(1)证明略(2)45°
(1) 设
=a,
=b,
=c,正四面体的棱长为1,
则
=
(a+b+c),
=
(b+c-5a),
=
(a+c-5b),
=
(a+b-5c)
∴
·
=
(b+c-5a)·(a+c-5b)
=
(18a·b-9|a| 2)
=
(18×1×1·cos60°-9)=0.
∴
⊥
,∴AO⊥BO,
同理
⊥
,BO⊥CO,
∴AO、BO、CO两两垂直.
(2)
=
+
=-
(a+b+c)+
=
(-2a-2b+c).
∴|
|=
=
,
|
|=
=
,
·
=
(-2a-2b+c)·
(b+c-5a)=
,
∴cos〈
,
〉=
=
,
∵〈
,
〉∈(0,
),∴〈
,
〉=45°.