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1 | 5918-5921 | 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 e |
1 | 5919-5922 | 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 e A (x – x1) + B (y – y1) + C (z – z1) = 0
Example 17 Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1 |
1 | 5920-5923 | 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 e A (x – x1) + B (y – y1) + C (z – z1) = 0
Example 17 Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1 Fig 11 |
1 | 5921-5924 | e A (x – x1) + B (y – y1) + C (z – z1) = 0
Example 17 Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1 Fig 11 12
Fig 11 |
1 | 5922-5925 | A (x – x1) + B (y – y1) + C (z – z1) = 0
Example 17 Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1 Fig 11 12
Fig 11 13
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
483
Y
Z
O
a
r
R
P
bS
c
(RS RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as
ˆ
ˆ
ˆ
5
2
4
a
i
j
k
=
+
−
r
and the
normal vector N
r
perpendicular to the plane as
ˆ
ˆ
N =2 + 3ˆ
i
j
k−
r
Therefore, the vector equation of the plane is given by (
) |
1 | 5923-5926 | Fig 11 12
Fig 11 13
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
483
Y
Z
O
a
r
R
P
bS
c
(RS RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as
ˆ
ˆ
ˆ
5
2
4
a
i
j
k
=
+
−
r
and the
normal vector N
r
perpendicular to the plane as
ˆ
ˆ
N =2 + 3ˆ
i
j
k−
r
Therefore, the vector equation of the plane is given by (
) N
0
r
a−
=
r
r
r
or
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(5
2
4 )] (2
3
)
0
r
i
j
k
i
j
k
−
+
−
⋅
+
−
=
r |
1 | 5924-5927 | 12
Fig 11 13
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
483
Y
Z
O
a
r
R
P
bS
c
(RS RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as
ˆ
ˆ
ˆ
5
2
4
a
i
j
k
=
+
−
r
and the
normal vector N
r
perpendicular to the plane as
ˆ
ˆ
N =2 + 3ˆ
i
j
k−
r
Therefore, the vector equation of the plane is given by (
) N
0
r
a−
=
r
r
r
or
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(5
2
4 )] (2
3
)
0
r
i
j
k
i
j
k
−
+
−
⋅
+
−
=
r (1)
Transforming (1) into Cartesian form, we have
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[( –5)
(
2)
(
4) ] (2
3
)
0
x
i
y
j
z
k
i
j
k
+
−
+
+
⋅
+
−
=
or
2(
5) 3(
2) 1(
4) 0
x
y
z
− + − − + =
i |
1 | 5925-5928 | 13
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
483
Y
Z
O
a
r
R
P
bS
c
(RS RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as
ˆ
ˆ
ˆ
5
2
4
a
i
j
k
=
+
−
r
and the
normal vector N
r
perpendicular to the plane as
ˆ
ˆ
N =2 + 3ˆ
i
j
k−
r
Therefore, the vector equation of the plane is given by (
) N
0
r
a−
=
r
r
r
or
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(5
2
4 )] (2
3
)
0
r
i
j
k
i
j
k
−
+
−
⋅
+
−
=
r (1)
Transforming (1) into Cartesian form, we have
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[( –5)
(
2)
(
4) ] (2
3
)
0
x
i
y
j
z
k
i
j
k
+
−
+
+
⋅
+
−
=
or
2(
5) 3(
2) 1(
4) 0
x
y
z
− + − − + =
i e |
1 | 5926-5929 | N
0
r
a−
=
r
r
r
or
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(5
2
4 )] (2
3
)
0
r
i
j
k
i
j
k
−
+
−
⋅
+
−
=
r (1)
Transforming (1) into Cartesian form, we have
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[( –5)
(
2)
(
4) ] (2
3
)
0
x
i
y
j
z
k
i
j
k
+
−
+
+
⋅
+
−
=
or
2(
5) 3(
2) 1(
4) 0
x
y
z
− + − − + =
i e 2x + 3y – z = 20
which is the cartesian equation of the plane |
1 | 5927-5930 | (1)
Transforming (1) into Cartesian form, we have
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[( –5)
(
2)
(
4) ] (2
3
)
0
x
i
y
j
z
k
i
j
k
+
−
+
+
⋅
+
−
=
or
2(
5) 3(
2) 1(
4) 0
x
y
z
− + − − + =
i e 2x + 3y – z = 20
which is the cartesian equation of the plane 11 |
1 | 5928-5931 | e 2x + 3y – z = 20
which is the cartesian equation of the plane 11 6 |
1 | 5929-5932 | 2x + 3y – z = 20
which is the cartesian equation of the plane 11 6 3 Equation of a plane passing through three non collinear points
Let R, S and T be three non collinear points on the plane with position vectors ar ,b
r
and
crrespectively (Fig 11 |
1 | 5930-5933 | 11 6 3 Equation of a plane passing through three non collinear points
Let R, S and T be three non collinear points on the plane with position vectors ar ,b
r
and
crrespectively (Fig 11 14) |
1 | 5931-5934 | 6 3 Equation of a plane passing through three non collinear points
Let R, S and T be three non collinear points on the plane with position vectors ar ,b
r
and
crrespectively (Fig 11 14) Fig 11 |
1 | 5932-5935 | 3 Equation of a plane passing through three non collinear points
Let R, S and T be three non collinear points on the plane with position vectors ar ,b
r
and
crrespectively (Fig 11 14) Fig 11 14
The vectors RS
uuur
and RT
uuur
are in the given plane |
1 | 5933-5936 | 14) Fig 11 14
The vectors RS
uuur
and RT
uuur
