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1 | 6018-6021 | Solution The angle between two planes is the angle between their normals From the
equation of the planes, the normal vectors are
N1
ur
=
ˆ
ˆ
ˆ
2
2
i
j
k
+
−
and
2
ˆ
ˆ
ˆ
N
3
6
2
i
j
k
=
−
−
ur
Therefore
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
N
N
(2
2
) (3
6
2
)
| N | |N |
4
1
4
9
36
4
i
j
k
i
j
k
⋅
+
−
⋅
−
−
=
+
+
+
+
ur
ur
ur
ur
=
4
21
Hence
θ = cos – 1
21
4
© NCERT
not to be republished
MATHEMATI CS
490
Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5 Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
A2 = 2, B2 = 2, C2 = – 2
cos θ =
(
)
(
)
2
2
2
2
2
2
3
2
( 6) (2)
(2) ( 2)
3
( 6)
( 2)
2
2
( 2)
×
+ −
+
−
+ −
+ −
+
+ −
=
10
5
5 3
21
7
2 3
7 3
−
=
=
×
Therefore,
θ = cos-1 5 3
21
11 9 Distance of a Point from a Plane
Vector form
Consider a point P with position vector ar and a plane π 1 whose equation is
⋅rr nˆ
= d (Fig 11 |
1 | 6019-6022 | From the
equation of the planes, the normal vectors are
N1
ur
=
ˆ
ˆ
ˆ
2
2
i
j
k
+
−
and
2
ˆ
ˆ
ˆ
N
3
6
2
i
j
k
=
−
−
ur
Therefore
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
N
N
(2
2
) (3
6
2
)
| N | |N |
4
1
4
9
36
4
i
j
k
i
j
k
⋅
+
−
⋅
−
−
=
+
+
+
+
ur
ur
ur
ur
=
4
21
Hence
θ = cos – 1
21
4
© NCERT
not to be republished
MATHEMATI CS
490
Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5 Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
A2 = 2, B2 = 2, C2 = – 2
cos θ =
(
)
(
)
2
2
2
2
2
2
3
2
( 6) (2)
(2) ( 2)
3
( 6)
( 2)
2
2
( 2)
×
+ −
+
−
+ −
+ −
+
+ −
=
10
5
5 3
21
7
2 3
7 3
−
=
=
×
Therefore,
θ = cos-1 5 3
21
11 9 Distance of a Point from a Plane
Vector form
Consider a point P with position vector ar and a plane π 1 whose equation is
⋅rr nˆ
= d (Fig 11 19) |
1 | 6020-6023 | Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
A2 = 2, B2 = 2, C2 = – 2
cos θ =
(
)
(
)
2
2
2
2
2
2
3
2
( 6) (2)
(2) ( 2)
3
( 6)
( 2)
2
2
( 2)
×
+ −
+
−
+ −
+ −
+
+ −
=
10
5
5 3
21
7
2 3
7 3
−
=
=
×
Therefore,
θ = cos-1 5 3
21
11 9 Distance of a Point from a Plane
Vector form
Consider a point P with position vector ar and a plane π 1 whose equation is
⋅rr nˆ
= d (Fig 11 19) Consider a plane π2 through P parallel to the plane π1 |
1 | 6021-6024 | 9 Distance of a Point from a Plane
Vector form
Consider a point P with position vector ar and a plane π 1 whose equation is
⋅rr nˆ
= d (Fig 11 19) Consider a plane π2 through P parallel to the plane π1 The unit vector normal to
π2 is nˆ |
1 | 6022-6025 | 19) Consider a plane π2 through P parallel to the plane π1 The unit vector normal to
π2 is nˆ Hence, its equation is
ˆ
(
)
0
r
a
n
−
⋅
=
r
r
i |
1 | 6023-6026 | Consider a plane π2 through P parallel to the plane π1 The unit vector normal to
π2 is nˆ Hence, its equation is
ˆ
(
)
0
r
a
n
−
⋅
=
r
r
i e |
1 | 6024-6027 | The unit vector normal to
π2 is nˆ Hence, its equation is
ˆ
(
)
0
r
a
n
−
⋅
=
r
r
i e ,
⋅rr nˆ
=
a nˆ
⋅r
Thus, the distance ON′ of this plane from the origin is
ˆ
|
a n|
r⋅ |
1 | 6025-6028 | Hence, its equation is
ˆ
(
)
0
r
a
n
−
⋅
=
r
r
i e ,
⋅rr nˆ
=
a nˆ
⋅r
Thus, the distance ON′ of this plane from the origin is
ˆ
|
a n|
r⋅ Therefore, the distance
PQ from the plane π1 is (Fig |
1 | 6026-6029 | e ,
⋅rr nˆ
=
a nˆ
⋅r
Thus, the distance ON′ of this plane from the origin is
ˆ
|
a n|
r⋅ Therefore, the distance
PQ from the plane π1 is (Fig 11 |
1 | 6027-6030 | ,
⋅rr nˆ
=
a nˆ
⋅r
Thus, the distance ON′ of this plane from the origin is
ˆ
|
a n|
r⋅ Therefore, the distance
PQ from the plane π1 is (Fig 11 21 (a))
i |
1 | 6028-6031 | Therefore, the distance
PQ from the plane π1 is (Fig 11 21 (a))
i e |
1 | 6029-6032 | 11 21 (a))
i e ,
ON – ON′ = |d –
a n⋅ˆ |
r
Fig 11 |
1 | 6030-6033 | 21 (a))
i e ,
ON – ON′ = |d –
a n⋅ˆ |
r
Fig 11 19
(a)
a
Z
X
Y
d
N’
P
O
π1
π2
N
Q
π1
(b)
P
X
Z
Y
O
d
N’
N
π2
a
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
491
which is the length of the perpendicular from a point to the given plane |
1 | 6031-6034 | e ,
ON – ON′ = |d –
a n⋅ˆ |
r
Fig 11 19
(a)
a
Z
X
Y
d
N’
P
O
π1
π2
N
Q
π1
(b)
P
X
Z
Y
O
d
N’
N
π2
a
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
491
which is the length of the perpendicular from a point to the given plane We may establish the similar results for (Fig 11 |
1 | 6032-6035 | ,
ON – ON′ = |d –
a n⋅ˆ |
r
Fig 11 19
(a)
a
Z
X
Y
d
N’
P
O
π1
π2
N
Q
π1
(b)
P
X
Z
Y
O
d
N’
N
π2
a
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
491
which is the length of the perpendicular from a point to the given plane We may establish the similar results for (Fig 11 19 (b)) |
1 | 6033-6036 | 19
(a)
a
Z
X
Y
d
N’
P
O
π1
π2
N
Q
π1
(b)
P
X
Z
Y
O
d
N’
N
π2
a
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
491
which is the length of the perpendicular from a point to the given plane We may establish the similar results for (Fig 11 19 (b)) �Note
1 |
1 | 6034-6037 | We may establish the similar results for (Fig 11 19 (b)) �Note
1 If the equation of the plane π2 is in the form
rN
⋅ =d
rur
, where N
ur
