Chapter
stringclasses 18
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1 | 5818-5821 | ) We have
1ar =
ˆ
ˆ
2ˆ
4
i
j
k
+
−
,
2ar
=
ˆ
ˆ
ˆ
3
3
5
i
j
k
+
−
and b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
Therefore, the distance between the lines is given by
d =
2
1
(
)
| |
b
a
a
b
×
−
r
r
r
r
=
ˆ
ˆ
ˆ
2
3
6
2
1
1
4
9
36
i
j
k
−
+
+
or
=
ˆ
ˆ
ˆ
|
9
14
4
|
293
7293
49
49
i
j
k
−
+
−
=
=
EXERCISE 11 2
1 Show that the three lines with direction cosines
12
3
4
4
12
3
3
4
12
,
,
;
,
,
;
,
,
13
13
13
13
13
13
13
13
13
−
−
−
are mutually perpendicular 2 |
1 | 5819-5822 | 2
1 Show that the three lines with direction cosines
12
3
4
4
12
3
3
4
12
,
,
;
,
,
;
,
,
13
13
13
13
13
13
13
13
13
−
−
−
are mutually perpendicular 2 Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6) |
1 | 5820-5823 | Show that the three lines with direction cosines
12
3
4
4
12
3
3
4
12
,
,
;
,
,
;
,
,
13
13
13
13
13
13
13
13
13
−
−
−
are mutually perpendicular 2 Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6) 3 |
1 | 5821-5824 | 2 Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6) 3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5) |
1 | 5822-5825 | Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6) 3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5) 4 |
1 | 5823-5826 | 3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5) 4 Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector
ˆ
ˆ
ˆ
3
2
2
i
j
k
+
− |
1 | 5824-5827 | Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5) 4 Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector
ˆ
ˆ
ˆ
3
2
2
i
j
k
+
− 5 |
1 | 5825-5828 | 4 Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector
ˆ
ˆ
ˆ
3
2
2
i
j
k
+
− 5 Find the equation of the line in vector and in cartesian form that passes through
the point with position vector
ˆ
2ˆ
4
i
j
k
−
+
and is in the direction
ˆ
ˆ
2ˆ
i
j
k
+
− |
1 | 5826-5829 | Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector
ˆ
ˆ
ˆ
3
2
2
i
j
k
+
− 5 Find the equation of the line in vector and in cartesian form that passes through
the point with position vector
ˆ
2ˆ
4
i
j
k
−
+
and is in the direction
ˆ
ˆ
2ˆ
i
j
k
+
− 6 |
1 | 5827-5830 | 5 Find the equation of the line in vector and in cartesian form that passes through
the point with position vector
ˆ
2ˆ
4
i
j
k
−
+
and is in the direction
ˆ
ˆ
2ˆ
i
j
k
+
− 6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
3
4
8
3
5
6
x
y
z
+
−
+
=
= |
1 | 5828-5831 | Find the equation of the line in vector and in cartesian form that passes through
the point with position vector
ˆ
2ˆ
4
i
j
k
−
+
and is in the direction
ˆ
ˆ
2ˆ
i
j
k
+
− 6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
3
4
8
3
5
6
x
y
z
+
−
+
=
= 7 |
1 | 5829-5832 | 6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
3
4
8
3
5
6
x
y
z
+
−
+
=
= 7 The cartesian equation of a line is
5
4
6
3
7
2
x
y
z
−
+
−
=
= |
1 | 5830-5833 | Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
3
4
8
3
5
6
x
y
z
+
−
+
=
= 7 The cartesian equation of a line is
5
4
6
3
7
2
x
y
z
−
+
−
=
= Write its vector form |
1 | 5831-5834 | 7 The cartesian equation of a line is
5
4
6
3
7
2
x
y
z
−
+
−
=
= Write its vector form 8 |
1 | 5832-5835 | The cartesian equation of a line is
5
4
6
3
7
2
x
y
z
−
+
−
=
= Write its vector form 8 Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3) |
1 | 5833-5836 | Write its vector form 8 Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3) © NCERT
not to be republished
MATHEMATI CS
478
9 |
1 | 5834-5837 | 8 Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3) © NCERT
not to be republished
MATHEMATI CS
478
9 Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6) |
1 | 5835-5838 | Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3) © NCERT
not to be republished
MATHEMATI CS
478
9 Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6) 10 |