are in the given plane Therefore, the vector RS
RT
×
uuur
uuur
is perpendicular to the plane containing points R, S and T |
1 | 5934-5937 | Fig 11 14
The vectors RS
uuur
and RT
uuur
are in the given plane Therefore, the vector RS
RT
×
uuur
uuur
is perpendicular to the plane containing points R, S and T Let rr be the position vector
of any point P in the plane |
1 | 5935-5938 | 14
The vectors RS
uuur
and RT
uuur
are in the given plane Therefore, the vector RS
RT
×
uuur
uuur
is perpendicular to the plane containing points R, S and T Let rr be the position vector
of any point P in the plane Therefore, the equation of the plane passing through R and
perpendicular to the vector RS
RT
×
uuur
uuur
is
(
) (RS
RT)
r
−a
⋅
×
uuur
uuur
r
r
= 0
or
r
r
r
r
r
r
(
) |
1 | 5936-5939 | Therefore, the vector RS
RT
×
uuur
uuur
is perpendicular to the plane containing points R, S and T Let rr be the position vector
of any point P in the plane Therefore, the equation of the plane passing through R and
perpendicular to the vector RS
RT
×
uuur
uuur
is
(
) (RS
RT)
r
−a
⋅
×
uuur
uuur
r
r
= 0
or
r
r
r
r
r
r
(
) [(
)×(
)]
r –a
b –a
c –a
= 0 … (1)
© NCERT
not to be republished
MATHEMATI CS
484
Fig 11 |
1 | 5937-5940 | Let rr be the position vector
of any point P in the plane Therefore, the equation of the plane passing through R and
perpendicular to the vector RS
RT
×
uuur
uuur
is
(
) (RS
RT)
r
−a
⋅
×
uuur
uuur
r
r
= 0
or
r
r
r
r
r
r
(
) [(
)×(
)]
r –a
b –a
c –a
= 0 … (1)
© NCERT
not to be republished
MATHEMATI CS
484
Fig 11 15
This is the equation of the plane in vector form passing through three noncollinear
points |
1 | 5938-5941 | Therefore, the equation of the plane passing through R and
perpendicular to the vector RS
RT
×
uuur
uuur
is
(
) (RS
RT)
r
−a
⋅
×
uuur
uuur
r
r
= 0
or
r
r
r
r
r
r
(
) [(
)×(
)]
r –a
b –a
c –a
= 0 … (1)
© NCERT
not to be republished
MATHEMATI CS
484
Fig 11 15
This is the equation of the plane in vector form passing through three noncollinear
points �Note Why was it necessary to say that the three points
had to be non collinear |
1 | 5939-5942 | [(
)×(
)]
r –a
b –a
c –a
= 0 … (1)
© NCERT
not to be republished
MATHEMATI CS
484
Fig 11 15
This is the equation of the plane in vector form passing through three noncollinear
points �Note Why was it necessary to say that the three points
had to be non collinear If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11 |
1 | 5940-5943 | 15
This is the equation of the plane in vector form passing through three noncollinear
points �Note Why was it necessary to say that the three points
had to be non collinear If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11 15) |
1 | 5941-5944 | �Note Why was it necessary to say that the three points
had to be non collinear If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11 15) These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book |
1 | 5942-5945 | If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11 15) These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T
respectively |
1 | 5943-5946 | 15) These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T
respectively Let (x, y, z) be the coordinates of any point P on the plane with position
vector rr |
1 | 5944-5947 | These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T
respectively Let (x, y, z) be the coordinates of any point P on the plane with position
vector rr Then
RP
uuur = (x – x1) ˆi + (y – y1) ˆj + (z – z1) ˆk
RS
uuur = (x2 – x1) ˆi + (y2 – y1) ˆj + (z2 – z1) ˆk
RT
uuur = (x3 – x1) ˆi + (y3 – y1) ˆj + (z3 – z1) ˆk
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
1
1
2
1
2
1
2
1
3
1
3
1
3
1
0
x
x
y
y
z
z
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
−
−
−
=
−
−
−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) |
1 | 5945-5948 | Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T
respectively Let (x, y, z) be the coordinates of any point P on the plane with position
vector rr Then
RP
uuur = (x – x1) ˆi + (y – y1) ˆj + (z – z1) ˆk
RS
uuur = (x2 – x1) ˆi + (y2 – y1) ˆj + (z2 – z1) ˆk
RT
uuur = (x3 – x1) ˆi + (y3 – y1) ˆj + (z3 – z1) ˆk
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
1
1
2
1
2
1
2
1
3
1
3
1
3
1
0
x
x
y
y
z
z
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
−
−
−
=
−
−
−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) Example 18 Find the vector equations of the plane passing through the points
R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3) |
1 | 5946-5949 | Let (x, y, z) be the coordinates of any point P on the plane with position
vector rr Then
RP
uuur = (x – x1) ˆi + (y – y1) ˆj + (z – z1) ˆk
RS
uuur = (x2 – x1) ˆi + (y2 – y1) ˆj + (z2 – z1) ˆk
RT
uuur = (x3 – x1) ˆi + (y3 – y1) ˆj + (z3 – z1) ˆk
