Solution Let θ be the angle between the line and the normal to the plane Converting the
given equations into vector form, we have
rr  =
ˆ
ˆ
ˆ
ˆ
ˆ
( –
3 )
( 2
3
6
)
i
k
i
j
k
+
+ λ
+
+
and
ˆ
ˆ
ˆ
( 10
2
11
)
r
i
j
k
⋅
+
−
r
 = 3
Here
b
r
 =
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
   and  
k
j
i
n
11ˆ
2ˆ
10ˆ
−
+
=
r
sin φ =
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(2
3
6
) (10
2
11 )
2
3
6
10
2
11
i
j
k
i
j
k
+
+
⋅
+
−
+
+
+
+
=
40
7
15
−
×
 = 
−218
 = 8
21   or φ = 
1
8
sin
21
− 





Fig 11 20
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
493
EXERCISE 11 3
1