(3)
Then
PQ
uuur
 = d nˆ
where, d is the magnitude of the shortest distance vector Let θ be the angle between
ST
uur
 and PQ
uuur 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
(
)