问题
解答题
在△ABC中,点D在BC上,设
(1)若BD=2DC,求
(2)若∠BAC=120°,AB=2,AC=1,AD⊥BC,
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答案
(1)由题意,
=AD
+AB
=BD
+AB 2 3
=BC
+AB
(2 3
-AC
)=AB 1 3
+AB 2 3
=AC 1 3
+a 2 3
;b
(2)∵
=λBD
,∴BC
•BC
=λBD
2,BC
∴
•(BC
-AD
)=λAB
2,BC
∵AD⊥BC,∴-
•BC
=λ(AB
-AC
)2AB
∴-(
-AC
)•AB
=λ(AB
-AC
)2AB
∴
2-a
•b
=λ(a
2-2b
•b
+a
2)a
∴λ=
2-a
•b a
2-2b
•b
+a
2a
∵∠BAC=120°,AB=2,AC=1,
∴λ=
=4-2cos120° 1-2×2×cos120°+4
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