is normal
to the plane, then the perpendicular distance is |
N
|
| N |
a
⋅ −d
ur
r
ur |
1 | 6035-6038 | 19 (b)) �Note
1 If the equation of the plane π2 is in the form
rN
⋅ =d
rur
, where N
ur
is normal
to the plane, then the perpendicular distance is |
N
|
| N |
a
⋅ −d
ur
r
ur 2 |
1 | 6036-6039 | �Note
1 If the equation of the plane π2 is in the form
rN
⋅ =d
rur
, where N
ur
is normal
to the plane, then the perpendicular distance is |
N
|
| N |
a
⋅ −d
ur
r
ur 2 The length of the perpendicular from origin O to the plane
rrurN⋅ =
d is |
|
| N |
dur
(since ar = 0) |
1 | 6037-6040 | If the equation of the plane π2 is in the form
rN
⋅ =d
rur
, where N
ur
is normal
to the plane, then the perpendicular distance is |
N
|
| N |
a
⋅ −d
ur
r
ur 2 The length of the perpendicular from origin O to the plane
rrurN⋅ =
d is |
|
| N |
dur
(since ar = 0) Cartesian form
Let P(x1, y1, z1) be the given point with position vector ar and
Ax + By + Cz = D
be the Cartesian equation of the given plane |
1 | 6038-6041 | 2 The length of the perpendicular from origin O to the plane
rrurN⋅ =
d is |
|
| N |
dur
(since ar = 0) Cartesian form
Let P(x1, y1, z1) be the given point with position vector ar and
Ax + By + Cz = D
be the Cartesian equation of the given plane Then
ar =
1
1
1 ˆ
ˆ
ˆ
x i
y
j
z k
+
+
N
ur
=
ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
Hence, from Note 1, the perpendicular from P to the plane is
1
1
1
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) ( A
B
C
)
D
A
B
C
x i
y
j
z k
i
j
k
+
+
⋅
+
+
−
+
+
=
1
1
1
2
2
2
A
B
C
D
A
B
C
x
y
z
+
+
−
+ +
Example 24 Find the distance of a point (2, 5, – 3) from the plane
ˆ
ˆ
ˆ
( 6
3
2
)
r
i
j
k
⋅
−
+
r
= 4
Solution Here,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
3
, N
6
3
2
= + −
= − +
ur
ra
i
j
k
i
j
k and d = 4 |
1 | 6039-6042 | The length of the perpendicular from origin O to the plane
rrurN⋅ =
d is |
|
| N |
dur
(since ar = 0) Cartesian form
Let P(x1, y1, z1) be the given point with position vector ar and
Ax + By + Cz = D
be the Cartesian equation of the given plane Then
ar =
1
1
1 ˆ
ˆ
ˆ
x i
y
j
z k
+
+
N
ur
=
ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
Hence, from Note 1, the perpendicular from P to the plane is
1
1
1
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) ( A
B
C
)
D
A
B
C
x i
y
j
z k
i
j
k
+
+
⋅
+
+
−
+
+
=
1
1
1
2
2
2
A
B
C
D
A
B
C
x
y
z
+
+
−
+ +
Example 24 Find the distance of a point (2, 5, – 3) from the plane
ˆ
ˆ
ˆ
( 6
3
2
)
r
i
j
k
⋅
−
+
r
= 4
Solution Here,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
3
, N
6
3
2
= + −
= − +
ur
ra
i
j
k
i
j
k and d = 4 Therefore, the distance of the point (2, 5, – 3) from the given plane is
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
| (2
5
3 ) (6
3
2
)
4|
ˆ
ˆ
ˆ
| 6
3
2
|
i
j
k
i
j
k
i
j
k
+
−
⋅
−
+
−
−
+
=
| 12
15
6
4 |
713
36
9
4
−
−
−
=
+
+
© NCERT
not to be republished
MATHEMATI CS
492
11 |
1 | 6040-6043 | Cartesian form
Let P(x1, y1, z1) be the given point with position vector ar and
Ax + By + Cz = D
be the Cartesian equation of the given plane Then
ar =
1
1
1 ˆ
ˆ
ˆ
x i
y
j
z k
+
+
N
ur
=
ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
Hence, from Note 1, the perpendicular from P to the plane is
1
1
1
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) ( A
B
C
)
D
A
B
C
x i
y
j
z k
i
j
k
+
+
⋅
+
+
−
+
+
=
1
1
1
2
2
2
A
B
C
D
A
B
C
x
y
z
+
+
−
+ +
Example 24 Find the distance of a point (2, 5, – 3) from the plane
ˆ
ˆ
ˆ
( 6
3
2
)
r
i
j
k
⋅
−
+
r
= 4
Solution Here,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
3
, N
6
3
2
= + −
= − +
ur
ra
i
j
k
i
j
k and d = 4 Therefore, the distance of the point (2, 5, – 3) from the given plane is
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
| (2
5
3 ) (6
3
2
)
4|
ˆ
ˆ
ˆ
| 6
3
2
|
i
j
k
i
j
k
i
j
k
+
−
⋅
−
+
−
−
+
=
| 12
15
6
4 |
713
36
9
4
−
−
−
=
+
+
© NCERT
not to be republished
MATHEMATI CS
492
11 10 Angle between a Line and a Plane
Definition 3 The angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11 |
1 | 6041-6044 | Then
ar =
1
1
1 ˆ
ˆ
ˆ
x i
y
j
z k
+
+
N
ur
=
ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
Hence, from Note 1, the perpendicular from P to the plane is
1
1
1
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) ( A
B
C
)
D
A
B
C
x i
y
j
z k
i
j
k
+
+
⋅
+
+
−
+
+
=
1
1
1
2
2
2
A
B
C
D
A
B
C
x
y
z
+
+
−
+ +
Example 24 Find the distance of a point (2, 5, – 3) from the plane
ˆ
ˆ
ˆ
( 6
3
2
)
r
i
j
k
⋅
−
+
r
= 4
Solution Here,
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
5
3
, N
6
3
2
= + −
= − +
ur
ra
i
j
k
i
j
k and d = 4 Therefore, the distance of the point (2, 5, – 3) from the given plane is
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
| (2
5
3 ) (6
3
2
)
4|
ˆ
ˆ
ˆ
| 6
3
2
|
i
j
k
i
j
k
i
j
k
+
−
⋅
−
+
−
−
+
=
| 12
15
6
4 |
713
36
9
4
−
−
−
=
+
+
© NCERT
not to be republished
MATHEMATI CS
492
11 10 Angle between a Line and a Plane
Definition 3 The angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11 20) |