1 | 5836-5839 | © NCERT
not to be republished
MATHEMATI CS
478
9 Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6) 10 Find the angle between the following pairs of lines:
(i)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
(3
2
6 )
r
i
j
k
i
j
k
=
−
+
+ λ
+
+
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
7
6
(
2
2 )
r
i
k
i
j
k
=
−
+ µ
+
+
r
(ii)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
(
2 )
r
i
j
k
i
j
k
=
+
−
+ λ
−
−
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
56
(3
5
4 )
r
i
j
k
i
j
k
=
−
−
+ µ
−
−
r
11 |
1 | 5837-5840 | Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6) 10 Find the angle between the following pairs of lines:
(i)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
(3
2
6 )
r
i
j
k
i
j
k
=
−
+
+ λ
+
+
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
7
6
(
2
2 )
r
i
k
i
j
k
=
−
+ µ
+
+
r
(ii)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
(
2 )
r
i
j
k
i
j
k
=
+
−
+ λ
−
−
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
56
(3
5
4 )
r
i
j
k
i
j
k
=
−
−
+ µ
−
−
r
11 Find the angle between the following pair of lines:
(i)
2
1
3
2
4
5
and
2
5
3
1
8
4
x
y
z
x
y
z
−
− +
+
−
−
=
=
=
=
−
−
(ii)
5
2
3
and
2
2
1
4
1
8
x
y
z
x
y
z
−
−
−
=
=
=
=
12 |
1 | 5838-5841 | 10 Find the angle between the following pairs of lines:
(i)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
(3
2
6 )
r
i
j
k
i
j
k
=
−
+
+ λ
+
+
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
7
6
(
2
2 )
r
i
k
i
j
k
=
−
+ µ
+
+
r
(ii)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
(
2 )
r
i
j
k
i
j
k
=
+
−
+ λ
−
−
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
56
(3
5
4 )
r
i
j
k
i
j
k
=
−
−
+ µ
−
−
r
11 Find the angle between the following pair of lines:
(i)
2
1
3
2
4
5
and
2
5
3
1
8
4
x
y
z
x
y
z
−
− +
+
−
−
=
=
=
=
−
−
(ii)
5
2
3
and
2
2
1
4
1
8
x
y
z
x
y
z
−
−
−
=
=
=
=
12 Find the values of p so that the lines 1
7
14
3
3
2
2
x
y
z
p
−
−
−
=
=
and 7
7
5
6
3
1
5
x
y
z
p
−
−
−
=
=
are at right angles |
1 | 5839-5842 | Find the angle between the following pairs of lines:
(i)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
(3
2
6 )
r
i
j
k
i
j
k
=
−
+
+ λ
+
+
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
7
6
(
2
2 )
r
i
k
i
j
k
=
−
+ µ
+
+
r
(ii)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
(
2 )
r
i
j
k
i
j
k
=
+
−
+ λ
−
−
r
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
56
(3
5
4 )
r
i
j
k
i
j
k
=
−
−
+ µ
−
−
r
11 Find the angle between the following pair of lines:
(i)
2
1
3
2
4
5
and
2
5
3
1
8
4
x
y
z
x
y
z
−
− +
+
−
−
=
=
=
=
−
−
(ii)
5
2
3
and
2
2
1
4
1
8
x
y
z
x
y
z
−
−
−
=
=
=
=
12 Find the values of p so that the lines 1
7
14
3
3
2
2
x
y
z
p
−
−
−
=
=
and 7
7
5
6
3
1
5
x
y
z
p
−
−
−
=
=
are at right angles 13 |
1 | 5840-5843 | Find the angle between the following pair of lines:
(i)
2
1
3
2
4
5
and
2
5
3
1
8
4
x
y
z
x
y
z
−
− +
+
−
−
=
=
=
=
−
−
(ii)
5
2
3
and
2
2
1
4
1
8
x
y
z
x
y
z
−
−
−
=
=
=
=
12 Find the values of p so that the lines 1
7
14
3
3
2
2
x
y
z
p
−
−
−
=
=
and 7
7
5
6
3
1
5
x
y
z
p
−
−
−
=
=
are at right angles 13 Show that the lines
5
2
7
5
1
x
y
z
−
+
=
=
−
and
1
2
3
x
y
z
=
=
are perpendicular to
each other |
1 | 5841-5844 | Find the values of p so that the lines 1
7
14
3
3
2
2
x
y
z
p
−
−
−
=
=
and 7
7
5
6
3
1
5
x
y
z
p
−
−
−
=
=
are at right angles 13 Show that the lines
5
2
7
5
1
x
y
z
−
+
=
=
−
and
1
2
3
x
y
z
=
=
are perpendicular to
each other 14 |
1 | 5842-5845 | 13 Show that the lines
5
2
7
5
1
x
y
z
−
+
=
=
−
and
1
2
3
x
y
z
=
=
are perpendicular to
each other 14 Find the shortest distance between the lines
ˆ
ˆ
ˆ
(
2
)
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
)
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(2
2 )
r
i
j
k
i
j
k
=
−
−
+ µ
+
+
r
15 |