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
1
1
2
1
2
1
2
1
3
1
3
1
3
1
0
x
x
y
y
z
z
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
−
−
−
=
−
−
−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) Example 18 Find the vector equations of the plane passing through the points
R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3) Solution Let
ˆ
ˆ
ˆ
2
5
3
a
i
j
r= + −k
,
ˆ
ˆ
ˆ
2
3
5
b
i
j
k
= −
−
+
r
,
ˆ
ˆ
ˆ
5
3
3
c
i
j
= + −k
r
Then the vector equation of the plane passing through ar , b
r
and crand is
given by
(
) (RS RT)
r
−a
⋅
×
uuur uuur
r
r
= 0 (Why |
1 | 5947-5950 | Then
RP
uuur = (x – x1) ˆi + (y – y1) ˆj + (z – z1) ˆk
RS
uuur = (x2 – x1) ˆi + (y2 – y1) ˆj + (z2 – z1) ˆk
RT
uuur = (x3 – x1) ˆi + (y3 – y1) ˆj + (z3 – z1) ˆk
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
1
1
2
1
2
1
2
1
3
1
3
1
3
1
0
x
x
y
y
z
z
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
−
−
−
=
−
−
−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) Example 18 Find the vector equations of the plane passing through the points
R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3) Solution Let
ˆ
ˆ
ˆ
2
5
3
a
i
j
r= + −k
,
ˆ
ˆ
ˆ
2
3
5
b
i
j
k
= −
−
+
r
,
ˆ
ˆ
ˆ
5
3
3
c
i
j
= + −k
r
Then the vector equation of the plane passing through ar , b
r
and crand is
given by
(
) (RS RT)
r
−a
⋅
×
uuur uuur
r
r
= 0 (Why )
or
(
) [(
) (
)]
r
a
b
a
c
a
−
⋅
−
×
−
r
r
r
r
r
r
= 0
i |
1 | 5948-5951 | Example 18 Find the vector equations of the plane passing through the points
R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3) Solution Let
ˆ
ˆ
ˆ
2
5
3
a
i
j
r= + −k
,
ˆ
ˆ
ˆ
2
3
5
b
i
j
k
= −
−
+
r
,
ˆ
ˆ
ˆ
5
3
3
c
i
j
= + −k
r
Then the vector equation of the plane passing through ar , b
r
and crand is
given by
(
) (RS RT)
r
−a
⋅
×
uuur uuur
r
r
= 0 (Why )
or
(
) [(
) (
)]
r
a
b
a
c
a
−
⋅
−
×
−
r
r
r
r
r
r
= 0
i e |
1 | 5949-5952 | Solution Let
ˆ
ˆ
ˆ
2
5
3
a
i
j
r= + −k
,
ˆ
ˆ
ˆ
2
3
5
b
i
j
k
= −
−
+
r
,
ˆ
ˆ
ˆ
5
3
3
c
i
j
= + −k
r
Then the vector equation of the plane passing through ar , b
r
and crand is
given by
(
) (RS RT)
r
−a
⋅
×
uuur uuur
r
r
= 0 (Why )
or
(
) [(
) (
)]
r
a
b
a
c
a
−
⋅
−
×
−
r
r
r
r
r
r
= 0
i e ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
5
3 )] [( 4
8
8 )
(3
2 )]
0
r
i
j
k
i
j
k
i
j
− + − ⋅ − − + × − =
r
R
S
T
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
485
11 |
1 | 5950-5953 | )
or
(
) [(
) (
)]
r
a
b
a
c
a
−
⋅
−
×
−
r
r
r
r
r
r
= 0
i e ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
5
3 )] [( 4
8
8 )
(3
2 )]
0
r
i
j
k
i
j
k
i
j
− + − ⋅ − − + × − =
r
R
S
T
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
485
11 6 |
1 | 5951-5954 | e ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
5
3 )] [( 4
8
8 )
(3
2 )]
0
r
i
j
k
i
j
k
i
j
− + − ⋅ − − + × − =
r
R
S
T
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
485
11 6 4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes |
1 | 5952-5955 | ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
5
3 )] [( 4
8
8 )
(3
2 )]
0
r
i
j
k
i
j
k
i
j
− + − ⋅ − − + × − =
r
R
S
T
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
485
11 6 4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes Let the equation of the plane be
Ax + By + Cz + D = 0 (D ≠ 0) |
1 | 5953-5956 | 6 4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes Let the equation of the plane be
Ax + By + Cz + D = 0 (D ≠ 0) (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11 |
1 | 5954-5957 | 4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes Let the equation of the plane be
Ax + By + Cz + D = 0 (D ≠ 0) (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11 16) |
1 | 5955-5958 | Let the equation of the plane be
Ax + By + Cz + D = 0 (D ≠ 0) (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11 16) Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively |
1 | 5956-5959 | (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11 16) Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively Therefore
Aa + D = 0 or A =
D
a−
Bb + D = 0 or B =
D
b−
Cc + D = 0 or C =
D
c−
Substituting these values in the equation (1) of the
plane and simplifying, we get
x
y
z
a
b
c
+
+
= 1 |
1 | 5957-5960 | 16) Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively Therefore
Aa + D = 0 or A =
D
a−
Bb + D = 0 or B =
D
b−
Cc + D = 0 or C =
D
c−
Substituting these values in the equation (1) of the
plane and simplifying, we get
x
y
z
a
b
c
+
+
= 1 (1)
which is the required equation of the plane in the intercept form |
1 | 5958-5961 | Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively Therefore
Aa + D = 0 or A =
D
a−
Bb + D = 0 or B =
D
b−
Cc + D = 0 or C =
D
c−
Substituting these values in the equation (1) of the
plane and simplifying, we get
x
y
z
a
b
c
+
+
= 1 (1)
