题目内容
已知非零向量








【答案】分析:由已知中非零向量
,
,|
|=2|
|,若关于x的方程x2+|
|x+
•
=0有实根,我们可以构造一个关于
与
的夹角θ的三角形不等式,解不等式可以确定cosθ的范围,进而得到
与
的夹角的最小值.
解答:解:∵关于x的方程x2+|
|x+
•
=0有实根,
∴|
|2-4
•
≥0
即|
|2-4|
|•|
|cosθ=|
|2-2|
|2cosθ≥0
∴cosθ≤
故
与
的夹角的最小值为
故答案为:
点评:本题考查的知识点是数量积表示两个向量的夹角,一元二次方程根的个数与系数的关系,其中根据已知条件,构造关于
与
的夹角θ的三角形不等式,是解答本题的关键.











解答:解:∵关于x的方程x2+|



∴|



即|





∴cosθ≤

故



故答案为:

点评:本题考查的知识点是数量积表示两个向量的夹角,一元二次方程根的个数与系数的关系,其中根据已知条件,构造关于



练习册系列答案
相关题目
已知非零向量
,
,若2
+3
与2
-3
互相垂直,则|
|=( )
a |
b |
a |
b |
a |
b |
| ||
|
A、
| ||
B、
| ||
C、
| ||
D、
|
已知非零向量
,
,
满足
+
+
=0,且
与
的夹角为60°,|
|=
|
|,则
与
的夹角为( )
a |
b |
c |
a |
b |
c |
a |
c |
b |
3 |
a |
a |
b |
A、30° | B、150° |
C、60° | D、120° |
给定命题p:若x2≥0,则x≥0;命题q:已知非零向量
,
,则“
⊥
”是“|
-
|=|
+
|”的充要条件.则下列各命题中,假命题的是( )
a |
b |
a |
b |
a |
b |
a |
b |
A、p∨q |
B、(?p)∨q |
C、(?p)∧q |
D、(?p)∧(?q) |