY-plane to meet at R and S |
1 | 5663-5666 | 3
r
Z
X
Y
R
S
P ( , , )
x y z
A
Oα
A
O
P
x
α
α
Fig 11 2
© NCERT
not to be republished
MATHEMATI CS
466
Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ
with the x, y and z-axis, respectively Draw perpendiculars from P and Q to XY-plane to meet at R and S Draw a
perpendicular from P to QS to meet at N |
1 | 5664-5667 | 2
© NCERT
not to be republished
MATHEMATI CS
466
Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ
with the x, y and z-axis, respectively Draw perpendiculars from P and Q to XY-plane to meet at R and S Draw a
perpendicular from P to QS to meet at N Now, in right angle triangle PNQ, ∠PQN=
γ (Fig 11 |
1 | 5665-5668 | Draw perpendiculars from P and Q to XY-plane to meet at R and S Draw a
perpendicular from P to QS to meet at N Now, in right angle triangle PNQ, ∠PQN=
γ (Fig 11 3 (b) |
1 | 5666-5669 | Draw a
perpendicular from P to QS to meet at N Now, in right angle triangle PNQ, ∠PQN=
γ (Fig 11 3 (b) Therefore,
cosγ =
2
1
NQ
PQ
zPQ
−z
=
Similarly
cosα =
2
1
2
1
and cos
PQ
PQ
x
x
y
y
−
−
β =
Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and
Q(x2, y2, z2) are
2
1
PQ
x
−x
,
2
1
PQ
y
−y
,
2
1
PQ
z
z
−
where
PQ =
(
)
2
2
2
2
1
2
1
2
1
(
)
(
)
x
x
y
y
z
z
−
+ −
+ −
�Note The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)
may be taken as
x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2
Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and
z-axis respectively, find its direction cosines |
1 | 5667-5670 | Now, in right angle triangle PNQ, ∠PQN=
γ (Fig 11 3 (b) Therefore,
cosγ =
2
1
NQ
PQ
zPQ
−z
=
Similarly
cosα =
2
1
2
1
and cos
PQ
PQ
x
x
y
y
−
−
β =
Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and
Q(x2, y2, z2) are
2
1
PQ
x
−x
,
2
1
PQ
y
−y
,
2
1
PQ
z
z
−
where
PQ =
(
)
2
2
2
2
1
2
1
2
1
(
)
(
)
x
x
y
y
z
z
−
+ −
+ −
�Note The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)
may be taken as
x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2
Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and
z-axis respectively, find its direction cosines Solution Let the d |
1 | 5668-5671 | 3 (b) Therefore,
cosγ =
2
1
NQ
PQ
zPQ
−z
=
Similarly
cosα =
2
1
2
1
and cos
PQ
PQ
x
x
y
y
−
−
β =
Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and
Q(x2, y2, z2) are
2
1
PQ
x
−x
,
2
1
PQ
y
−y
,
2
1
PQ
z
z
−
where
PQ =
(
)
2
2
2
2
1
2
1
2
1
(
)
(
)
x
x
y
y
z
z
−
+ −
+ −
�Note The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)
may be taken as
x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2
Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and
z-axis respectively, find its direction cosines Solution Let the d c |
1 | 5669-5672 | Therefore,
cosγ =
2
1
NQ
PQ
zPQ
−z
=
Similarly
cosα =
2
1
2
1
and cos
PQ
PQ
x
x
y
y
−
−
β =
Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and
Q(x2, y2, z2) are
2
1
PQ
x
−x
,
2
1
PQ
y
−y
,
2
1
PQ
z
z
−
where
PQ =
(
)
2
2
2
2
1
2
1
2
1
(
)
(
)
x
x
y
y
z
z
−
+ −
+ −
�Note The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)
may be taken as
x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2
Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and
z-axis respectively, find its direction cosines Solution Let the d c 's of the lines be l , m, n |
1 | 5670-5673 | Solution Let the d c 's of the lines be l , m, n Then l = cos 900 = 0, m = cos 600 = 1
2,
n = cos 300 = 2
3 |
1 | 5671-5674 | c 's of the lines be l , m, n Then l = cos 900 = 0, m = cos 600 = 1
2,
n = cos 300 = 2
3 Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines |
1 | 5672-5675 | 's of the lines be l , m, n Then l = cos 900 = 0, m = cos 600 = 1
2,
n = cos 300 = 2
3 Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines Solution Direction cosines are
2
2
2
)2
(
)1
(
2
2
+−
+−
,
2
2
2
(2)
)1
(
2
1
+−
−
+
−
,
(
)
2
2
2
)2
(
1
2
2
+−
−
+
−
or
2
1
2
,
,
3
3
3
−
−
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3) |
1 | 5673-5676 | Then l = cos 900 = 0, m = cos 600 = 1
2,
n = cos 300 = 2
3 Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines Solution Direction cosines are
2
2
2
)2
(
)1
(
2
2
+−
+−
,
2
2
2
(2)
)1
(
2
1
+−
−
+
−
,
(
)
2
2
2
)2
(
1
2
2
+−
−
+
−
or
2
1
2
,
,
3
3
3
−
−
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3) © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
467
Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
2
1
2
1
2
1
,
,
PQ
PQ
PQ
x
x
y
y
z
z
−
−
−
where
PQ =
(
)
12
2
12
2
2
1
2
)
(
)
(
z
z
y
y
x
x
−
+
−
+
−
Here P is (– 2, 4, – 5) and Q is (1, 2, 3) |
1 | 5674-5677 | Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines Solution Direction cosines are
2
2
2
)2
(
)1
(
2
2
+−
+−
,
2
2
2
(2)
)1
(
2
1
+−
−
+
−
,
(
)
2
2
2
)2
(
1
2
2
+−
−
+
−
or
2
1
2
,
,
3
3
3
−
−
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3) © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
