已知向量\(\boldsymbol{\alpha}\),\(\boldsymbol{\beta},\boldsymbol{\gamma}\)满足\(\left|\boldsymbol{\alpha}\right|=1\),\(\left| \boldsymbol{\alpha}-\boldsymbol{\beta}\right|=\left| \boldsymbol{\beta}\right|\),\(\left(\boldsymbol{\alpha}-\boldsymbol{\gamma}\right)\left(\boldsymbol{\beta}-\boldsymbol{\gamma}\right)=0\).若对每一确定的\(\boldsymbol{\beta}\),\(\left| \boldsymbol{\gamma}\right|\)的最大值和最小值分别为\(m,n\),则对任意\(\boldsymbol{\beta}\),\(m-n\)的最小值是\(\underline{\qquad\qquad}\).
解析:
如图,设\[ (\boldsymbol{\alpha},\boldsymbol{\beta},\boldsymbol{\gamma})=\left(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}\right).\]
由题可知\(OB=AB\),所以若固定\(OA\),则\(B\)的轨迹为线段\(OA\)的中垂线.
又\(CB\perp CA\),所以当\(\boldsymbol{\beta}\),也即点\(B\)确定时,\(C\)点的轨迹为以\(AB\)为直径的圆.

记\(AB\)中点为\(M\),则\[m-n=2MC=AB=OB.\]因此我们可知原题等价于求\(OB\)的最小值,显然\(OB\)最小值为\(\dfrac12\),当且仅当\(B\)位于\(OA\)的中点取得.