1 | 5843-5846 | Show that the lines
5
2
7
5
1
x
y
z
−
+
=
=
−
and
1
2
3
x
y
z
=
=
are perpendicular to
each other 14 Find the shortest distance between the lines
ˆ
ˆ
ˆ
(
2
)
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
)
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(2
2 )
r
i
j
k
i
j
k
=
−
−
+ µ
+
+
r
15 Find the shortest distance between the lines
1
1
1
7
6
1
x
y
z
+
+
+
=
=
−
and
3
5
7
1
2
1
x
y
z
−
−
−
=
=
−
16 |
1 | 5844-5847 | 14 Find the shortest distance between the lines
ˆ
ˆ
ˆ
(
2
)
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
)
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(2
2 )
r
i
j
k
i
j
k
=
−
−
+ µ
+
+
r
15 Find the shortest distance between the lines
1
1
1
7
6
1
x
y
z
+
+
+
=
=
−
and
3
5
7
1
2
1
x
y
z
−
−
−
=
=
−
16 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(
2
3 )
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
3
2 )
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
5
6
(2
3
)
r
i
j
k
i
j
k
=
+
+
+ µ
+
+
r
17 |
1 | 5845-5848 | Find the shortest distance between the lines
ˆ
ˆ
ˆ
(
2
)
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
)
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(2
2 )
r
i
j
k
i
j
k
=
−
−
+ µ
+
+
r
15 Find the shortest distance between the lines
1
1
1
7
6
1
x
y
z
+
+
+
=
=
−
and
3
5
7
1
2
1
x
y
z
−
−
−
=
=
−
16 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(
2
3 )
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
3
2 )
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
5
6
(2
3
)
r
i
j
k
i
j
k
=
+
+
+ µ
+
+
r
17 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(1
)
(
2)
(3
2 )
r
t i
t
j
t k
=
−
+
−
+
−
r
and
ˆ
ˆ
ˆ
(
1)
(2
1)
(2
1)
r
s
i
s
j
s
k
=
+
+
−
−
+
r
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
479
11 |
1 | 5846-5849 | Find the shortest distance between the lines
1
1
1
7
6
1
x
y
z
+
+
+
=
=
−
and
3
5
7
1
2
1
x
y
z
−
−
−
=
=
−
16 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(
2
3 )
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
3
2 )
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
5
6
(2
3
)
r
i
j
k
i
j
k
=
+
+
+ µ
+
+
r
17 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(1
)
(
2)
(3
2 )
r
t i
t
j
t k
=
−
+
−
+
−
r
and
ˆ
ˆ
ˆ
(
1)
(2
1)
(2
1)
r
s
i
s
j
s
k
=
+
+
−
−
+
r
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
479
11 6 Plane
A plane is determined uniquely if any one of the following is known:
(i)
the normal to the plane and its distance from the origin is given, i |
1 | 5847-5850 | Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(
2
3 )
r
i
j
k
=
+
+
r
+
ˆ
ˆ
ˆ
(
3
2 )
i
j
k
λ
−
+
and
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
5
6
(2
3
)
r
i
j
k
i
j
k
=
+
+
+ µ
+
+
r
17 Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(1
)
(
2)
(3
2 )
r
t i
t
j
t k
=
−
+
−
+
−
r
and
ˆ
ˆ
ˆ
(
1)
(2
1)
(2
1)
r
s
i
s
j
s
k
=
+
+
−
−
+
r
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
479
11 6 Plane
A plane is determined uniquely if any one of the following is known:
(i)
the normal to the plane and its distance from the origin is given, i e |
1 | 5848-5851 | Find the shortest distance between the lines whose vector equations are
ˆ
ˆ
ˆ
(1
)
(
2)
(3
2 )
r
t i
t
j
t k
=
−
+
−
+
−
r
and
ˆ
ˆ
ˆ
(
1)
(2
1)
(2
1)
r
s
i
s
j
s
k
=
+
+
−
−
+
r
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
479
11 6 Plane
A plane is determined uniquely if any one of the following is known:
(i)
the normal to the plane and its distance from the origin is given, i e , equation of
a plane in normal form |
1 | 5849-5852 | 6 Plane
A plane is determined uniquely if any one of the following is known:
(i)
the normal to the plane and its distance from the origin is given, i e , equation of
a plane in normal form (ii)
it passes through a point and is perpendicular to a given direction |
