题目内容
已知向量
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【答案】分析:由
可知
,由
与
垂直,则
•
=0,即-4+2y2=0,解方程即得y值.
解答:解:∵
∴
,
又∵
⊥
垂直,
∴
•
=0,
∴-4+2y2=0,
解得y=±
故答案为:±
点评:垂直问题是一个重要的知识点,在高考题中常常出现,常与向量的模、向量的坐标表示等联系在一起,要特别注意垂直与平行的区别.若
=(a1,a2),
=(b1,b2),则
⊥
?a1a2+b1b2=0,
∥
?a1b2-a2b1=0.
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解答:解:∵

∴

又∵


∴


∴-4+2y2=0,
解得y=±

故答案为:±

点评:垂直问题是一个重要的知识点,在高考题中常常出现,常与向量的模、向量的坐标表示等联系在一起,要特别注意垂直与平行的区别.若
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