which is the required equation of the plane in the intercept form Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively |
1 | 5959-5962 | Therefore
Aa + D = 0 or A =
D
a−
Bb + D = 0 or B =
D
b−
Cc + D = 0 or C =
D
c−
Substituting these values in the equation (1) of the
plane and simplifying, we get
x
y
z
a
b
c
+
+
= 1 (1)
which is the required equation of the plane in the intercept form Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively Solution Let the equation of the plane be
x
y
z
a
b
c
+
+
= 1 |
1 | 5960-5963 | (1)
which is the required equation of the plane in the intercept form Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively Solution Let the equation of the plane be
x
y
z
a
b
c
+
+
= 1 (1)
Here
a = 2, b = 3, c = 4 |
1 | 5961-5964 | Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively Solution Let the equation of the plane be
x
y
z
a
b
c
+
+
= 1 (1)
Here
a = 2, b = 3, c = 4 Substituting the values of a, b and c in (1), we get the required equation of the
plane as
1
2
3
4
x
y
z
+
+
= or 6x + 4y + 3z = 12 |
1 | 5962-5965 | Solution Let the equation of the plane be
x
y
z
a
b
c
+
+
= 1 (1)
Here
a = 2, b = 3, c = 4 Substituting the values of a, b and c in (1), we get the required equation of the
plane as
1
2
3
4
x
y
z
+
+
= or 6x + 4y + 3z = 12 11 |
1 | 5963-5966 | (1)
Here
a = 2, b = 3, c = 4 Substituting the values of a, b and c in (1), we get the required equation of the
plane as
1
2
3
4
x
y
z
+
+
= or 6x + 4y + 3z = 12 11 6 |
1 | 5964-5967 | Substituting the values of a, b and c in (1), we get the required equation of the
plane as
1
2
3
4
x
y
z
+
+
= or 6x + 4y + 3z = 12 11 6 5 Plane passing through the intersection
of two given planes
Let π1 and π 2 be two planes with equat ions
rr n⋅1ˆ
= d1 and
rr n⋅2ˆ
= d2 respectively |
1 | 5965-5968 | 11 6 5 Plane passing through the intersection
of two given planes
Let π1 and π 2 be two planes with equat ions
rr n⋅1ˆ
= d1 and
rr n⋅2ˆ
= d2 respectively The position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11 |
1 | 5966-5969 | 6 5 Plane passing through the intersection
of two given planes
Let π1 and π 2 be two planes with equat ions
rr n⋅1ˆ
= d1 and
rr n⋅2ˆ
= d2 respectively The position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11 17) |
1 | 5967-5970 | 5 Plane passing through the intersection
of two given planes
Let π1 and π 2 be two planes with equat ions
rr n⋅1ˆ
= d1 and
rr n⋅2ˆ
= d2 respectively The position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11 17) Fig 11 |
1 | 5968-5971 | The position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11 17) Fig 11 16
Fig 11 |
1 | 5969-5972 | 17) Fig 11 16
Fig 11 17
© NCERT
not to be republished
MATHEMATI CS
486
If tr is the position vector of a point on the line, then
⋅rt n1ˆ
= d1 and
⋅rt n2ˆ
= d2
Therefore, for all real values of λ, we have
1
2
ˆ
ˆ
(
)
t
n
n
⋅
+λ
r
=
1
2
d
d
+λ
Since tr is arbitrary, it satisfies for any point on the line |
1 | 5970-5973 | Fig 11 16
Fig 11 17
© NCERT
not to be republished
MATHEMATI CS
486
If tr is the position vector of a point on the line, then
⋅rt n1ˆ
= d1 and
⋅rt n2ˆ
= d2
Therefore, for all real values of λ, we have
1
2
ˆ
ˆ
(
)
t
n
n
⋅
+λ
r
=
1
2
d
d
+λ
Since tr is arbitrary, it satisfies for any point on the line Hence, the equation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
represents a plane π3 which is such
that if any vector rr satisfies both the equations π1 and π2, it also satisfies the equation
π3 i |
1 | 5971-5974 | 16
Fig 11 17
© NCERT
not to be republished
MATHEMATI CS
486
If tr is the position vector of a point on the line, then
⋅rt n1ˆ
= d1 and
⋅rt n2ˆ
= d2
Therefore, for all real values of λ, we have
1
2
ˆ
ˆ
(
)
t
n
n
⋅
+λ
r
=
1
2
d
d
+λ
Since tr is arbitrary, it satisfies for any point on the line Hence, the equation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
represents a plane π3 which is such
that if any vector rr satisfies both the equations π1 and π2, it also satisfies the equation
π3 i e |
1 | 5972-5975 | 17
© NCERT
not to be republished
MATHEMATI CS
486
If tr is the position vector of a point on the line, then
⋅rt n1ˆ
= d1 and
⋅rt n2ˆ
= d2
Therefore, for all real values of λ, we have
1
2
ˆ
ˆ
(
)
t
n
n
⋅
+λ
r
=
1
2
d
d
+λ
Since tr is arbitrary, it satisfies for any point on the line Hence, the equation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
represents a plane π3 which is such
that if any vector rr satisfies both the equations π1 and π2, it also satisfies the equation
π3 i e , any plane passing through the intersection of the planes
r n1
⋅r r =
1
2
2
dand
r n
d
⋅
=
r r
has the equation
1
2
(
)
r
n
n
⋅
+ λ
r
r
r
= d1 + λd2 |
1 | 5973-5976 | Hence, the equation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
