Known vector \ (\ boldsymbol {\ Alpha} \) , \ (\ boldsymbol {\ Beta}, \ boldsymbol {\ Gamma} \) satisfies \ (\ 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 \) If for each determination. \ (\ boldsymbol {\ Beta} \) , \ (\ left | \ boldsymbol { \ gamma} \ right | \) maximum and minimum values are \ (m, n-\) , then for any \ (\ boldsymbol {\ Beta} \) , \ (Mn \ ) minimum value is \ (\ underline {\ qquad \ qquad} \) .
Analysis:
As shown, provided\ [(\ Boldsymbol {\ alpha }, \ boldsymbol {\ beta}, \ boldsymbol {\ gamma}) = \ left (\ overrightarrow {OA}, \ overrightarrow {OB}, \ overrightarrow {OC} \ right). \ ]
apparent from the title \ (the OB = AB \) , so if a fixed \ (OA \) , then \ (B \) of the trajectory line \ (OA \) in perpendicular.
and \ (CB \ perp CA \) so when \ (\ boldsymbol {\ Beta} \) , i.e. the point \ (B \) is determined, \ (C \) locus of points is to \ (AB \) diameter circle.
Note \ (AB \) midpoint \ (M \) , then \ [mn = 2MC = AB = OB. \] We found that the original title is equivalent to finding \ (the OB \) minimum, obviously \ (the OB \) the minimum value is \ (\ dfrac12 \) , if and only if \ (B \) located \ (OA \) midpoint achieved.