在△ABC中,
(1)求
(2)判断
|
(1)因为
•AB
=0,故AB⊥AC,又|AC
|=12,|AB
|=15,可知|BC
|=9.AC
由已知可得
=AD
(1 2
+AB
),AC
=CB
-AB
,AC
∴
•AD
=CB
(1 2
+AB
)(AC
-AB
)AC
=
(1 2
2-AB
2)=AC
(141-81)=1 2
.…(4分)63 2
(2)
•AE
的值为一个常数.CB
∵L为线段BC的垂直平分线,L与BC交与点D,E为L上异于D的任意一点,
∴
•DE
=0,CB
故
•AE
=(CB
+AD
)•DE
=CB
•AD
+CB
•DE
=CB
•AD
=CB
…(9分)63 2