represents a plane π3 which is such
that if any vector rr satisfies both the equations π1 and π2, it also satisfies the equation
π3 i e , any plane passing through the intersection of the planes
r n1
⋅r r =
1
2
2
dand
r n
d
⋅
=
r r
has the equation
1
2
(
)
r
n
n
⋅
+ λ
r
r
r
= d1 + λd2 (1)
Cartesian form
In Cartesian system, let
1nr =
1
2
1 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
2nr =
2
2
2 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
and
rr =
ˆ
ˆ
ˆ
xi
y j
z k
+
+
Then (1) becomes
x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2
or
(A1x + B1y + C 1z – d1) + λ(A2x + B2 y + C 2z – d2) = 0 |
1 | 5974-5977 | e , any plane passing through the intersection of the planes
r n1
⋅r r =
1
2
2
dand
r n
d
⋅
=
r r
has the equation
1
2
(
)
r
n
n
⋅
+ λ
r
r
r
= d1 + λd2 (1)
Cartesian form
In Cartesian system, let
1nr =
1
2
1 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
2nr =
2
2
2 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
and
rr =
ˆ
ˆ
ˆ
xi
y j
z k
+
+
Then (1) becomes
x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2
or
(A1x + B1y + C 1z – d1) + λ(A2x + B2 y + C 2z – d2) = 0 (2)
which is the required Cartesian form of the equation of the plane passing through the
intersection of the given planes for each value of λ |
1 | 5975-5978 | , any plane passing through the intersection of the planes
r n1
⋅r r =
1
2
2
dand
r n
d
⋅
=
r r
has the equation
1
2
(
)
r
n
n
⋅
+ λ
r
r
r
= d1 + λd2 (1)
Cartesian form
In Cartesian system, let
1nr =
1
2
1 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
2nr =
2
2
2 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
and
rr =
ˆ
ˆ
ˆ
xi
y j
z k
+
+
Then (1) becomes
x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2
or
(A1x + B1y + C 1z – d1) + λ(A2x + B2 y + C 2z – d2) = 0 (2)
which is the required Cartesian form of the equation of the plane passing through the
intersection of the given planes for each value of λ Example 20 Find the vector equation of the plane passing through the intersection of
the planes
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) 6 and
(2
3
4 )
5,
r
i
j
k
r
i
j
k
⋅
+ +
=
⋅
+
+
= −
r
r
and the point (1, 1, 1) |
1 | 5976-5979 | (1)
Cartesian form
In Cartesian system, let
1nr =
1
2
1 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
2nr =
2
2
2 ˆ
ˆ
ˆ
A
B
C
i
j
k
+
+
and
rr =
ˆ
ˆ
ˆ
xi
y j
z k
+
+
Then (1) becomes
x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2
or
(A1x + B1y + C 1z – d1) + λ(A2x + B2 y + C 2z – d2) = 0 (2)
which is the required Cartesian form of the equation of the plane passing through the
intersection of the given planes for each value of λ Example 20 Find the vector equation of the plane passing through the intersection of
the planes
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) 6 and
(2
3
4 )
5,
r
i
j
k
r
i
j
k
⋅
+ +
=
⋅
+
+
= −
r
r
and the point (1, 1, 1) Solution Here,
1
ˆ
ˆ
ˆ
n
i
j
k
=
+ +
r
and
2
nr =
ˆ
ˆ
ˆ
2
3
4 ;
i
j
k
+
+
and
d1 = 6 and d2 = –5
Hence, using the relation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
3
4 )]
r
i
j
k
i
j
k
⋅
+ +
+λ
+
+
r
= 6 5
− λ
or
ˆ
ˆ
ˆ
[(1 2 )
(1 3 )
(1 4 ) ]
r
i
j
k
⋅
+ λ
+
+ λ
+
+ λ
r
= 6 5
− λ … (1)
where, λ is some real number |
1 | 5977-5980 | (2)
which is the required Cartesian form of the equation of the plane passing through the
intersection of the given planes for each value of λ Example 20 Find the vector equation of the plane passing through the intersection of
the planes
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) 6 and
(2
3
4 )
5,
r
i
j
k
r
i
j
k
⋅
+ +
=
⋅
+
+
= −
r
r
and the point (1, 1, 1) Solution Here,
1
ˆ
ˆ
ˆ
n
i
j
k
=
+ +
r
and
2
nr =
ˆ
ˆ
ˆ
2
3
4 ;
i
j
k
+
+
and
d1 = 6 and d2 = –5
Hence, using the relation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
3
4 )]
r
i
j
k
i
j
k
⋅
+ +
+λ
+
+
r
= 6 5
− λ
or
ˆ
ˆ
ˆ
[(1 2 )
(1 3 )
(1 4 ) ]
r
i
j
k
⋅
+ λ
+
+ λ
+
+ λ
r
= 6 5
− λ … (1)
where, λ is some real number © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
487
Taking
ˆ
ˆ
ˆ
r
xi
y j
r= + +z k
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) [(1 2 )
(1 3 )
(1 4 ) ] 6 5
xi
y j
z k
i
j
k
+
+
⋅
+ λ
+
+ λ
+
+ λ
= − λ
or
(1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or
(x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 |
1 | 5978-5981 | Example 20 Find the vector equation of the plane passing through the intersection of
the planes
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) 6 and
(2
3
4 )
5,
r
i
j
k
r
i
j
k
⋅
+ +
=
⋅
+
+
= −
r
r
and the point (1, 1, 1) Solution Here,
1
ˆ
ˆ
ˆ
n
i
j
k
=
+ +
r
and
2
nr =
ˆ
ˆ
ˆ
2
3
4 ;
i
j
k
+
+
and
d1 = 6 and d2 = –5
Hence, using the relation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
3
