平面向量oa=(1.7).ob=(5.1).oc=(2.1).m是直线上ob上一动点,求ma乘以mb的
1个回答

设m(x,y)

直线ob过原点,则直线ob解析式为正比例函数

则设直线ob:y=kx

又b=(0,0)+(5,1)=(5,1)

则将b(5,1)代入y=kx

得k=1/5,则y=1/5*x

又m在直线上ob上

则m(x,y)也满足y=1/5*x

又ma=(1,7)-(x,y)=(1-x,7-y)

mb=(5,1)-(x,y)=(5-x,1-y)

则ma*mb=(1-x,7-y)*(5-x,1-y)

=(1-x)*(5-x)+(7-y)*(1-y)

又y=1/5*x

则ma*mb=(1-x)*(5-x)+(7-y)*(1-y)

=x^2-6x+5+(7- x/5)*(1- x/5)

=x^2-6x+5+ x^2/25- 8x/5 +7

=26x^2/25 -38x/5 +12

则当x=-(-38/5)/[2*(26/25)]=85/26

ma*mb最小=1026/13

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