(i,j)为基向者禅量即点A(2,3)B(4,-1)设P(x,0)则由题意可知,向量AP=(x-2,-3),向量BP=(x-4,1)可得AP*BP=(x-2)(x-4)-3=x^2-6x+5=x^2-6x+9-4=(x-3)-4即当x=3时,满足首扰尘题意,即P(3,0),设其夹角w则cosw=向量BP*向量AP/BP的模*AP的模李纤=-4/2根号5=-2根号5/5,即角w=arccos-2根号5/5,,,,,,,,加油