0
 
a b
a
b
⋅
=
⇔
⊥
r
r
r
r
3 If θ = 0, then 
|
| |
|
a b
a
b
⋅
r=
r
r
r
In particular, 
2
|
| ,
a a
a
⋅
r r=
r
 as θ in this case is 0 4 If θ = π, then 
|
| |
|
a b
a
b
⋅
r= −
r
r
r
In particular, 
2
(
)
|
|
a
a
a
  
  
r
r
r
, as θ in this case is π