Then
ST
b ×
uur
r
 =
ˆ
(|
b|| ST| sin )
n
θ
uur
r (3)
where nˆ  is the unit vector perpendicular to the plane of the lines l1 and l2 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