Question

丨`∵。t√、

丨OA丨=2,丨OB丨=2,向量OC=xOA+yOB且x+y=1,∠AOB是钝角,f（t）=丨OA-tOB丨的最小值为根号3,则丨OC丨的

Answer 1

OC= xOA+yOB
f(t) = |OA-tOB|
[f(t)]^2= |OA|^2+t^2|OB|^2-2tOA.OB
= 4t^2-8tcos∠AOB + 4
([f(t)]^2)' = 8t -8cos∠AOB =0
t = cos∠AOB
min f(t) at t= cos∠AOB
f(cos∠AOB) =√[4-4(cos∠AOB)^2] =√3
4-4(cos∠AOB)^2=3
cos∠AOB = 1/2 or -1/2 (rejected)
∠AOB= π/3
|OC|^2 = x^2|OA|^2 +y^2|OB|^2 + 2xy|OA||OB|cos∠AOB
= 4x^2 +4y^2+4xy
= 4(x+y)^2-4xy
= 4- 4xy
>= 4- 4((x+y)/2)^2
= 4- 1
=3
min |OC| = √3