1 | 6042-6045 | Therefore, the distance of the point (2, 5, – 3) from the given plane is
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
| (2
5
3 ) (6
3
2
)
4|
ˆ
ˆ
ˆ
| 6
3
2
|
i
j
k
i
j
k
i
j
k
+
−
⋅
−
+
−
−
+
=
| 12
15
6
4 |
713
36
9
4
−
−
−
=
+
+
© NCERT
not to be republished
MATHEMATI CS
492
11 10 Angle between a Line and a Plane
Definition 3 The angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11 20) Vector form If the equation of the line is
b
a
r
r
r
r
+λ
=
and the equation of the plane is
r n
d
⋅
r r= |
1 | 6043-6046 | 10 Angle between a Line and a Plane
Definition 3 The angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11 20) Vector form If the equation of the line is
b
a
r
r
r
r
+λ
=
and the equation of the plane is
r n
d
⋅
r r= Then the angle θ between the line and the
normal to the plane is
cos θ =
|
| |
|
bb n
n
⋅
⋅
rr r
r
and so the angle φ between the line and the plane is given by 90 – θ, i |
1 | 6044-6047 | 20) Vector form If the equation of the line is
b
a
r
r
r
r
+λ
=
and the equation of the plane is
r n
d
⋅
r r= Then the angle θ between the line and the
normal to the plane is
cos θ =
|
| |
|
bb n
n
⋅
⋅
rr r
r
and so the angle φ between the line and the plane is given by 90 – θ, i e |
1 | 6045-6048 | Vector form If the equation of the line is
b
a
r
r
r
r
+λ
=
and the equation of the plane is
r n
d
⋅
r r= Then the angle θ between the line and the
normal to the plane is
cos θ =
|
| |
|
bb n
n
⋅
⋅
rr r
r
and so the angle φ between the line and the plane is given by 90 – θ, i e ,
sin (90 – θ) = cos θ
i |
1 | 6046-6049 | Then the angle θ between the line and the
normal to the plane is
cos θ =
|
| |
|
bb n
n
⋅
⋅
rr r
r
and so the angle φ between the line and the plane is given by 90 – θ, i e ,
sin (90 – θ) = cos θ
i e |
1 | 6047-6050 | e ,
sin (90 – θ) = cos θ
i e sin φ =
|
| |
|
bb n
n
⋅
rr r
r
or φ =
sin–1
b n
b n
⋅
Example 25 Find the angle between the line
1
2
x +
=
3
3
6
y
z −
=
and the plane 10 x + 2y – 11 z = 3 |
1 | 6048-6051 | ,
sin (90 – θ) = cos θ
i e sin φ =
|
| |
|
bb n
n
⋅
rr r
r
or φ =
sin–1
b n
b n
⋅
Example 25 Find the angle between the line
1
2
x +
=
3
3
6
y
z −
=
and the plane 10 x + 2y – 11 z = 3 Solution Let θ be the angle between the line and the normal to the plane |
1 | 6049-6052 | e sin φ =
|
| |
|
bb n
n
⋅
rr r
r
or φ =
sin–1
b n
b n
⋅
Example 25 Find the angle between the line
1
2
x +
=
3
3
6
y
z −
=
and the plane 10 x + 2y – 11 z = 3 Solution Let θ be the angle between the line and the normal to the plane Converting the
given equations into vector form, we have
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
( –
3 )
( 2
3
6
)
i
k
i
j
k
+
+ λ
+
+
and
ˆ
ˆ
ˆ
( 10
2
11
)
r
i
j
k
⋅
+
−
r
= 3
Here
b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
and
k
j
i
n
11ˆ
2ˆ
10ˆ
−
+
=
r
sin φ =
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6
) (10
2
11 )
2
3
6
10
2
11
i
j
k
i
j
k
+
+
⋅
+
−
+
+
+
+
=
40
7
15
−
×
=
−218
= 8
21 or φ =
1
8
sin
21
−
Fig 11 |
1 | 6050-6053 | sin φ =
|
| |
|
bb n
n
⋅
rr r
r
or φ =
sin–1
b n
b n
⋅
Example 25 Find the angle between the line
1
2
x +
=
3
3
6
y
z −
=
and the plane 10 x + 2y – 11 z = 3 Solution Let θ be the angle between the line and the normal to the plane Converting the
given equations into vector form, we have
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
( –
3 )
( 2
3
6
)
i
k
i
j
k
+
+ λ
+
+
and
ˆ
ˆ
ˆ
( 10
2
11
)
r
i
j
k
⋅
+
−
r
= 3
Here
b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
and
k
j
i
n
11ˆ
2ˆ
10ˆ
−
+
=
r
sin φ =
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6
) (10
2
11 )
2
3
6
10
2
11
i
j
k
i
j
k
+
+
⋅
+
−
+
+
+
+
=
40
7
15
−
×
=
−218
= 8
21 or φ =
1
8
sin
21
−
Fig 11 20
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
493
EXERCISE 11 |
1 | 6051-6054 | Solution Let θ be the angle between the line and the normal to the plane Converting the
given equations into vector form, we have
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
( –
3 )
( 2
3
6
)
i
k
i
j
k
+
+ λ
+
+
and
ˆ
ˆ
ˆ
( 10
2
11
)
r
i
j
k
⋅
+
−
r
= 3
Here
b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
and
k
j
i
n
11ˆ
2ˆ
10ˆ
−
+
=
r
sin φ =
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6
) (10
2
11 )
2
3
6
10
2
11
i
j
k
i
j
k
+
+
⋅
+
−
+
+
+
+
=
40
7
15
−
×
=
−218
= 8
21 or φ =
1
8
sin
21
−
Fig 11 20
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
493
EXERCISE 11 3
1 |
1 | 6052-6055 | Converting the
given equations into vector form, we have
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
( –
3 )
( 2
3
6
)
i
k
i
j
k
+
+ λ
+
+
and
ˆ
ˆ
ˆ
( 10
2
11
)
r
i
j
k
⋅
+
−
r
= 3
Here
b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
and
k
j
i
n
11ˆ
2ˆ
10ˆ
−
+
=
r
sin φ =
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6
) (10
2
11 )
2
3
6
10
2
11
i
j
k
i
j
k
+
+
⋅
+
−
+
+
+
+
=
40
7
15
−
×
=
−218
= 8
21 or φ =
1
8
sin
21
−
Fig 11 20
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
493
EXERCISE 11 3