1 | 5850-5853 | e , equation of
a plane in normal form (ii)
it passes through a point and is perpendicular to a given direction (iii)
it passes through three given non collinear points |
1 | 5851-5854 | , equation of
a plane in normal form (ii)
it passes through a point and is perpendicular to a given direction (iii)
it passes through three given non collinear points Now we shall find vector and Cartesian equations of the planes |
1 | 5852-5855 | (ii)
it passes through a point and is perpendicular to a given direction (iii)
it passes through three given non collinear points Now we shall find vector and Cartesian equations of the planes 11 |
1 | 5853-5856 | (iii)
it passes through three given non collinear points Now we shall find vector and Cartesian equations of the planes 11 6 |
1 | 5854-5857 | Now we shall find vector and Cartesian equations of the planes 11 6 1Equation of a plane in normal form
Consider a plane whose perpendicular distance from the origin is d (d ≠ 0) |
1 | 5855-5858 | 11 6 1Equation of a plane in normal form
Consider a plane whose perpendicular distance from the origin is d (d ≠ 0) Fig 11 |
1 | 5856-5859 | 6 1Equation of a plane in normal form
Consider a plane whose perpendicular distance from the origin is d (d ≠ 0) Fig 11 10 |
1 | 5857-5860 | 1Equation of a plane in normal form
Consider a plane whose perpendicular distance from the origin is d (d ≠ 0) Fig 11 10 If ON
uuur
is the normal from the origin to the plane, and nˆ is the unit normal vector
along ON
uuur |
1 | 5858-5861 | Fig 11 10 If ON
uuur
is the normal from the origin to the plane, and nˆ is the unit normal vector
along ON
uuur Then ON
uuur
= d nˆ |
1 | 5859-5862 | 10 If ON
uuur
is the normal from the origin to the plane, and nˆ is the unit normal vector
along ON
uuur Then ON
uuur
= d nˆ Let P be any
point on the plane |
1 | 5860-5863 | If ON
uuur
is the normal from the origin to the plane, and nˆ is the unit normal vector
along ON
uuur Then ON
uuur
= d nˆ Let P be any
point on the plane T herefore, NP
uuur
is
perpendicular to ON
uuur
Therefore, NP ON |
1 | 5861-5864 | Then ON
uuur
= d nˆ Let P be any
point on the plane T herefore, NP
uuur
is
perpendicular to ON
uuur
Therefore, NP ON ⋅
uuur uuur
= 0 |
1 | 5862-5865 | Let P be any
point on the plane T herefore, NP
uuur
is
perpendicular to ON
uuur
Therefore, NP ON ⋅
uuur uuur
= 0 (1)
Let rr be the position vector of the point P,,
then NP
uuur
=
dn
r
ˆ
r−
(as ON
NP
OP
+ =
uuur
uuur
uuur
)
Therefore, (1) becomes
(
)
r
d n
d n
∧
∧
−
⋅
r
= 0
or
(
)
r
d n
n
∧
∧
−
⋅
r
= 0
(d ≠ 0)
or
r n
d n n
∧
∧
∧
⋅
−
⋅
r
= 0
i |
1 | 5863-5866 | T herefore, NP
uuur
is
perpendicular to ON
uuur
Therefore, NP ON ⋅
uuur uuur
= 0 (1)
Let rr be the position vector of the point P,,
then NP
uuur
=
dn
r
ˆ
r−
(as ON
NP
OP
+ =
uuur
uuur
uuur
)
Therefore, (1) becomes
(
)
r
d n
d n
∧
∧
−
⋅
r
= 0
or
(
)
r
d n
n
∧
∧
−
⋅
r
= 0
(d ≠ 0)
or
r n
d n n
∧
∧
∧
⋅
−
⋅
r
= 0
i e |
1 | 5864-5867 | ⋅
uuur uuur
= 0 (1)
Let rr be the position vector of the point P,,
then NP
uuur
=
dn
r
ˆ
r−
(as ON
NP
OP
+ =
uuur
uuur
uuur
)
Therefore, (1) becomes
(
)
r
d n
d n
∧
∧
−
⋅
r
= 0
or
(
)
r
d n
n
∧
∧
−
⋅
r
= 0
(d ≠ 0)
or
r n
d n n
∧
∧
∧
⋅
−
⋅
r
= 0
i e ,
r n
r⋅∧
= d
(as
1)
∧n n
∧⋅
=
… (2)
This is the vector form of the equation of the plane |
1 | 5865-5868 | (1)
Let rr be the position vector of the point P,,
then NP
uuur
=
dn
r
ˆ
r−
(as ON
NP
OP
+ =
uuur
uuur
uuur
)
Therefore, (1) becomes
(
)
r
d n
d n
∧
∧
−
⋅
r
= 0
or
(
)
r
d n
n
∧
∧
−
⋅
r
= 0
(d ≠ 0)
or
r n
d n n
∧
∧
∧
⋅
−
⋅
r
= 0
i e ,
r n
r⋅∧
= d
(as
1)
∧n n
∧⋅
=
… (2)
This is the vector form of the equation of the plane Cartesian form
Equation (2) gives the vector equation of a plane, where nˆ is the unit vector normal to
the plane |
1 | 5866-5869 | e ,
r n
r⋅∧
= d
(as
1)
∧n n
∧⋅
=
… (2)
