Solution Given 
(
)
a
b
c
⋅
r+
r
r  = 0,
(
)
0,
(
)
0 b
c
a
c
a
b
⋅
+
=
⋅
+
=
r
r
r
r
r
r
Now
2
|
|
a
b
c
+
+
r
r
r
 =
2
(
)
(
) (
)
a
b
c
a
b
c
a
b
c
+
+
=
+
+
⋅
+
+
r
r
r
r
r
r
r
r
r
=
(
)
(
)
a a
a
b
c
b b
b
a
c
⋅
+
⋅
+
+
⋅
+
⋅
+
r
r r
r
r r
r
r
r
r
+ (
) c a
b
c c
+
r+
r r
r r
=
2
2
2
|
|
|
|
|
|
a
b
c
+
+
r
r
r
= 9 + 16 + 25 = 50
Therefore
|
|
a
b
c
+
r+
r
r  =
50
5 2
=
© NCERT
not to be republished
VECTOR ALGEBRA
457
Example 29 Three vectors ,    and  
a
b
c
r
r
r  satisfy the condition 
0
a
b
c
+
+
=
r
r
r
r