1试题呈现,立意分析(2013年高考数学浙江卷理科第7题)设△ABC,P_0是边AB上一定点,满足P_0B=1/4AB,且对于边AB上任一点P,恒有PB·PC≥P_0B·P_0C,则()A.∠ABC=90°B.∠BAC=90°C.AB=AC D.AC=BC命题立意:本题以平面向量在三角形中的应用为载体,考查平面向量的内容,包括平面向量的坐标运算,数量积的定义及其几何意义,不等式恒成立问题等基本知识。本题考查学生上述基本知识的综合运用能力,以 1 questions presented, the intention analysis (2013 college entrance exam Zhejiang Criminology Question 7) set △ ABC, P_0 is a certain point on the edge AB, to meet P_0B = 1 / 4AB, and for any point AB AB, constant PB · PC ≥ P_0B · P_0C, then () A.ABC = 90 ° B.∠BAC = 90 ° C.AB = AC D.AC = BC Proposition Conception: This question uses the plane vector in the triangle as the carrier, examines the plane vector The content includes the coordinate vector of plane vector, the definition of quantity and its geometric meanings, the basic knowledge of inequality constant establishment problem and so on. This question examines students’ comprehensive utilization ability of the above basic knowledge