This is the vector form of the equation of the plane Cartesian form
Equation (2) gives the vector equation of a plane, where nˆ is the unit vector normal to
the plane Let P(x, y, z) be any point on the plane |
1 | 5867-5870 | ,
r n
r⋅∧
= d
(as
1)
∧n n
∧⋅
=
… (2)
This is the vector form of the equation of the plane Cartesian form
Equation (2) gives the vector equation of a plane, where nˆ is the unit vector normal to
the plane Let P(x, y, z) be any point on the plane Then
OP
uuur
=
ˆ
ˆ
ˆ
r
x i
y j
= + +z k
r
Let l, m, n be the direction cosines of nˆ |
1 | 5868-5871 | Cartesian form
Equation (2) gives the vector equation of a plane, where nˆ is the unit vector normal to
the plane Let P(x, y, z) be any point on the plane Then
OP
uuur
=
ˆ
ˆ
ˆ
r
x i
y j
= + +z k
r
Let l, m, n be the direction cosines of nˆ Then
ˆn =
ˆ
ˆ
ˆ
l i
m j
n k
+ +
X
Y
Z
N
P(
)
x, y,z
r
d
O
Fig 11 |
1 | 5869-5872 | Let P(x, y, z) be any point on the plane Then
OP
uuur
=
ˆ
ˆ
ˆ
r
x i
y j
= + +z k
r
Let l, m, n be the direction cosines of nˆ Then
ˆn =
ˆ
ˆ
ˆ
l i
m j
n k
+ +
X
Y
Z
N
P(
)
x, y,z
r
d
O
Fig 11 10
© NCERT
not to be republished
MATHEMATI CS
480
Therefore, (2) gives
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
x i
y j
z k
l i
m j
n k
d
+
+
⋅
+
+
=
i |
1 | 5870-5873 | Then
OP
uuur
=
ˆ
ˆ
ˆ
r
x i
y j
= + +z k
r
Let l, m, n be the direction cosines of nˆ Then
ˆn =
ˆ
ˆ
ˆ
l i
m j
n k
+ +
X
Y
Z
N
P(
)
x, y,z
r
d
O
Fig 11 10
© NCERT
not to be republished
MATHEMATI CS
480
Therefore, (2) gives
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
x i
y j
z k
l i
m j
n k
d
+
+
⋅
+
+
=
i e |
1 | 5871-5874 | Then
ˆn =
ˆ
ˆ
ˆ
l i
m j
n k
+ +
X
Y
Z
N
P(
)
x, y,z
r
d
O
Fig 11 10
© NCERT
not to be republished
MATHEMATI CS
480
Therefore, (2) gives
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
x i
y j
z k
l i
m j
n k
d
+
+
⋅
+
+
=
i e ,
lx + my + nz = d |
1 | 5872-5875 | 10
© NCERT
not to be republished
MATHEMATI CS
480
Therefore, (2) gives
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) (
)
x i
y j
z k
l i
m j
n k
d
+
+
⋅
+
+
=
i e ,
lx + my + nz = d (3)
This is the cartesian equation of the plane in the normal form |
1 | 5873-5876 | e ,
lx + my + nz = d (3)
This is the cartesian equation of the plane in the normal form �Note Equation (3) shows that if
ˆ
ˆ
ˆ
(
)
r
a i
b j
c k
⋅
+
+
r
= d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane |
1 | 5874-5877 | ,
lx + my + nz = d (3)
This is the cartesian equation of the plane in the normal form �Note Equation (3) shows that if
ˆ
ˆ
ˆ
(
)
r
a i
b j
c k
⋅
+
+
r
= d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane Example 13 Find the vector equation of the plane which is at a distance of
29
6
from the origin and its normal vector from the origin is
ˆ
ˆ
ˆ
2
3
4
i
j
− +k |
1 | 5875-5878 | (3)
This is the cartesian equation of the plane in the normal form �Note Equation (3) shows that if
ˆ
ˆ
ˆ
(
)
r
a i
b j
c k
⋅
+
+
r
= d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane Example 13 Find the vector equation of the plane which is at a distance of
29
6
from the origin and its normal vector from the origin is
ˆ
ˆ
ˆ
2
3
4
i
j
− +k Solution Let
k
j
i
n
4ˆ
3ˆ
2ˆ
+
−
r= |
1 | 5876-5879 | �Note Equation (3) shows that if
ˆ
ˆ
ˆ
(
)
r
a i
b j
c k
⋅
+
+
r
= d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane Example 13 Find the vector equation of the plane which is at a distance of
29
6
from the origin and its normal vector from the origin is
ˆ
ˆ
ˆ
2
3
4
i
j
− +k Solution Let
k
j
i
n
4ˆ
3ˆ
2ˆ
+
−
r= Then
|
|
ˆ
nn
n
r
r
=
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
3
4
4
9
16
29
i
j
k
i
j
k
−
+
−
+
=
+
+
Hence, the required equation of the plane is
2
3
4
6
ˆ
ˆ
ˆ
29
29
29
29
r
i
j
k
−
⋅
+
+
=
r
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆ
ˆ
(6
3
2 )
1
r
i
j
k
⋅
−
−
+
r
= 0 passing through the origin |
