From the
equation of the planes, the normal vectors are
N1
ur
 =
ˆ
ˆ
ˆ
2
2
i
j
k
+
−
 and  
2
ˆ
ˆ
ˆ
N
3
6
2
i
j
k
=
−
−
ur
Therefore
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
N
N
(2
2
) (3
6
2
)
| N | |N |
4
1
4
9
36
4
i
j
k
i
j
k
⋅
+
−
⋅
−
−
=
+
+
+
+
ur
ur
ur
ur
 = 
4
21






Hence
θ = cos – 1


21
4
© NCERT
not to be republished
 MATHEMATI CS
490
Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5 Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0  and  A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
 A2 = 2, B2 = 2, C2 = – 2
cos θ =
(
)
(
)
2
2
2
2
2
2
3
2
( 6) (2)
(2) ( 2)
3
( 6)
( 2)
2
2
( 2)
×
+ −
+
−
+ −
+ −
+
+ −
=
10
5
5 3
21
7
2 3
7 3
−
=
=
×
Therefore,
θ = cos-1  5 3
21








11 9  Distance of a Point from a Plane
Vector form
Consider a point P with position vector ar  and a plane π 1 whose equation is
⋅rr nˆ
 = d (Fig 11 19)