Question

已知不共线的两个向量oa ，ob ，oa的模=ob的模=3 ， 若oc=xoa+（1-x）ob（0＜x＜1）且 oc的模=根号3

已知不共线的两个向量oa ，ob ，oa的模=ob的模=3 ， 若oc=xoa+（1-x）ob（0＜x＜1）且 oc的模=根号3，则ab的模的最小值为

Answer 1

|OA|=|OB|=3
OC = xOA+(1-x)OB (0<x<1)
|OC|= √3
To find:
min |AB|
OC = xOA+(1-x)OB
|OC|^2= x^2|OA|^2 +(1-x)^2|OB|^2 + 2x(1-x)OB.OA
3 = 9x^2+9(1-x)^2+2x(1-x)OA.OB
OA.OB = 3(3x^2-3x+1) /[x(x-1)]

AB = OB-OA
|AB|^2 = |OB|^2 + |OA|^2 - 2OB.OA
= 18-6(3x^2-3x+1) /[x(x-1)]
(|AB|^2)' = -6{ [x(x-1)](6x-3) -(3x^2-3x+1)[2x-1] } /[x(x-1)]^2 =0
[x(x-1)](6x-3) -(3x^2-3x+1)[2x-1] =0
(2x-1)[3x(x-1)- (3x^2-3x+1)] =0
(2x-1)=0
x = 1/2
min |AB|^2 at x =1/2
min |AB|^2
=18+24( 3/4-3/2+1)
=18+6=24
min |AB| = 2√6