Solution We have 
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
and
2
a
b
i
j
k
a
b
j
k
+
=
+
+
−
= − −
r
r
r
r
A vector which is perpendicular to both 
and
a
b
a
b
+
−
r
r
r
r
 is given by
(
)
(
)
a
b
a
b
+
×
−
r
r
r
r
 =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
4
2
(
, say)
0
1
2
i
j
k
i
j
k
c
= −
+
−
=
−
−
r
Now
|
|
cr  =
4
16
4
24
2 6
+
+
=
=
Therefore, the required unit vector is
|
c|
c
r
r  =
1
2
1 ˆ
ˆ
ˆ
6
6
6
i
j
k
−
+
−
�Note   There are two perpendicular directions to any plane Thus, another unit
vector perpendicular to 
and
a
b
a
b
+
−
r
r
r
r
 will be 1
2
1 ˆ
ˆ
ˆ 6
6
6
i
j
k
−
+
 But that will
be a consequence of (
)
(
)
a
b
a
b
−
×
+
r
r
r
r Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3)
and C(2, 3, 1) as its vertices