Question

已知平面向量α,β（α≠0,α≠β）满足︳β︳=1,且α与β-α的夹角为120°,︳α︳取值范围?

Answer 1

let α与β的夹角=x
| β-α|^2 = | β|^2+|α|^2 - 2| β||α|cosx
= 1+ |α|^2-2|α|cosx
α.( β-α)= |α|| β-α|cos120°
|α||β|cosx - |α|^2 = (-1/2)|α|√(1+ |α|^2-2|α|cosx)
cosx - |α| = (-1/2)√(1+ |α|^2-2|α|cosx)
4(cosx - |α|)^2 = 1+ |α|^2-2|α|cosx
3|α|^2-6cosx|α|+(4(cosx)^2-1) =0
|α|= {6cosx+√[(6cosx)^2+12(4(cosx)^2-1)]}/6
= {2cosx+√[21(cosx)^2-3]}/3
max |α| at cosx=1
max|α| = (2+√18)/3