(
) 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 Evaluate
the quantity 
,  if  |
| 1, |
| 4  and |
| 2
a b
b c
c a
a
b
c
μ =
⋅
+
⋅
+ ⋅
=
=
=
r
r
r
r
r
r r
r
r Solution Since 
0
a
b
c
+
+
=
r
r
r
r
, we have
(
)
a
a
b
c
 
 
r
r
r
r  = 0
or
a a
a b
a c
⋅
+
⋅
+
⋅
r
r
r r
r
r
 = 0
Therefore
a b
a c
⋅
+
⋅
r
r
r
r
 =
2
1
−a
= −
r