1 | 5877-5880 | Example 13 Find the vector equation of the plane which is at a distance of
29
6
from the origin and its normal vector from the origin is
ˆ
ˆ
ˆ
2
3
4
i
j
− +k Solution Let
k
j
i
n
4ˆ
3ˆ
2ˆ
+
−
r= Then
|
|
ˆ
nn
n
r
r
=
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
3
4
4
9
16
29
i
j
k
i
j
k
−
+
−
+
=
+
+
Hence, the required equation of the plane is
2
3
4
6
ˆ
ˆ
ˆ
29
29
29
29
r
i
j
k
−
⋅
+
+
=
r
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆ
ˆ
(6
3
2 )
1
r
i
j
k
⋅
−
−
+
r
= 0 passing through the origin Solution The given equation can be written as
ˆ
ˆ
ˆ
(
6
3
2
⋅
−
+
+
rr
i
j
k ) = 1 |
1 | 5878-5881 | Solution Let
k
j
i
n
4ˆ
3ˆ
2ˆ
+
−
r= Then
|
|
ˆ
nn
n
r
r
=
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
3
4
4
9
16
29
i
j
k
i
j
k
−
+
−
+
=
+
+
Hence, the required equation of the plane is
2
3
4
6
ˆ
ˆ
ˆ
29
29
29
29
r
i
j
k
−
⋅
+
+
=
r
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆ
ˆ
(6
3
2 )
1
r
i
j
k
⋅
−
−
+
r
= 0 passing through the origin Solution The given equation can be written as
ˆ
ˆ
ˆ
(
6
3
2
⋅
−
+
+
rr
i
j
k ) = 1 (1)
Now
ˆ
ˆ
ˆ
|
6
3
2
|
i
j
k
−
+
+
=
36
9
4
7
+ + =
Therefore, dividing both sides of (1) by 7, we get
6
3
2 ˆ
ˆ
ˆ
7
7
7
r
i
j
k
⋅ −
+
+
r
= 1
7
which is the equation of the plane in the form
r nˆ
r⋅ =d |
1 | 5879-5882 | Then
|
|
ˆ
nn
n
r
r
=
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
3
4
4
9
16
29
i
j
k
i
j
k
−
+
−
+
=
+
+
Hence, the required equation of the plane is
2
3
4
6
ˆ
ˆ
ˆ
29
29
29
29
r
i
j
k
−
⋅
+
+
=
r
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆ
ˆ
(6
3
2 )
1
r
i
j
k
⋅
−
−
+
r
= 0 passing through the origin Solution The given equation can be written as
ˆ
ˆ
ˆ
(
6
3
2
⋅
−
+
+
rr
i
j
k ) = 1 (1)
Now
ˆ
ˆ
ˆ
|
6
3
2
|
i
j
k
−
+
+
=
36
9
4
7
+ + =
Therefore, dividing both sides of (1) by 7, we get
6
3
2 ˆ
ˆ
ˆ
7
7
7
r
i
j
k
⋅ −
+
+
r
= 1
7
which is the equation of the plane in the form
r nˆ
r⋅ =d This shows that
k
j
i
n
ˆ
27
37ˆ
67ˆ
ˆ
+
+
=−
is a unit vector perpendicular to the
plane through the origin |
1 | 5880-5883 | Solution The given equation can be written as
ˆ
ˆ
ˆ
(
6
3
2
⋅
−
+
+
rr
i
j
k ) = 1 (1)
Now
ˆ
ˆ
ˆ
|
6
3
2
|
i
j
k
−
+
+
=
36
9
4
7
+ + =
Therefore, dividing both sides of (1) by 7, we get
6
3
2 ˆ
ˆ
ˆ
7
7
7
r
i
j
k
⋅ −
+
+
r
= 1
7
which is the equation of the plane in the form
r nˆ
r⋅ =d This shows that
k
j
i
n
ˆ
27
37ˆ
67ˆ
ˆ
+
+
=−
is a unit vector perpendicular to the
plane through the origin Hence, the direction cosines of nˆ are
27
,
37
,
7
−6 |
1 | 5881-5884 | (1)
Now
ˆ
ˆ
ˆ
|
6
3
2
|
i
j
k
−
+
+
=
36
9
4
7
+ + =
Therefore, dividing both sides of (1) by 7, we get
6
3
2 ˆ
ˆ
ˆ
7
7
7
r
i
j
k
⋅ −
+
+
r
= 1
7
which is the equation of the plane in the form
r nˆ
r⋅ =d This shows that
k
j
i
n
ˆ
27
37ˆ
67ˆ
ˆ
+
+
=−
is a unit vector perpendicular to the
plane through the origin Hence, the direction cosines of nˆ are
27
,
37
,
7
−6 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
481
Z
Y
X
O
P(x1, y1, z1)
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin |
1 | 5882-5885 | This shows that
k
j
i
n
ˆ
27
37ˆ
67ˆ
ˆ
+
+
=−
is a unit vector perpendicular to the
plane through the origin Hence, the direction cosines of nˆ are
27
,
37
,
7
−6 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
481
Z
Y
X
O
P(x1, y1, z1)
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2
2
2
2
2
2
2
2
2
2
3
4
,
,
2
( 3)
4
2
( 3)
4
2
( 3)
4
−
+ −
+
+ −
+
+ −
+
, i |
1 | 5883-5886 | Hence, the direction cosines of nˆ are
27
,
37
,
7
−6 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
481
Z
Y
X
O
P(x1, y1, z1)
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2
2
2
2
2
2
2
2
2
2
3
4
,
,
2
( 3)
4
2
( 3)
4
2
( 3)
4
−
+ −
+
+ −
+
+ −
+
, i e |
1 | 5884-5887 | © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
481
Z
Y
X
O
P(x1, y1, z1)
