So, let |
|
0
| |
a
b
 
 
r
r
r Then,
2
|
|
a
+b
r
r
 =
2
(
)
(
) (
)
a
b
a
b
a
b
+
=
+
⋅
+
r
r
r
r
r
r
= a a
a b
b a
b b
⋅
+
⋅
+
⋅
+
⋅
r
r
r r
r r
r
r
=
2
2
|
|
2
|
|
a
a b
b
+
⋅ +
r
r
r
r
 (scalar product is commutative)
≤
2
2
|
|
2|
|
|
|
a
a b
b
+
⋅
r+
r
r
r
(since 
|
|
x
x
x
≤
∀ ∈ R )
≤
2
2
|
|
2|
||
|
|
|
a
a b
b
+
+
r
r
r
r
(from Example 19)
=
2
(|
|
|
|)
a
b
 
r
r
Fig 10 21
© NCERT
not to be republished
VECTOR ALGEBRA
447
Hence
|
|
a
b
 
r
r
≤ |
|
|
|
a
b
 
r
r
Remark If the equality holds in triangle inequality (in the above Example 20), i e