题目内容

已知
OA
=(3,-4)
OB
=(6,-3),
OC
=(5-m,-3-m)
(1)求向量
OA
OB
的夹角θ的余弦值;
(2)若A、B、C三点共线,求实数m的值.
分析:(1)由条件求得
OA
OB
=18+12=30,|
OA
|=5,|
OB
|=3
5
,设向量
OA
OB
的夹角为θ,则由cosθ=
OA
OB
|
OA
|•|
OB
|
,运算求得结果.
(2)若A、B、C三点共线,则有
AB
AC
,再由
AB
=(3,1),
AC
=(2-m,1-m),可得 3(1-m)-1×(2-m)=0,由此求得m的值.
解答:解:(1)∵已知
OA
=(3,-4),
OB
=(6,-3),
OC
=(5-m,-3-m),∴
OA
OB
=18+12=30,|
OA
|=5,|
OB
|=3
5

设向量
OA
OB
的夹角为θ,则cosθ=
OA
OB
|
OA
|•|
OB
|
=
30
5×3
5
=
2
5
5

(2)若A、B、C三点共线,则有
AB
AC
,∵
AB
=(3,1),
AC
=(2-m,1-m),∴3(1-m)-1×(2-m)=0,解得m=
1
2
点评:本题主要考查用两个向量的数量积表示两个向量的夹角,两个向量共线的性质,两个向量坐标形式的运算,属于中档题.
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