1 In each of the following cases, determine the direction cosines of the normal to
the plane and the distance from the origin |
1 | 6053-6056 | 20
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
493
EXERCISE 11 3
1 In each of the following cases, determine the direction cosines of the normal to
the plane and the distance from the origin (a)
z = 2
(b)
x + y + z = 1
(c)
2x + 3y – z = 5
(d)
5y + 8 = 0
2 |
1 | 6054-6057 | 3
1 In each of the following cases, determine the direction cosines of the normal to
the plane and the distance from the origin (a)
z = 2
(b)
x + y + z = 1
(c)
2x + 3y – z = 5
(d)
5y + 8 = 0
2 Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector
k
j
i
6ˆ
5ˆ
3ˆ
−
+ |
1 | 6055-6058 | In each of the following cases, determine the direction cosines of the normal to
the plane and the distance from the origin (a)
z = 2
(b)
x + y + z = 1
(c)
2x + 3y – z = 5
(d)
5y + 8 = 0
2 Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector
k
j
i
6ˆ
5ˆ
3ˆ
−
+ 3 |
1 | 6056-6059 | (a)
z = 2
(b)
x + y + z = 1
(c)
2x + 3y – z = 5
(d)
5y + 8 = 0
2 Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector
k
j
i
6ˆ
5ˆ
3ˆ
−
+ 3 Find the Cartesian equation of the following planes:
(a)
ˆ
ˆ
ˆ
(
)
2
r
i
j
k
⋅
+
−
=
r
(b)
ˆ
ˆ
ˆ
(2
3
4 )
1
r
i
j
k
⋅
+
−
=
r
(c)
ˆ
ˆ
ˆ
[(
2 )
(3
)
(2
)
]
15
r
s
t i
t
j
s
t k
⋅
−
+
−
+
+
=
r
4 |
1 | 6057-6060 | Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector
k
j
i
6ˆ
5ˆ
3ˆ
−
+ 3 Find the Cartesian equation of the following planes:
(a)
ˆ
ˆ
ˆ
(
)
2
r
i
j
k
⋅
+
−
=
r
(b)
ˆ
ˆ
ˆ
(2
3
4 )
1
r
i
j
k
⋅
+
−
=
r
(c)
ˆ
ˆ
ˆ
[(
2 )
(3
)
(2
)
]
15
r
s
t i
t
j
s
t k
⋅
−
+
−
+
+
=
r
4 In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin |
1 | 6058-6061 | 3 Find the Cartesian equation of the following planes:
(a)
ˆ
ˆ
ˆ
(
)
2
r
i
j
k
⋅
+
−
=
r
(b)
ˆ
ˆ
ˆ
(2
3
4 )
1
r
i
j
k
⋅
+
−
=
r
(c)
ˆ
ˆ
ˆ
[(
2 )
(3
)
(2
)
]
15
r
s
t i
t
j
s
t k
⋅
−
+
−
+
+
=
r
4 In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin (a)
2x + 3y + 4z – 12 = 0
(b)
3y + 4z – 6 = 0
(c)
x + y + z = 1
(d)
5y + 8 = 0
5 |
1 | 6059-6062 | Find the Cartesian equation of the following planes:
(a)
ˆ
ˆ
ˆ
(
)
2
r
i
j
k
⋅
+
−
=
r
(b)
ˆ
ˆ
ˆ
(2
3
4 )
1
r
i
j
k
⋅
+
−
=
r
(c)
ˆ
ˆ
ˆ
[(
2 )
(3
)
(2
)
]
15
r
s
t i
t
j
s
t k
⋅
−
+
−
+
+
=
r
4 In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin (a)
2x + 3y + 4z – 12 = 0
(b)
3y + 4z – 6 = 0
(c)
x + y + z = 1
(d)
5y + 8 = 0
5 Find the vector and cartesian equations of the planes
(a)
that passes through the point (1, 0, – 2) and the normal to the plane is
ˆ
ˆ
ˆ |
1 | 6060-6063 | In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin (a)
2x + 3y + 4z – 12 = 0
(b)
3y + 4z – 6 = 0
(c)
x + y + z = 1
(d)
5y + 8 = 0
5 Find the vector and cartesian equations of the planes
(a)
that passes through the point (1, 0, – 2) and the normal to the plane is
ˆ
ˆ
ˆ i
j
k
+ −
(b)
that passes through the point (1,4, 6) and the normal vector to the plane is
ˆ
ˆ
2ˆ |
1 | 6061-6064 | (a)
2x + 3y + 4z – 12 = 0
(b)
3y + 4z – 6 = 0
(c)
x + y + z = 1
(d)
5y + 8 = 0
5 Find the vector and cartesian equations of the planes
(a)
that passes through the point (1, 0, – 2) and the normal to the plane is
ˆ
ˆ
ˆ i
j
k
+ −
(b)
that passes through the point (1,4, 6) and the normal vector to the plane is
ˆ
ˆ
2ˆ i
j
k
− +
6 |
1 | 6062-6065 | Find the vector and cartesian equations of the planes
(a)
that passes through the point (1, 0, – 2) and the normal to the plane is
ˆ
ˆ
ˆ i
j
k
+ −
(b)
that passes through the point (1,4, 6) and the normal vector to the plane is
ˆ
ˆ
2ˆ i
j
k
− +
6 Find the equations of the planes that passes through three points |
1 | 6063-6066 | i
j
k
+ −
(b)
that passes through the point (1,4, 6) and the normal vector to the plane is
ˆ
ˆ
2ˆ i
j
k
− +
6 Find the equations of the planes that passes through three points (a)
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b)
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7 |
1 | 6064-6067 | i
j
k
− +
6 Find the equations of the planes that passes through three points (a)
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b)
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7 Find the intercepts cut off by the plane 2x + y – z = 5 |
1 | 6065-6068 | Find the equations of the planes that passes through three points (a)
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b)
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7 Find the intercepts cut off by the plane 2x + y – z = 5 8 |
1 | 6066-6069 | (a)
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b)
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7 Find the intercepts cut off by the plane 2x + y – z = 5 8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane |
1 | 6067-6070 | Find the intercepts cut off by the plane 2x + y – z = 5 8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane 9 |
1 | 6068-6071 | 8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane 9 Find the equation of t he plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1) |
