Let P be any
point  on the plane T herefore, NP
uuur
is
perpendicular to  ON
uuur
Therefore,  NP ON ⋅
uuur uuur
 = 0 (1)
Let rr be the position vector of  the point P,,
then NP
uuur
= 
dn
r
ˆ
r−
 (as ON
NP
OP
+ =
uuur
uuur
uuur
)
Therefore, (1) becomes
(
)
r
d n
d n
∧
∧
−
⋅
r
 = 0
or
(
)
r
d n
n
∧
∧
−
⋅
r
 = 0
(d ≠ 0)
or
r n
d n n
∧
∧
∧
⋅
−
⋅
r
 = 0
i