12
Fig 11 13
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
483
Y
Z
O
a
r
R
P
bS
c
(RS  RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as 
ˆ
ˆ
ˆ
5
2
4
a
i
j
k
=
+
−
r
 and the
normal vector N
r
 perpendicular to the plane as 
ˆ
ˆ
N =2 + 3ˆ
i
j
k−
r
Therefore, the vector equation of the plane is given by  (
) N
0
r
a−
=
r
r
r
or
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[
(5
2
4 )] (2
3
)
0
r
i
j
k
i
j
k
−
+
−
⋅
+
−
=
r (1)
Transforming (1) into Cartesian form, we have
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
[( –5)
(
2)
(
4) ] (2
3
)
0
x
i
y
j
z
k
i
j
k
+
−
+
+
⋅
+
−
=
or
2(
5) 3(
2) 1(
4) 0
x
y
z
− + − − + =
i