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2
2
2
2
2
2
2
2
2
2
3
4
,
,
2
( 3)
4
2
( 3)
4
2
( 3)
4
−
+ −
+
+ −
+
+ −
+
, i e ,
2
3
4
,
,
29
29
29
−
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i |
1 | 5885-5888 | Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2
2
2
2
2
2
2
2
2
2
3
4
,
,
2
( 3)
4
2
( 3)
4
2
( 3)
4
−
+ −
+
+ −
+
+ −
+
, i e ,
2
3
4
,
,
29
29
29
−
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i e |
1 | 5886-5889 | e ,
2
3
4
,
,
29
29
29
−
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i e , 2x – 3y + 4z = 6 throughout by
29 , we get
2
3
4
6
29
29
29
29
x
y
z
−
+
+
=
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin |
1 | 5887-5890 | ,
2
3
4
,
,
29
29
29
−
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i e , 2x – 3y + 4z = 6 throughout by
29 , we get
2
3
4
6
29
29
29
29
x
y
z
−
+
+
=
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin So, the distance of the plane from the origin is
29
6 |
1 | 5888-5891 | e , 2x – 3y + 4z = 6 throughout by
29 , we get
2
3
4
6
29
29
29
29
x
y
z
−
+
+
=
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin So, the distance of the plane from the origin is
29
6 Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0 |
1 | 5889-5892 | , 2x – 3y + 4z = 6 throughout by
29 , we get
2
3
4
6
29
29
29
29
x
y
z
−
+
+
=
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin So, the distance of the plane from the origin is
29
6 Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0 Solution Let the coordinates of the foot of the perpendicular P from the origin to the
plane is (x1, y1, z1) (Fig 11 |
1 | 5890-5893 | So, the distance of the plane from the origin is
29
6 Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0 Solution Let the coordinates of the foot of the perpendicular P from the origin to the
plane is (x1, y1, z1) (Fig 11 11) |
1 | 5891-5894 | Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0 Solution Let the coordinates of the foot of the perpendicular P from the origin to the
plane is (x1, y1, z1) (Fig 11 11) Then, the direction ratios of the line OP are
x1, y1, z1 |
1 | 5892-5895 | Solution Let the coordinates of the foot of the perpendicular P from the origin to the
plane is (x1, y1, z1) (Fig 11 11) Then, the direction ratios of the line OP are
x1, y1, z1 Writing the equation of the plane in the normal
form, we have
2
3
4
6
29
29
29
29
x
y
z
−
+
=
where,
2
3
4
,
,
29
29
29
−
are the direct ion
cosines of the OP |
1 | 5893-5896 | 11) Then, the direction ratios of the line OP are
x1, y1, z1 Writing the equation of the plane in the normal
form, we have
2
3
4
6
29
29
29
29
x
y
z
−
+
=
where,
2
3
4
,
,
29
29
29
−
are the direct ion
cosines of the OP Since d |
1 | 5894-5897 | Then, the direction ratios of the line OP are
x1, y1, z1 Writing the equation of the plane in the normal
form, we have
2
3
4
6
29
29
29
29
x
y
z
−
+
=
where,
2
3
4
,
,
29
29
29
−
are the direct ion
cosines of the OP Since d c |
1 | 5895-5898 | Writing the equation of the plane in the normal
form, we have
2
3
4
6
29
29
29
29
x
y
z
−
+
=
where,
2
3
4
,
,
29
29
29
−
are the direct ion
cosines of the OP Since d c ’s and direction ratios of a line are proportional, we have
21
29
x
=
1
1
3
4
29
29
y
z
=
−
= k
i |
1 | 5896-5899 | Since d c ’s and direction ratios of a line are proportional, we have
21
29
x
=
1
1
3
4
29
29
y
z
=
−
= k
i e |
1 | 5897-5900 | c ’s and direction ratios of a line are proportional, we have
21
29
x
=
1
1
3
4
29
29
y
z
=
−
= k
i e ,
x1 =
29
2k , y1 =
1
3
4
,
29
29
k
k
z
−
=
Fig 11 |
1 | 5898-5901 | ’s and direction ratios of a line are proportional, we have
21
29
x
=
1
1
3
4
29
29
y
z
=
−
= k
i e ,
x1 =
29
2k , y1 =
1
3
4
,
29
29
k
k
z
−
=
Fig 11 11
© NCERT
not to be republished
MATHEMATI CS
482
Substituting these in the equation of the plane, we get k =
