问题 解答题
OA
=(2,-1),
OB
=(3,0),
OC
=(m,3)

(1)当m=8时,将
OC
OA
OB
表示;
(2)若A、B、C三点能构成三角形,求实数m应满足的条件.
答案

(1)当m=8时,

OC
=(8,3).

OC
OA
OB
,则(8,3)=λ(2,-1)+μ(3,0)=(2λ+3μ,-λ),

2λ+3μ=8
-λ=3
,解得
λ=-3
μ=
14
3

所以

OC
=-3
OA
+
14
3
OB

(2)由

OA
=(2,-1),
OB
=(3,0),
OC
=(m,3).

AB
=
OB
-
OA
=(3,0)-(2,-1)=(1,1),

AC
=
OC
-
OA
=(m,3)-(2,-1)=(m-2,4),

若A、B、C三点能构成三角形,

AB
AC
不共线.由1×4-1×(m-2)=0得:m=6.

所以A、B、C三点能构成三角形的实数m应满足m≠6.

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