Then
PQ = ST |cos θ|
But
cos θ =
PQ ST
| PQ | | ST |
⋅
uuur uur
uuuur
uur
 =
2
1
ˆ (
)
ST
d n
a
a
⋅d
−
r
r
(since  
2
1
ST
)
a
a
=
−
uur
r
r
=
1
2
2
1
1
2
(
) (
)
ST
b
b
a
a
b
b
×
⋅
−
×
r
r
r
r
r
r
[From (3)]
Fig 11 8
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or
d =
1
2
2
1
1
2
(
) (
)
|
|
b
b
a
a
b
b
×
−
×
r
r
r
r
r
r
Cartesian form
The shortest distance between the lines
l1 : 
1
1
x
x
−a
 =
1
1
1
1
y
y
z
z
b
c
−
−
=
and
l2  : 
2
2
x
x
−a
 =
2
2
2
2
y
y
z
z
b
c
−
−
=
is
2
1
2
1
2
1
1
1
1
2
2
2
2
2
2
1 2
2 1
1
2
2 1
1 2
2 1
(
)
(
)
(
)
x
x
y
y
z
z
a
b
c
a
b
c
b c
b c
c a
c a
a b
a b
−
−
−
−
+
−
+
−
11 5