4 )]
r
i
j
k
i
j
k
⋅
+ +
+λ
+
+
r
= 6 5
− λ
or
ˆ
ˆ
ˆ
[(1 2 )
(1 3 )
(1 4 ) ]
r
i
j
k
⋅
+ λ
+
+ λ
+
+ λ
r
= 6 5
− λ … (1)
where, λ is some real number © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
487
Taking
ˆ
ˆ
ˆ
r
xi
y j
r= + +z k
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) [(1 2 )
(1 3 )
(1 4 ) ] 6 5
xi
y j
z k
i
j
k
+
+
⋅
+ λ
+
+ λ
+
+ λ
= − λ
or
(1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or
(x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i |
1 | 5979-5982 | Solution Here,
1
ˆ
ˆ
ˆ
n
i
j
k
=
+ +
r
and
2
nr =
ˆ
ˆ
ˆ
2
3
4 ;
i
j
k
+
+
and
d1 = 6 and d2 = –5
Hence, using the relation
1
2
1
2
(
)
r
n
n
d
d
⋅
+ λ
=
+λ
r
r
r
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(2
3
4 )]
r
i
j
k
i
j
k
⋅
+ +
+λ
+
+
r
= 6 5
− λ
or
ˆ
ˆ
ˆ
[(1 2 )
(1 3 )
(1 4 ) ]
r
i
j
k
⋅
+ λ
+
+ λ
+
+ λ
r
= 6 5
− λ … (1)
where, λ is some real number © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
487
Taking
ˆ
ˆ
ˆ
r
xi
y j
r= + +z k
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) [(1 2 )
(1 3 )
(1 4 ) ] 6 5
xi
y j
z k
i
j
k
+
+
⋅
+ λ
+
+ λ
+
+ λ
= − λ
or
(1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or
(x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i e |
1 | 5980-5983 | © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
487
Taking
ˆ
ˆ
ˆ
r
xi
y j
r= + +z k
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) [(1 2 )
(1 3 )
(1 4 ) ] 6 5
xi
y j
z k
i
j
k
+
+
⋅
+ λ
+
+ λ
+
+ λ
= − λ
or
(1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or
(x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i e (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or
λ = 3
14
Putting the values of λ in (1), we get
3
9
6 ˆ
ˆ
ˆ
1
1
1
7
14
7
r
i
j
k
+
+
+
+
+
r
=
15
6 14
−
or
10
23
13 ˆ
ˆ
ˆ
7
14
7
r
i
j
k
+
+
r
= 69
14
or
ˆ
ˆ
ˆ
(20
23
26 )
r
i
j
k
⋅
+
+
r
= 69
which is the required vector equation of the plane |
1 | 5981-5984 | (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i e (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or
λ = 3
14
Putting the values of λ in (1), we get
3
9
6 ˆ
ˆ
ˆ
1
1
1
7
14
7
r
i
j
k
+
+
+
+
+
r
=
15
6 14
−
or
10
23
13 ˆ
ˆ
ˆ
7
14
7
r
i
j
k
+
+
r
= 69
14
or
ˆ
ˆ
ˆ
(20
23
26 )
r
i
j
k
⋅
+
+
r
= 69
which is the required vector equation of the plane 11 |
1 | 5982-5985 | e (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or
λ = 3
14
Putting the values of λ in (1), we get
3
9
6 ˆ
ˆ
ˆ
1
1
1
7
14
7
r
i
j
k
+
+
+
+
+
r
=
15
6 14
−
or
10
23
13 ˆ
ˆ
ˆ
7
14
7
r
i
j
k
+
+
r
= 69
14
or
ˆ
ˆ
ˆ
(20
23
26 )
r
i
j
k
⋅
+
+
r
= 69
which is the required vector equation of the plane 11 7 Coplanarity of Two Lines
Let the given lines be
rr =
1
1
a
b
+λ
r
r |
1 | 5983-5986 | (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or
λ = 3
14
Putting the values of λ in (1), we get
3
9
6 ˆ
ˆ
ˆ
1
1
1
7
14
7
r
i
j
k
+
+
+
+
+
r
=
15
6 14
−
or
10
23
13 ˆ
ˆ
ˆ
7
14
7
r
i
j
k
+
+
r
= 69
14
or
ˆ
ˆ
ˆ
(20
23
26 )
r
i
j
k
⋅
+
+
r
= 69
which is the required vector equation of the plane 11 7 Coplanarity of Two Lines
Let the given lines be
rr =
1
1
a
b
+λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r |
1 | 5984-5987 | 11 7 Coplanarity of Two Lines
Let the given lines be
rr =
1
1
a
b
+λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
The line (1) passes through the point, say A, with position vector
1
ar and is parallel
to
1b
r |
1 | 5985-5988 | 7 Coplanarity of Two Lines
Let the given lines be
rr =
1
1
a
b
+λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
The line (1) passes through the point, say A, with position vector
1
ar and is parallel
to
1b
r The line (2) passes through the point, say B with position vector
2
ar and is parallel
to
2b
r |
1 | 5986-5989 | (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
The line (1) passes through the point, say A, with position vector
1
ar and is parallel
to
1b
r The line (2) passes through the point, say B with position vector
2
ar and is parallel
to
2b
r Thus,
AB
uuur
=
2
1
a
a−
r
r
The given lines are coplanar if and only if AB
uuur
is perpendicular to 1
2
b
b
×
r
r |
1 | 5987-5990 | (2)
The line (1) passes through the point, say A, with position vector
1
ar and is parallel
to
1b
r The line (2) passes through the point, say B with position vector
2
ar and is parallel
to
2b
r Thus,
AB
uuur
=
2
1
a
a−
r
r
The given lines are coplanar if and only if AB
uuur
is perpendicular to 1
2
b
b
×
r
r i |
1 | 5988-5991 | The line (2) passes through the point, say B with position vector
2
ar and is parallel
to
2b
r Thus,
AB
uuur
=
2
1
a
a−
r
r
The given lines are coplanar if and only if AB
uuur
is perpendicular to 1
2
b
b
×
r
r i e |
1 | 5989-5992 | Thus,
AB
uuur
=
2
1
a
a−
r
r
The given lines are coplanar if and only if AB
uuur
