Chapter 11 Three Dimensional Geometry Worksheet Page 14
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MATHEMATICS
Therefore, from (3), we get
H
H
H
H
×
−
(since PT = ST sin θ)
b
(
a
a
)
=
ˆ
| | PT
b
n
2
1
H
H
H
H
×
−
⋅
|
b
(
a
a
)|
i.e.,
=
(as |
| ˆ
= 1)
n
| | PT 1
b
2
1
Hence, the distance between the given parallel lines is
H
H
H
KKK H
×
−
b
(
a
a
)
=
H
2
1
| PT |
d =
| |
b
Example 11
Find the shortest distance between the lines l
and l
whose vector
1
2
equations are
H
+
+ λ
−
+
ˆ
ˆ
ˆ
ˆ
ˆ
=
... (1)
i
j
(2
i
j
k
)
r
H
+
− + µ
−
+
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
=
... (2)
2
i
j
k
(3
i
5
j
2 )
k
r
H
H
H
H
H
H
=
+
µ
+ λ
Solution
Comparing (1) and (2) with
=
and
respectively,
a
b
r
a
b
r
1
1
2
2
H
H
+
=
−
+
ˆ
ˆ
ˆ
ˆ
ˆ
we get
=
a
i
j
,
b
2
i
j
k
1
1
H
H
ˆ j
ˆ j
ˆ k
ˆ k
a
ˆ i
ˆ i
b
= 2
+
–
and
= 3
– 5
+ 2
2
2
H
H
−
−
ˆ
ˆ i
Therefore
=
a
a
k
2
1
H
H
×
−
+
×
−
+
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
and
=
b
b
( 2
i
j
k
) ( 3
i
5
j
2
k
)
1
2
ˆ
ˆ
ˆ
i
j
k
−
=
−
−
ˆ
ˆ
ˆ
2
1 1
3
i
j
7
k
=
−
3
5 2
H
H
+ +
=
×
So
=
9 1
49
59
|
b
b
|
1
2
Hence, the shortest distance between the given lines is given by
H
H
H
H
−
+
×
−
|
3
0
| 7
10
(
b
b
)
( .
a
a
)
=
=
H
H
1
2
2
1
d =
×
59
59
|
b
b
|
1
2
Example 12
Find the distance between the lines l
and l
given by
1
2
H
+
−
+ λ
+
+
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
i
2
j
4
k
( 2
i
3
j
6
k
)
=
r
H
+
−
+ µ
+
+
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
i
3
j
5
k
( 2
i
3
j
6
k
)
and
=
r
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