,
2
1
|
(
)|
b
a
a
×
−
r
r
r
 = |
b| PT 1
⋅
r
       (as |
|
ˆn  = 1)
Hence, the distance between the given parallel lines is
d =
2
1
(
)
| PT |
|
|
b
a
a
b
×
−
=
r
r
r
uuur
r
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
rr  =
ˆ
ˆ
ˆ
ˆ
ˆ
(2
)
i
j
i
j
k
+
+ λ
−
+ (1)
and
rr  =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(3
5
2
)
i
j
k
i
j
k
+
−
+ µ
−
+ (2)
Solution Comparing (1) and (2) with rr  = 
1
1
a
b
+ λ
r
r
 and  
2
2
b
a
r
r
r
r
+µ
=
 respectively,,
we get
1ar  =
1
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
i
j
b
i
j
k
+
=
−
+
r
2
ar
 = 2 ˆi  + ˆj – ˆk  and 
2b
r
 = 3 ˆi  – 5 ˆj  + 2 ˆk
Therefore
2
1
a
r−a
r  =
ˆ
ˆi
k
−
and
1
2
b
b
×
r
r
 =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( 2
)
( 3
5
2
)
i
j
k
i
j
k
−
+
×
−
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
1
1
3
7
3
5
2
i
j
k
i
j
k
−
=
−
−
−
So
1
2
|
|
b
b
×
r
r
 =
9
1
49
59
+ +
=
Hence, the shortest distance between the given lines is given by
d  =
|
|
)
) (
(
2
1
1
2
2
1
b
b
a
a
b
b
r
r
r
r
r
r
×
−
×
  
59
10
59
7|
0
|3
=
+
−
=
Example 12 Find the distance between the lines l1 and l2 given by
rr  =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
( 2
3
6
)
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr  =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
3
5
( 2
3
6
)
i
j
k
i
j
k
+
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
477
Solution The two lines are parallel (Why