467
Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
2
1
2
1
2
1
,
,
PQ
PQ
PQ
x
x
y
y
z
z
−
−
−
where
PQ =
(
)
12
2
12
2
2
1
2
)
(
)
(
z
z
y
y
x
x
−
+
−
+
−
Here P is (– 2, 4, – 5) and Q is (1, 2, 3) So
PQ =
2
2
2
(1
( 2))
(2
4)
(3
( 5))
− −
+
−
+
− −
=
77
Thus, the direction cosines of the line joining two points is
3
2
8
,
,
77
77
77
−
Example 4 Find the direction cosines of x, y and z-axis |
1 | 5675-5678 | Solution Direction cosines are
2
2
2
)2
(
)1
(
2
2
+−
+−
,
2
2
2
(2)
)1
(
2
1
+−
−
+
−
,
(
)
2
2
2
)2
(
1
2
2
+−
−
+
−
or
2
1
2
,
,
3
3
3
−
−
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3) © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
467
Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
2
1
2
1
2
1
,
,
PQ
PQ
PQ
x
x
y
y
z
z
−
−
−
where
PQ =
(
)
12
2
12
2
2
1
2
)
(
)
(
z
z
y
y
x
x
−
+
−
+
−
Here P is (– 2, 4, – 5) and Q is (1, 2, 3) So
PQ =
2
2
2
(1
( 2))
(2
4)
(3
( 5))
− −
+
−
+
− −
=
77
Thus, the direction cosines of the line joining two points is
3
2
8
,
,
77
77
77
−
Example 4 Find the direction cosines of x, y and z-axis Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis |
1 | 5676-5679 | © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
467
Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
2
1
2
1
2
1
,
,
PQ
PQ
PQ
x
x
y
y
z
z
−
−
−
where
PQ =
(
)
12
2
12
2
2
1
2
)
(
)
(
z
z
y
y
x
x
−
+
−
+
−
Here P is (– 2, 4, – 5) and Q is (1, 2, 3) So
PQ =
2
2
2
(1
( 2))
(2
4)
(3
( 5))
− −
+
−
+
− −
=
77
Thus, the direction cosines of the line joining two points is
3
2
8
,
,
77
77
77
−
Example 4 Find the direction cosines of x, y and z-axis Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i |
1 | 5677-5680 | So
PQ =
2
2
2
(1
( 2))
(2
4)
(3
( 5))
− −
+
−
+
− −
=
77
Thus, the direction cosines of the line joining two points is
3
2
8
,
,
77
77
77
−
Example 4 Find the direction cosines of x, y and z-axis Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i e |
1 | 5678-5681 | Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i e , 1,0,0 |
1 | 5679-5682 | Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i e , 1,0,0 Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively |
1 | 5680-5683 | e , 1,0,0 Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are
collinear |
1 | 5681-5684 | , 1,0,0 Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are
collinear Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i |
1 | 5682-5685 | Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are
collinear Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i e |
1 | 5683-5686 | Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are
collinear Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i e , – 1, – 5, 7 |
1 | 5684-5687 | Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i e , – 1, – 5, 7 The direction ratios of line joining B and C are
3 –1, 8 + 2, – 11 – 3, i |
1 | 5685-5688 | e , – 1, – 5, 7 The direction ratios of line joining B and C are
3 –1, 8 + 2, – 11 – 3, i e |
1 | 5686-5689 | , – 1, – 5, 7 The direction ratios of line joining B and C are
3 –1, 8 + 2, – 11 – 3, i e , 2, 10, – 14 |
1 | 5687-5690 | The direction ratios of line joining B and C are
3 –1, 8 + 2, – 11 – 3, i e , 2, 10, – 14 It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC |
1 | 5688-5691 | e , 2, 10, – 14 It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC But point B is common to both AB and BC |
1 | 5689-5692 | , 2, 10, – 14 It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC But point B is common to both AB and BC Therefore, A, B, C are
collinear points |
1 | 5690-5693 | It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC But point B is common to both AB and BC Therefore, A, B, C are
collinear points EXERCISE 11 |
1 | 5691-5694 | But point B is common to both AB and BC Therefore, A, B, C are
collinear points EXERCISE 11 1
1 |
1 | 5692-5695 | Therefore, A, B, C are
collinear points EXERCISE 11 1
1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its
direction cosines |
1 | 5693-5696 | EXERCISE 11 1
1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its
direction cosines 2 |
1 | 5694-5697 | 1
1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its
direction cosines 2 Find the direction cosines of a line which makes equal angles with the coordinate
axes |
1 | 5695-5698 | If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its
direction cosines 2 Find the direction cosines of a line which makes equal angles with the coordinate
axes 3 |
1 | 5696-5699 | 2 Find the direction cosines of a line which makes equal angles with the coordinate
axes 3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines |
1 | 5697-5700 | Find the direction cosines of a line which makes equal angles with the coordinate
