题目内容
设

(1)若



(2)若




【答案】分析:(1)由三点A,B,C共线,结合向量的共线定理可知,必存在一个常数λ使得
,由此等式建立起关于λ,t的方程求出t的值;
(2)由题设条件,可以 把
表示成关于实数x的函数,根据所得的函数判断出它取出最小值时的x的值.
解答:解:(1)由三点A,B,C共线,必存在一个常数λ使得
,
则有
又
∴
=
,又
、
是两个不共线的非零向量
∴
解得 
故存在
时,A、B、C三点共线
(2)∵
且
两向量的夹角是120°
∴
=
=1+m+m2=(m+
)2+
∴当m=-
时,
的值最小为 
点评:本题考查平面向量的综合题,解题的关键是熟练掌握向量共线的坐标表示,向量的模的坐标表示,理解题设条件,正确转化.本题把三点共线转化为了向量共线,将模的最小值求参数的问题转化为求函数的最小值,解题时要注意恰当地运用转化、化归这一数学思想

(2)由题设条件,可以 把

解答:解:(1)由三点A,B,C共线,必存在一个常数λ使得

则有

又

∴




∴


故存在

(2)∵


∴




∴当m=-



点评:本题考查平面向量的综合题,解题的关键是熟练掌握向量共线的坐标表示,向量的模的坐标表示,理解题设条件,正确转化.本题把三点共线转化为了向量共线,将模的最小值求参数的问题转化为求函数的最小值,解题时要注意恰当地运用转化、化归这一数学思想

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