Example 13 Find the vector equation of the plane which is at a distance of 
29
6
from the origin and its normal vector from the origin is 
ˆ
ˆ
ˆ
2
3
4
i
j
− +k Solution Let 
k
j
i
n
4ˆ
3ˆ
2ˆ
+
−
r= Then
|
|
ˆ
nn
n
r
r
=
 =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
3
4
2
3
4
4
9
16
29
i
j
k
i
j
k
−
+
−
+
=
+
+
Hence, the required equation of the plane is
2
3
4
6
ˆ
ˆ
ˆ
29
29
29
29
r
i
j
k
−


⋅
+
+
=




r
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆ
ˆ
(6
3
2 )
1
r
i
j
k
⋅
−
−
+
r
 = 0  passing through the origin Solution The given equation can be written as
ˆ
ˆ
ˆ
(
6
3
2
⋅
−
+
+
rr
i
j
k ) = 1