1 | 6069-6072 | Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane 9 Find the equation of t he plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1) 10 |
1 | 6070-6073 | 9 Find the equation of t he plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1) 10 Find the vector equation of the plane passing through the intersection of the
planes
ˆ
ˆ
ˆ |
1 | 6071-6074 | Find the equation of t he plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1) 10 Find the vector equation of the plane passing through the intersection of the
planes
ˆ
ˆ
ˆ (2
2
3
)
7
r
i
j
k
+
−
=
r
,
ˆ
ˆ
ˆ |
1 | 6072-6075 | 10 Find the vector equation of the plane passing through the intersection of the
planes
ˆ
ˆ
ˆ (2
2
3
)
7
r
i
j
k
+
−
=
r
,
ˆ
ˆ
ˆ (2
5
3
)
9
r
i
j
k
+
+
=
r
and through the point
(2, 1, 3) |
1 | 6073-6076 | Find the vector equation of the plane passing through the intersection of the
planes
ˆ
ˆ
ˆ (2
2
3
)
7
r
i
j
k
+
−
=
r
,
ˆ
ˆ
ˆ (2
5
3
)
9
r
i
j
k
+
+
=
r
and through the point
(2, 1, 3) 11 |
1 | 6074-6077 | (2
2
3
)
7
r
i
j
k
+
−
=
r
,
ˆ
ˆ
ˆ (2
5
3
)
9
r
i
j
k
+
+
=
r
and through the point
(2, 1, 3) 11 Find the equation of the plane through the line of intersection of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0 |
1 | 6075-6078 | (2
5
3
)
9
r
i
j
k
+
+
=
r
and through the point
(2, 1, 3) 11 Find the equation of the plane through the line of intersection of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0 © NCERT
not to be republished
MATHEMATI CS
494
12 |
1 | 6076-6079 | 11 Find the equation of the plane through the line of intersection of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0 © NCERT
not to be republished
MATHEMATI CS
494
12 Find the angle between the planes whose vector equations are
ˆ
ˆ
ˆ
(2
2
3 )
5
r
i
j
k
⋅
+
−
=
r
and
ˆ
ˆ
ˆ
(3
3
5 )
3
r
i
j
k
⋅
−
+
=
r |
1 | 6077-6080 | Find the equation of the plane through the line of intersection of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0 © NCERT
not to be republished
MATHEMATI CS
494
12 Find the angle between the planes whose vector equations are
ˆ
ˆ
ˆ
(2
2
3 )
5
r
i
j
k
⋅
+
−
=
r
and
ˆ
ˆ
ˆ
(3
3
5 )
3
r
i
j
k
⋅
−
+
=
r 13 |
1 | 6078-6081 | © NCERT
not to be republished
MATHEMATI CS
494
12 Find the angle between the planes whose vector equations are
ˆ
ˆ
ˆ
(2
2
3 )
5
r
i
j
k
⋅
+
−
=
r
and
ˆ
ˆ
ˆ
(3
3
5 )
3
r
i
j
k
⋅
−
+
=
r 13 In the following cases, determine whether the given planes are parallel or
perpendicular, and in case they are neither, find the angles between them |
1 | 6079-6082 | Find the angle between the planes whose vector equations are
ˆ
ˆ
ˆ
(2
2
3 )
5
r
i
j
k
⋅
+
−
=
r
and
ˆ
ˆ
ˆ
(3
3
5 )
3
r
i
j
k
⋅
−
+
=
r 13 In the following cases, determine whether the given planes are parallel or
perpendicular, and in case they are neither, find the angles between them (a)
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b)
2x + y + 3z – 2 = 0
and x – 2y + 5 = 0
(c)
2x – 2y + 4z + 5 = 0
and 3x – 3y + 6z – 1 = 0
(d)
2x – y + 3z – 1 = 0
and 2x – y + 3z + 3 = 0
(e)
4x + 8y + z – 8 = 0
and y + z – 4 = 0
14 |
1 | 6080-6083 | 13 In the following cases, determine whether the given planes are parallel or
perpendicular, and in case they are neither, find the angles between them (a)
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b)
2x + y + 3z – 2 = 0
and x – 2y + 5 = 0
(c)
2x – 2y + 4z + 5 = 0
and 3x – 3y + 6z – 1 = 0
(d)
2x – y + 3z – 1 = 0
and 2x – y + 3z + 3 = 0
(e)
4x + 8y + z – 8 = 0
and y + z – 4 = 0
14 In the following cases, find the distance of each of the given points from the
corresponding given plane |
1 | 6081-6084 | In the following cases, determine whether the given planes are parallel or
perpendicular, and in case they are neither, find the angles between them (a)
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b)
2x + y + 3z – 2 = 0
and x – 2y + 5 = 0
(c)
2x – 2y + 4z + 5 = 0
and 3x – 3y + 6z – 1 = 0
(d)
2x – y + 3z – 1 = 0
and 2x – y + 3z + 3 = 0
(e)
4x + 8y + z – 8 = 0
and y + z – 4 = 0
14 In the following cases, find the distance of each of the given points from the
corresponding given plane Point
Plane
(a)
(0, 0, 0)
3x – 4y + 12 z = 3
(b)
(3, – 2, 1)
2x – y + 2z + 3 = 0
(c)
(2, 3, – 5)
x + 2y – 2z = 9
(d)
(– 6, 0, 0)
2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that
cos2 α + cos2 β + cos2 γ + cos2 δ = 4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height |
1 | 6082-6085 | (a)
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b)
2x + y + 3z – 2 = 0
and x – 2y + 5 = 0
(c)
2x – 2y + 4z + 5 = 0
and 3x – 3y + 6z – 1 = 0
(d)
2x – y + 3z – 1 = 0
and 2x – y + 3z + 3 = 0
(e)
4x + 8y + z – 8 = 0
and y + z – 4 = 0
14 In the following cases, find the distance of each of the given points from the
corresponding given plane Point
Plane
(a)
(0, 0, 0)
3x – 4y + 12 z = 3
(b)
(3, – 2, 1)
2x – y + 2z + 3 = 0
(c)
(2, 3, – 5)
x + 2y – 2z = 9
(d)
(– 6, 0, 0)
2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that