29
6 |
1 | 5899-5902 | e ,
x1 =
29
2k , y1 =
1
3
4
,
29
29
k
k
z
−
=
Fig 11 11
© NCERT
not to be republished
MATHEMATI CS
482
Substituting these in the equation of the plane, we get k =
29
6 Hence, the foot of the perpendicular is 12
18 24
,
,
29
29
29
−
|
1 | 5900-5903 | ,
x1 =
29
2k , y1 =
1
3
4
,
29
29
k
k
z
−
=
Fig 11 11
© NCERT
not to be republished
MATHEMATI CS
482
Substituting these in the equation of the plane, we get k =
29
6 Hence, the foot of the perpendicular is 12
18 24
,
,
29
29
29
−
�Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd) |
1 | 5901-5904 | 11
© NCERT
not to be republished
MATHEMATI CS
482
Substituting these in the equation of the plane, we get k =
29
6 Hence, the foot of the perpendicular is 12
18 24
,
,
29
29
29
−
�Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd) 11 |
1 | 5902-5905 | Hence, the foot of the perpendicular is 12
18 24
,
,
29
29
29
−
�Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd) 11 6 |
1 | 5903-5906 | �Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd) 11 6 2Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x1, y1, z1), only one such plane exists (see
Fig 11 |
1 | 5904-5907 | 11 6 2Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x1, y1, z1), only one such plane exists (see
Fig 11 12) |
1 | 5905-5908 | 6 2Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x1, y1, z1), only one such plane exists (see
Fig 11 12) Let a plane pass through a point A with position
vector ar and perpendicular to the vector N
ur |
1 | 5906-5909 | 2Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x1, y1, z1), only one such plane exists (see
Fig 11 12) Let a plane pass through a point A with position
vector ar and perpendicular to the vector N
ur Let rr be the position vector of any point P(x, y, z) in the plane |
1 | 5907-5910 | 12) Let a plane pass through a point A with position
vector ar and perpendicular to the vector N
ur Let rr be the position vector of any point P(x, y, z) in the plane (Fig 11 |
1 | 5908-5911 | Let a plane pass through a point A with position
vector ar and perpendicular to the vector N
ur Let rr be the position vector of any point P(x, y, z) in the plane (Fig 11 13) |
1 | 5909-5912 | Let rr be the position vector of any point P(x, y, z) in the plane (Fig 11 13) Then the point P lies in the plane if and only if
AP
uuur
is perpendicular to N
ur |
1 | 5910-5913 | (Fig 11 13) Then the point P lies in the plane if and only if
AP
uuur
is perpendicular to N
ur i |
1 | 5911-5914 | 13) Then the point P lies in the plane if and only if
AP
uuur
is perpendicular to N
ur i e |
1 | 5912-5915 | Then the point P lies in the plane if and only if
AP
uuur
is perpendicular to N
ur i e , AP
uuur |
1 | 5913-5916 | i e , AP
uuur N
ur
= 0 |
1 | 5914-5917 | e , AP
uuur N
ur
= 0 But
AP
r
a
=
−
uuur
r
r |
1 | 5915-5918 | , AP
uuur N
ur
= 0 But
AP
r
a
=
−
uuur
r
r Therefore, (
) N
0
r
−a
⋅
=
r
r
r
… (1)
This is the vector equation of the plane |
1 | 5916-5919 | N
ur
= 0 But
AP
r
a
=
−
uuur
r
r Therefore, (
) N
0
r
−a
⋅
=
r
r
r
… (1)
This is the vector equation of the plane Cartesian form
Let the given point A be (x1, y1, z1), P be (x, y, z)
and direction ratios of N
ur
are A, B and C |
1 | 5917-5920 | But
AP
r
a
=
−
uuur
r
r Therefore, (
) N
0
r
−a
⋅
=
r
r
r
… (1)
This is the vector equation of the plane Cartesian form
Let the given point A be (x1, y1, z1), P be (x, y, z)
and direction ratios of N
ur
are A, B and C Then,
1
1
1 ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
a
x i
y j
z k
r
xi
y j
z k
=
+
+
=
+
+
r
r
and
ˆ
ˆ
ˆ
N
A
B
C
i
j
k
= + +
r
Now
r( – ) N= 0
a ⋅
r
r
r
So
(
)
(
)
(
)
1
1
1 ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(A
B
C ) 0
x
x i
y
y
j
z
z
k
i
j
k
−
+
−
+
−
⋅
+
+
=
i |
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