Chapter 11 Three Dimensional Geometry Worksheet Page 24
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41486
MATHEMATICS
H
If
t
is the position vector of a point on the line, then
H
H
t n ⋅
t n ⋅
ˆ
= d
and
ˆ
= d
1
2
1
2
Therefore, for all real values of λ, we have
H
⋅
+ λ
+ λ
ˆ
ˆ
=
t
(
n
n
)
d
d
1
2
1
2
H
Since
t
is arbitrary, it satisfies for any point on the line.
H H
H
represents a plane π
⋅
+ λ
=
+ λ
Hence, the equation
which is such
r n
(
n
)
d
d
1
2
1
2
H
3
satisfies both the equations π
and π
that if any vector
, it also satisfies the equation
r
1
2
π
i.e., any plane passing through the intersection of the planes
H H
3
H H
⋅
=
r n ⋅
d
and
r n
d
=
1
2
2
1
H
H
H
+ λ λ λ λ λ d
⋅
+ λ
has the equation
= d
... (1)
r
(
n
n
)
1
2
1
2
Cartesian form
In Cartesian system, let
H
+
+
ˆ
ˆ
ˆ
=
n
A
i
B
j
C
k
1
1
2
1
H
+
+
ˆ
ˆ
ˆ
=
n
A
i
B
j
C
k
2
2
2
2
H
+
+
ˆ
ˆ
ˆ
xi y j z k
and
=
r
Then (1) becomes
+ λA
+ λB
+ λC
+ λd
x (A
) + y (B
) + z (C
) = d
1
2
1
2
1
2
1
2
) + λ λ λ λ λ (A
or
(A
x + B
y + C
z – d
x + B
y + C
z – d
) = 0
... (2)
1
1
1
1
2
2
2
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
H
H
⋅ + +
=
⋅
+
+
= −
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
the planes
and the point (1, 1, 1).
r i
(
j k
) 6 and
r
(2
i
3
j
4 )
k
5,
H
H
= + +
+
+
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
Solution
Here,
and
=
n
n
i
j k
2
i
3
j
4 ;
k
2
1
and
d
= 6 and d
= –5
1
2
H H
H
⋅
+ λ
=
+ λ
Hence, using the relation
, we get
r n
(
n
)
d
d
1
2
1
2
H
− λ
⋅ + + + λ
+
+
ˆ
ˆ
ˆ ˆ
ˆ
ˆ
=
6 5
r i
[
j k
(2
i
3
j
4 )]
k
H
− λ
⋅
+ λ + + λ + + λ
ˆ
ˆ
ˆ
or
=
6 5
… (1)
r
[(1 2 )
i
(1 3 )
j
(1 4 ) ]
k
where, λ is some real number.
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