cos2 α + cos2 β + cos2 γ + cos2 δ = 4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height Let OADBFEGC be the cube with each side of length a units |
1 | 6083-6086 | In the following cases, find the distance of each of the given points from the
corresponding given plane Point
Plane
(a)
(0, 0, 0)
3x – 4y + 12 z = 3
(b)
(3, – 2, 1)
2x – y + 2z + 3 = 0
(c)
(2, 3, – 5)
x + 2y – 2z = 9
(d)
(– 6, 0, 0)
2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that
cos2 α + cos2 β + cos2 γ + cos2 δ = 4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height Let OADBFEGC be the cube with each side of length a units (Fig 11 |
1 | 6084-6087 | Point
Plane
(a)
(0, 0, 0)
3x – 4y + 12 z = 3
(b)
(3, – 2, 1)
2x – y + 2z + 3 = 0
(c)
(2, 3, – 5)
x + 2y – 2z = 9
(d)
(– 6, 0, 0)
2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that
cos2 α + cos2 β + cos2 γ + cos2 δ = 4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height Let OADBFEGC be the cube with each side of length a units (Fig 11 21)
The four diagonals are OE, AF, BG and CD |
1 | 6085-6088 | Let OADBFEGC be the cube with each side of length a units (Fig 11 21)
The four diagonals are OE, AF, BG and CD The direction cosines of the diagonal OE which
is the line joining two points O and E are
2
2
2
2
2
2
2
2
2
0
0
0
,
,
a
a
a
a
a
a
a
a
a
a
a
a
−
−
−
+ +
+ +
+ +
i |
1 | 6086-6089 | (Fig 11 21)
The four diagonals are OE, AF, BG and CD The direction cosines of the diagonal OE which
is the line joining two points O and E are
2
2
2
2
2
2
2
2
2
0
0
0
,
,
a
a
a
a
a
a
a
a
a
a
a
a
−
−
−
+ +
+ +
+ +
i e |
1 | 6087-6090 | 21)
The four diagonals are OE, AF, BG and CD The direction cosines of the diagonal OE which
is the line joining two points O and E are
2
2
2
2
2
2
2
2
2
0
0
0
,
,
a
a
a
a
a
a
a
a
a
a
a
a
−
−
−
+ +
+ +
+ +
i e ,
3
1
,
3
1
,
3
1
B(0, , 0)
a
C(0, 0, )
a
a( , 0, )G
a
F(0, , )
a a
X
D( , , 0)
a
a
Y
Z
O
Fig 11 |
1 | 6088-6091 | The direction cosines of the diagonal OE which
is the line joining two points O and E are
2
2
2
2
2
2
2
2
2
0
0
0
,
,
a
a
a
a
a
a
a
a
a
a
a
a
−
−
−
+ +
+ +
+ +
i e ,
3
1
,
3
1
,
3
1
B(0, , 0)
a
C(0, 0, )
a
a( , 0, )G
a
F(0, , )
a a
X
D( , , 0)
a
a
Y
Z
O
Fig 11 21
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
495
Similarly, the direction cosines of AF, BG and CD are –1
3
,
3
1 ,
3
1 ;
3
1 ,
–1
3
,
3
1 and
3
1 ,
3
1 , –1
3
, respectively |
1 | 6089-6092 | e ,
3
1
,
3
1
,
3
1
B(0, , 0)
a
C(0, 0, )
a
a( , 0, )G
a
F(0, , )
a a
X
D( , , 0)
a
a
Y
Z
O
Fig 11 21
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
495
Similarly, the direction cosines of AF, BG and CD are –1
3
,
3
1 ,
3
1 ;
3
1 ,
–1
3
,
3
1 and
3
1 ,
3
1 , –1
3
, respectively Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ
with OE, AF, BG, CD, respectively |
1 | 6090-6093 | ,
3
1
,
3
1
,
3
1
B(0, , 0)
a
C(0, 0, )
a
a( , 0, )G
a
F(0, , )
a a
X
D( , , 0)
a
a
Y
Z
O
Fig 11 21
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
495
Similarly, the direction cosines of AF, BG and CD are –1
3
,
3
1 ,
3
1 ;
3
1 ,
–1
3
,
3
1 and
3
1 ,
3
1 , –1
3
, respectively Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ
with OE, AF, BG, CD, respectively Then
cosα = 1
3
(l + m+ n); cos β =
1
3 (– l + m + n);
cosγ =
1
3
(l – m + n); cos δ =
1
3 (l + m – n) (Why |
1 | 6091-6094 | 21
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
495
Similarly, the direction cosines of AF, BG and CD are –1
3
,
3
1 ,
3
1 ;
3
1 ,
–1
3
,
3
1 and
3
1 ,
3
1 , –1
3
, respectively Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ
with OE, AF, BG, CD, respectively Then
cosα = 1
3
(l + m+ n); cos β =
1
3 (– l + m + n);
cosγ =
1
3
(l – m + n); cos δ =
1
3 (l + m – n) (Why )
Squaring and adding, we get
cos2α + cos2 β + cos2 γ + cos2 δ
= 1
3 [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2]
= 1
3 [ 4 (l2 + m 2 + n2 ) ] = 3
4
(as l2 + m 2 + n2 = 1)
Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is
perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8 |
1 | 6092-6095 | Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ
with OE, AF, BG, CD, respectively Then
cosα = 1
3
(l + m+ n); cos β =
1
3 (– l + m + n);
cosγ =
1
3
(l – m + n); cos δ =
1
3 (l + m – n) (Why )
Squaring and adding, we get
cos2α + cos2 β + cos2 γ + cos2 δ
= 1
3 [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2]
= 1
3 [ 4 (l2 + m 2 + n2 ) ] = 3
4
(as l2 + m 2 + n2 = 1)
Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is
perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8 Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
Applying the condition of perpendicularly to the plane given in (1) with the planes |
1 | 6093-6096 | Then
cosα = 1
3
(l + m+ n); cos β =
1
3 (– l + m + n);
cosγ =
1
3
(l – m + n); cos δ =
1
3 (l + m – n) (Why )
Squaring and adding, we get
cos2α + cos2 β + cos2 γ + cos2 δ
= 1
3 [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2]
= 1
3 [ 4 (l2 + m 2 + n2 ) ] = 3
4
(as l2 + m 2 + n2 = 1)
Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is
perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8 Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
Applying the condition of perpendicularly to the plane given in (1) with the planes (1)
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C |
1 | 6094-6097 | )
Squaring and adding, we get
cos2α + cos2 β + cos2 γ + cos2 δ
= 1
3 [ (l + m + n )2 + (–l + m + n)2 ] + (l – m + n)2 + (l + m –n)2]
= 1
3 [ 4 (l2 + m 2 + n2 ) ] = 3
4
(as l2 + m 2 + n2 = 1)
Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is
perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8 Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
Applying the condition of perpendicularly to the plane given in (1) with the planes (1)
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i |
1 | 6095-6098 | Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
Applying the condition of perpendicularly to the plane given in (1) with the planes (1)
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i e |
1 | 6096-6099 | (1)
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i e 5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6) |
1 | 6097-6100 | Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i e 5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6) Solution Let A, B, C be the three points in the plane |
1 | 6098-6101 | e 5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6) Solution Let A, B, C be the three points in the plane D is the foot of the perpendicular
drawn from a point P to the plane |
1 | 6099-6102 | 5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6) Solution Let A, B, C be the three points in the plane D is the foot of the perpendicular
drawn from a point P to the plane PD is the required distance to be determined, which
is the projection of AP
uuur
on AB
AC
×
uuur
uuur |
1 | 6100-6103 | Solution Let A, B, C be the three points in the plane D is the foot of the perpendicular
drawn from a point P to the plane PD is the required distance to be determined, which
is the projection of AP
uuur
on AB
AC
×
uuur
uuur © NCERT
not to be republished
MATHEMATI CS
496
Hence, PD = the dot product of AP
uuur
with the unit vector along AB
AC
×
uuur
uuur |
1 | 6101-6104 | D is the foot of the perpendicular
drawn from a point P to the plane PD is the required distance to be determined, which
is the projection of AP
uuur
on AB
AC
×
uuur
uuur © NCERT
not to be republished
MATHEMATI CS
496
Hence, PD = the dot product of AP
uuur
with the unit vector along AB
AC
×
uuur
uuur So
AP
uuur
= 3
k
j
i
7ˆ
6ˆ
ˆ
+
+
and
AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
2
12
16
12
4
0
4
i
j
k
i
j
k
=
−
+
−
Unit vector along AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
Hence
PD = (
)ˆ
7
6ˆ
3ˆ
k
j
i
+
+ |
1 | 6102-6105 | PD is the required distance to be determined, which
is the projection of AP
uuur
on AB
AC
×
uuur
uuur © NCERT
not to be republished
MATHEMATI CS
496
Hence, PD = the dot product of AP
uuur
with the unit vector along AB
AC
×
uuur
uuur So
AP
uuur
= 3
k
j
i
7ˆ
6ˆ
ˆ
+
+
and
AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
2
12
16
12
4
0
4
i
j
k
i
j
k
=
−
+
−
Unit vector along AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
Hence
PD = (
)ˆ
7
6ˆ
3ˆ
k
j
i
+
+ ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
=
17
34
3
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane |
1 | 6103-6106 | © NCERT
not to be republished
MATHEMATI CS
496
Hence, PD = the dot product of AP
uuur
with the unit vector along AB
AC
×
uuur
uuur So
AP
uuur
= 3
k
j
i
7ˆ
6ˆ
ˆ
+
+
and
AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
2
12
16
12
4
0
4
i
j
k
i
j
k
=
−
+
−
Unit vector along AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
Hence
PD = (
)ˆ
7
6ˆ
3ˆ
k
j
i
+
+ ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
=
17
34
3
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane Example 29 Show that the lines
x
a
d
−
+
α − δ
= y
a
z
a
d
−
−
−
=
α
α + δ
and
x
b
c
−
+
β − γ
= y
b
z
b
c
−
−
−
=
β
β + γ
are coplanar |
1 | 6104-6107 | So
AP
uuur
= 3
k
j
i
7ˆ
6ˆ
ˆ
+
+
and
AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
2
12
16
12
4
0
4
i
j
k
i
j
k
=
−
+
−
Unit vector along AB
AC
×
uuur
uuur
=
ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
Hence
PD = (
)ˆ
7
6ˆ
3ˆ
k
j
i
+
+ ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
=
17
34
3
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane Example 29 Show that the lines
x
a
d
−
+
α − δ
= y
a
z
a
d
−
−
−
=
α
α + δ
and
x
b
c
−
+
β − γ
= y
b
z
b
c
−
−
−
=
β
β + γ
are coplanar Solution
Here
x1 = a – d
x2 = b – c
y1 = a
y2 = b
z1 = a + d
z2 = b + c
a1 = α – δ
a2 = β – γ
b1 = α
b2 = β
c1 = α + δ
c2 = β + γ
Now consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
x
x
y
y
z
z
a
b
c
a
b
c
−
−
−
=
b
c
a
d
b
a
b
c
a
d
− −
+
−
+
−
−
α − δ
α
α + δ
β − γ
β
β + γ
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
497
Adding third column to the first column, we get
2
b
a
b
a
b
c
a
d
−
−
+
−
−
α
α
α + δ
β
β
β + γ
= 0
