Let (x, y, z) be the coordinates of any point P on the plane with position
vector rr Then
RP
uuur  = (x – x1) ˆi  + (y – y1) ˆj  + (z – z1) ˆk
RS
uuur  = (x2 – x1) ˆi  + (y2 – y1) ˆj  + (z2 – z1) ˆk
RT
uuur  = (x3 – x1) ˆi + (y3 – y1) ˆj  + (z3 – z1) ˆk
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
1
1
2
1
2
1
2
1
3
1
3
1
3
1
0
x
x
y
y
z
z
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
−
−
−
=
−
−
−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) Example 18 Find the vector equations of the plane passing through the points
R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3) Solution Let 
ˆ
ˆ
ˆ
2
5
3
a
i
j
r= + −k
, 
ˆ
ˆ
ˆ
2
3
5
b
i
j
k
= −
−
+
r
, 
ˆ
ˆ
ˆ
5
3
3
c
i
j
= + −k
r
Then the vector equation of the plane passing through ar , b
r
 and crand  is
given by
(
) (RS RT)
r
−a
⋅
×
uuur uuur
r
r
 = 0     (Why