axes 3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines 4 |
1 | 5698-5701 | 3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines 4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear |
1 | 5699-5702 | If a line has the direction ratios –18, 12, – 4, then what are its direction cosines 4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear 5 |
1 | 5700-5703 | 4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear 5 Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2) |
1 | 5701-5704 | Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear 5 Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2) © NCERT
not to be republished
MATHEMATI CS
468
11 |
1 | 5702-5705 | 5 Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2) © NCERT
not to be republished
MATHEMATI CS
468
11 3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space |
1 | 5703-5706 | Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2) © NCERT
not to be republished
MATHEMATI CS
468
11 3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space (i)A line is uniquely determined if
it passes through a given point and has given direction, or
(ii)
it passes through two given points |
1 | 5704-5707 | © NCERT
not to be republished
MATHEMATI CS
468
11 3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space (i)A line is uniquely determined if
it passes through a given point and has given direction, or
(ii)
it passes through two given points 11 |
1 | 5705-5708 | 3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space (i)A line is uniquely determined if
it passes through a given point and has given direction, or
(ii)
it passes through two given points 11 3 |
1 | 5706-5709 | (i)A line is uniquely determined if
it passes through a given point and has given direction, or
(ii)
it passes through two given points 11 3 1Equation of a line through a given point and parallel to a given vector
r
b
Let ar be the position vector of the given point
A wit h respect to the origin O of t he
rectangular coordinate system |
1 | 5707-5710 | 11 3 1Equation of a line through a given point and parallel to a given vector
r
b
Let ar be the position vector of the given point
A wit h respect to the origin O of t he
rectangular coordinate system Let l be the
line which passes through the point A and is
parallel to a given vector b
r |
1 | 5708-5711 | 3 1Equation of a line through a given point and parallel to a given vector
r
b
Let ar be the position vector of the given point
A wit h respect to the origin O of t he
rectangular coordinate system Let l be the
line which passes through the point A and is
parallel to a given vector b
r Let rr be the
position vector of an arbitrary point P on the
line (Fig 11 |
1 | 5709-5712 | 1Equation of a line through a given point and parallel to a given vector
r
b
Let ar be the position vector of the given point
A wit h respect to the origin O of t he
rectangular coordinate system Let l be the
line which passes through the point A and is
parallel to a given vector b
r Let rr be the
position vector of an arbitrary point P on the
line (Fig 11 4) |
1 | 5710-5713 | Let l be the
line which passes through the point A and is
parallel to a given vector b
r Let rr be the
position vector of an arbitrary point P on the
line (Fig 11 4) Then AP
uuur
is parallel to the vector b
r
, i |
1 | 5711-5714 | Let rr be the
position vector of an arbitrary point P on the
line (Fig 11 4) Then AP
uuur
is parallel to the vector b
r
, i e |
1 | 5712-5715 | 4) Then AP
uuur
is parallel to the vector b
r
, i e ,
AP
uuur
= λb
r
, where λ is some real number |
1 | 5713-5716 | Then AP
uuur
is parallel to the vector b
r
, i e ,
AP
uuur
= λb
r
, where λ is some real number But
AP
uuur
= OP – OA
uuur
uuur
i |
1 | 5714-5717 | e ,
AP
uuur
= λb
r
, where λ is some real number But
AP
uuur
= OP – OA
uuur
uuur
i e |
1 | 5715-5718 | ,
AP
uuur
= λb
r
, where λ is some real number But
AP
uuur
= OP – OA
uuur
uuur
i e λb
r
= r
a−
r
r
Conversely, for each value of the parameter λ, this equation gives the position
vector of a point P on the line |
1 | 5716-5719 | But
AP
uuur
= OP – OA
uuur
uuur
i e λb
r
= r
a−
r
r
Conversely, for each value of the parameter λ, this equation gives the position
vector of a point P on the line Hence, the vector equation of the line is given by
rr =
λ
r
ra +
b |
1 | 5717-5720 | e λb
r
= r
a−
r
r
Conversely, for each value of the parameter λ, this equation gives the position
vector of a point P on the line Hence, the vector equation of the line is given by
rr =
λ
r
ra +
b (1)
Remark If
ˆ
ˆ
ˆ
b
ai
bj
ck
= + +
r
, then a, b, c are direction ratios of the line and conversely,,
if a, b, c are direction ratios of a line, then
ˆ
ˆ
ˆ
=
+
+
br
ai
bj
ck will be the parallel to
the line |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.