is perpendicular to 1
2
b
b
×
r
r i e 1
2
AB |
1 | 5990-5993 | i e 1
2
AB (
)
b
b
×
uuur r
r
= 0 or
2
1
1
2
(
) (
)
a
a
b
b
−
⋅
×
r
r
r
r
= 0
Cartesian form
Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively |
1 | 5991-5994 | e 1
2
AB (
)
b
b
×
uuur r
r
= 0 or
2
1
1
2
(
) (
)
a
a
b
b
−
⋅
×
r
r
r
r
= 0
Cartesian form
Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively © NCERT
not to be republished
MATHEMATI CS
488
Let a1, b1, c1 and a2, b2, c2 be the direction ratios of 1b
r
and
2b
r
, respectively |
1 | 5992-5995 | 1
2
AB (
)
b
b
×
uuur r
r
= 0 or
2
1
1
2
(
) (
)
a
a
b
b
−
⋅
×
r
r
r
r
= 0
Cartesian form
Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively © NCERT
not to be republished
MATHEMATI CS
488
Let a1, b1, c1 and a2, b2, c2 be the direction ratios of 1b
r
and
2b
r
, respectively Then
2
1
2
1
2
1 ˆ
ˆ
ˆ
AB (
)
(
)
(
)
x
x i
y
y
j
z
z k
= − + − + −
uuur
1
1
1
1
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
b
a i
b j
c k
b
a i
b j
c k
= + +
= + +
r
r
The given lines are coplanar if and only if
(
)
1
2
AB
0
⋅b b
×
=
uuur
r
r |
1 | 5993-5996 | (
)
b
b
×
uuur r
r
= 0 or
2
1
1
2
(
) (
)
a
a
b
b
−
⋅
×
r
r
r
r
= 0
Cartesian form
Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively © NCERT
not to be republished
MATHEMATI CS
488
Let a1, b1, c1 and a2, b2, c2 be the direction ratios of 1b
r
and
2b
r
, respectively Then
2
1
2
1
2
1 ˆ
ˆ
ˆ
AB (
)
(
)
(
)
x
x i
y
y
j
z
z k
= − + − + −
uuur
1
1
1
1
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
b
a i
b j
c k
b
a i
b j
c k
= + +
= + +
r
r
The given lines are coplanar if and only if
(
)
1
2
AB
0
⋅b b
×
=
uuur
r
r In the cartesian form,
it can be expressed as
2
1
2
1
2
1
1
1
1
2
2
2
0
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− = |
1 | 5994-5997 | © NCERT
not to be republished
MATHEMATI CS
488
Let a1, b1, c1 and a2, b2, c2 be the direction ratios of 1b
r
and
2b
r
, respectively Then
2
1
2
1
2
1 ˆ
ˆ
ˆ
AB (
)
(
)
(
)
x
x i
y
y
j
z
z k
= − + − + −
uuur
1
1
1
1
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
b
a i
b j
c k
b
a i
b j
c k
= + +
= + +
r
r
The given lines are coplanar if and only if
(
)
1
2
AB
0
⋅b b
×
=
uuur
r
r In the cartesian form,
it can be expressed as
2
1
2
1
2
1
1
1
1
2
2
2
0
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− = (4)
Example 21 Show that the lines
+3
1
5
–3
1
5
x
y
z
−
−
=
=
and
+1
2
5
–1
2
5
x
y
z
−
−
=
=
are coplanar |
1 | 5995-5998 | Then
2
1
2
1
2
1 ˆ
ˆ
ˆ
AB (
)
(
)
(
)
x
x i
y
y
j
z
z k
= − + − + −
uuur
1
1
1
1
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
b
a i
b j
c k
b
a i
b j
c k
= + +
= + +
r
r
The given lines are coplanar if and only if
(
)
1
2
AB
0
⋅b b
×
=
uuur
r
r In the cartesian form,
it can be expressed as
2
1
2
1
2
1
1
1
1
2
2
2
0
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− = (4)
Example 21 Show that the lines
+3
1
5
–3
1
5
x
y
z
−
−
=
=
and
+1
2
5
–1
2
5
x
y
z
−
−
=
=
are coplanar Solution Here,x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5
Now, consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
2 1
0
3 1 5
0
1
2
5
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− =−
=
−
Therefore, lines are coplanar |
1 | 5996-5999 | In the cartesian form,
it can be expressed as
2
1
2
1
2
1
1
1
1
2
2
2
0
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− = (4)
Example 21 Show that the lines
+3
1
5
–3
1
5
x
y
z
−
−
=
=
and
+1
2
5
–1
2
5
x
y
z
−
−
=
=
are coplanar Solution Here,x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5
Now, consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
2 1
0
3 1 5
0
1
2
5
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− =−
=
−
Therefore, lines are coplanar 11 |
1 | 5997-6000 | (4)
Example 21 Show that the lines
+3
1
5
–3
1
5
x
y
z
−
−
=
=
and
+1
2
5
–1
2
5
x
y
z
−
−
=
=
are coplanar Solution Here,x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5
Now, consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
2 1
0
3 1 5
0
1
2
5
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− =−
=
−
Therefore, lines are coplanar 11 8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11 |
1 | 5998-6001 | Solution Here,x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5
Now, consider the determinant
2
1
2
1
2
1
1
1
1
2
2
2
2 1
0
3 1 5
0
1
2
5
x
x
y
y
z
z
a
b
c
a
b
c
−
−
− =−
=
−
Therefore, lines are coplanar 11 8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11 18 (a)) |
1 | 5999-6002 | 11 8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11 18 (a)) Observe that if θ is an angle between the two planes, then so
is 180 – θ (Fig 11 |
1 | 6000-6003 | 8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11 18 (a)) Observe that if θ is an angle between the two planes, then so
is 180 – θ (Fig 11 18 (b)) |