Since the first and second columns are identical |
1 | 6105-6108 | ˆ
ˆ
ˆ
3
4
3
34
i
j
k
−
+
=
17
34
3
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane Example 29 Show that the lines
x
a
d
−
+
α − δ
= y
a
z
a
d
−
−
−
=
α
α + δ
and
x
b
c
−
+
β − γ
= y
b
z
b
c
−
−
−
=
β
β + γ
are coplanar Solution
Here
x1 = a – d
x2 = b – c
y1 = a
y2 = b
z1 = a + d
z2 = b + c
a1 = α – δ
a2 = β – γ
b1 = α
b2 = β
c1 = α + δ
c2 = β + γ
Now consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
x
x
y
y
z
z
a
b
c
a
b
c
−
−
−
=
b
c
a
d
b
a
b
c
a
d
− −
+
−
+
−
−
α − δ
α
α + δ
β − γ
β
β + γ
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
497
Adding third column to the first column, we get
2
b
a
b
a
b
c
a
d
−
−
+
−
−
α
α
α + δ
β
β
β + γ
= 0
Since the first and second columns are identical Hence, the given two lines are
coplanar |
1 | 6106-6109 | Example 29 Show that the lines
x
a
d
−
+
α − δ
= y
a
z
a
d
−
−
−
=
α
α + δ
and
x
b
c
−
+
β − γ
= y
b
z
b
c
−
−
−
=
β
β + γ
are coplanar Solution
Here
x1 = a – d
x2 = b – c
y1 = a
y2 = b
z1 = a + d
z2 = b + c
a1 = α – δ
a2 = β – γ
b1 = α
b2 = β
c1 = α + δ
c2 = β + γ
Now consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
x
x
y
y
z
z
a
b
c
a
b
c
−
−
−
=
b
c
a
d
b
a
b
c
a
d
− −
+
−
+
−
−
α − δ
α
α + δ
β − γ
β
β + γ
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
497
Adding third column to the first column, we get
2
b
a
b
a
b
c
a
d
−
−
+
−
−
α
α
α + δ
β
β
β + γ
= 0
Since the first and second columns are identical Hence, the given two lines are
coplanar Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane |
1 | 6107-6110 | Solution
Here
x1 = a – d
x2 = b – c
y1 = a
y2 = b
z1 = a + d
z2 = b + c
a1 = α – δ
a2 = β – γ
b1 = α
b2 = β
c1 = α + δ
c2 = β + γ
Now consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
x
x
y
y
z
z
a
b
c
a
b
c
−
−
−
=
b
c
a
d
b
a
b
c
a
d
− −
+
−
+
−
−
α − δ
α
α + δ
β − γ
β
β + γ
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
497
Adding third column to the first column, we get
2
b
a
b
a
b
c
a
d
−
−
+
−
−
α
α
α + δ
β
β
β + γ
= 0
Since the first and second columns are identical Hence, the given two lines are
coplanar Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane Solution The vector equation of the line through the points A and B is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
[ (5
3)
(1
4)
(6
1)
]
i
j
k
i
j
k
+
+
+ λ
−
+
−
+
−
i |
1 | 6108-6111 | Hence, the given two lines are
coplanar Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane Solution The vector equation of the line through the points A and B is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
[ (5
3)
(1
4)
(6
1)
]
i
j
k
i
j
k
+
+
+ λ
−
+
−
+
−
i e |
1 | 6109-6112 | Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane Solution The vector equation of the line through the points A and B is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
[ (5
3)
(1
4)
(6
1)
]
i
j
k
i
j
k
+
+
+ λ
−
+
−
+
−
i e rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
( 2
3
5
)
i
j
k
i
j
k
+
+
+ λ
−
+ |
1 | 6110-6113 | Solution The vector equation of the line through the points A and B is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
[ (5
3)
(1
4)
(6
1)
]
i
j
k
i
j
k
+
+
+ λ
−
+
−
+
−
i e rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
( 2
3
5
)
i
j
k
i
j
k
+
+
+ λ
−
+ (1)
Let P be the point where the line AB crosses the XY-plane |
1 | 6111-6114 | e rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
( 2
3
5
)
i
j
k
i
j
k
+
+
+ λ
−
+ (1)
Let P be the point where the line AB crosses the XY-plane Then the position
vector of the point P is of the form
yj
xi
ˆ
ˆ + |
1 | 6112-6115 | rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
4
( 2
3
5
)
i
j
k
i
j
k
+
+
+ λ
−
+ (1)
Let P be the point where the line AB crosses the XY-plane Then the position
vector of the point P is of the form
yj
xi
ˆ
ˆ + This point must satisfy the equation (1) |
1 | 6113-6116 | (1)
Let P be the point where the line AB crosses the XY-plane Then the position
vector of the point P is of the form
yj
xi
ˆ
ˆ + This point must satisfy the equation (1) (Why |
1 | 6114-6117 | Then the position
vector of the point P is of the form
yj
xi
ˆ
ˆ + This point must satisfy the equation (1) (Why )
i |
1 | 6115-6118 | This point must satisfy the equation (1) (Why )
i e |
1 | 6116-6119 | (Why )
i e ˆ
ˆ
x i
+y j
=
ˆ
ˆ
ˆ
(3
2
)
( 4
3 )
(1
5
)
i
j
k
+
λ
+
−
λ
+
+
λ
Equating the like coefficients of
ˆ
ˆ
,ˆ
and
i
j
k , we have
x = 3 + 2 λ
y = 4 – 3 λ
0 = 1 + 5 λ
Solving the above equations, we get
x = 13
23
and
5
5
y =
Hence, the coordinates of the required point are
5,0
135,23 |
1 | 6117-6120 | )
i e ˆ
ˆ
x i
+y j
=
ˆ
ˆ
ˆ
(3
2
)
( 4
3 )
(1
5
)
i
j
k
+
λ
+
−
λ
+
+
λ
Equating the like coefficients of
ˆ
ˆ
,ˆ
and
i
j
k , we have
x = 3 + 2 λ
y = 4 – 3 λ
0 = 1 + 5 λ
Solving the above equations, we get
x = 13
23
and
5
5
y =
Hence, the coordinates of the required point are
5,0
135,23 Miscellaneous Exercise on Chapter 11
1 |
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