But
ST
uur
=
2
1
a
a−
r
r
Fig 11 9
© NCERT
not to be republished
 MATHEMATI CS
476
Therefore, from (3), we get
2
1
(
)
b
a
a
×
−
r
r
r  =
ˆ
|
b| PT
n
r
        (since PT = ST sin θ)
i e ,
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
−
+