1 | 6001-6004 | 18 (a)) Observe that if θ is an angle between the two planes, then so
is 180 – θ (Fig 11 18 (b)) We shall take the acute angle as the angles between
two planes |
1 | 6002-6005 | Observe that if θ is an angle between the two planes, then so
is 180 – θ (Fig 11 18 (b)) We shall take the acute angle as the angles between
two planes Fig 11 |
1 | 6003-6006 | 18 (b)) We shall take the acute angle as the angles between
two planes Fig 11 18
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
489
If
1nr and
2nr are normals to the planes and θ be the angle between the planes
r n⋅1
r r = d1 and
2
2 |
1 | 6004-6007 | We shall take the acute angle as the angles between
two planes Fig 11 18
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
489
If
1nr and
2nr are normals to the planes and θ be the angle between the planes
r n⋅1
r r = d1 and
2
2 d
n
r
r=
r |
1 | 6005-6008 | Fig 11 18
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
489
If
1nr and
2nr are normals to the planes and θ be the angle between the planes
r n⋅1
r r = d1 and
2
2 d
n
r
r=
r Then θ is the angle between the normals to the planes drawn from some common
point |
1 | 6006-6009 | 18
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
489
If
1nr and
2nr are normals to the planes and θ be the angle between the planes
r n⋅1
r r = d1 and
2
2 d
n
r
r=
r Then θ is the angle between the normals to the planes drawn from some common
point We have,
cos θ =
1
2
1
2
|
| |
|
n
n
n
n
r⋅
r
r
r
�Note The planes are perpendicular to each other if
1nr |
1 | 6007-6010 | d
n
r
r=
r Then θ is the angle between the normals to the planes drawn from some common
point We have,
cos θ =
1
2
1
2
|
| |
|
n
n
n
n
r⋅
r
r
r
�Note The planes are perpendicular to each other if
1nr 2nr = 0 and parallel if
1nr is parallel to
2nr |
1 | 6008-6011 | Then θ is the angle between the normals to the planes drawn from some common
point We have,
cos θ =
1
2
1
2
|
| |
|
n
n
n
n
r⋅
r
r
r
�Note The planes are perpendicular to each other if
1nr 2nr = 0 and parallel if
1nr is parallel to
2nr Cartesian form Let θ be the angle between the planes,
A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0
The direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2
respectively |
1 | 6009-6012 | We have,
cos θ =
1
2
1
2
|
| |
|
n
n
n
n
r⋅
r
r
r
�Note The planes are perpendicular to each other if
1nr 2nr = 0 and parallel if
1nr is parallel to
2nr Cartesian form Let θ be the angle between the planes,
A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0
The direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2
respectively Therefore, cos θ =
1
2
1
2
1
2
2
2
2
2
2
2
1
1
1
2
2
2
A A
B B
C C
A
B
C
A
B
C
+
+
+
+
+
+
�Note
1 |
1 | 6010-6013 | 2nr = 0 and parallel if
1nr is parallel to
2nr Cartesian form Let θ be the angle between the planes,
A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0
The direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2
respectively Therefore, cos θ =
1
2
1
2
1
2
2
2
2
2
2
2
1
1
1
2
2
2
A A
B B
C C
A
B
C
A
B
C
+
+
+
+
+
+
�Note
1 If the planes are at right angles, t hen θ = 90o and so cos θ = 0 |
1 | 6011-6014 | Cartesian form Let θ be the angle between the planes,
A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0
The direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2
respectively Therefore, cos θ =
1
2
1
2
1
2
2
2
2
2
2
2
1
1
1
2
2
2
A A
B B
C C
A
B
C
A
B
C
+
+
+
+
+
+
�Note
1 If the planes are at right angles, t hen θ = 90o and so cos θ = 0 Hence, cos θ = A1A2 + B1B2 + C1C2 = 0 |
1 | 6012-6015 | Therefore, cos θ =
1
2
1
2
1
2
2
2
2
2
2
2
1
1
1
2
2
2
A A
B B
C C
A
B
C
A
B
C
+
+
+
+
+
+
�Note
1 If the planes are at right angles, t hen θ = 90o and so cos θ = 0 Hence, cos θ = A1A2 + B1B2 + C1C2 = 0 2 |
1 | 6013-6016 | If the planes are at right angles, t hen θ = 90o and so cos θ = 0 Hence, cos θ = A1A2 + B1B2 + C1C2 = 0 2 If the planes are parallel, then
1
1
1
2
2
2
A
B
C
A
B
= =C |
1 | 6014-6017 | Hence, cos θ = A1A2 + B1B2 + C1C2 = 0 2 If the planes are parallel, then
1
1
1
2
2
2
A
B
C
A
B
= =C Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method |
1 | 6015-6018 | 2 If the planes are parallel, then
1
1
1
2
2
2
A
B
C
A
B
= =C Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method Solution The angle between two planes is the angle between their normals |
1 | 6016-6019 | If the planes are parallel, then
1
1
1
2
2
2
A
B
C
A
B
= =C Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method 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 |
1 | 6017-6020 | Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method 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 |
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