每日一题_191016

若平面向量(left| oldsymbol{a} ight|=2),(left| oldsymbol{b} ight|=3),(left| oldsymbol{e} ight|=1),且(oldsymbol{a}cdotoldsymbol{b}-oldsymbol{e}cdotleft(oldsymbol{a}+oldsymbol{b} ight)+1=0).则(left| oldsymbol{a}-oldsymbol{b} ight|)的最小值是((qquad))
(mathrm{A}.1) (qquadmathrm{B}.2sqrt{3}-1) (qquadmathrm{C}.sqrt{12-4sqrt{3}}) (qquadmathrm{D}.sqrt{7})

解析: 法一 矩形的性质 由题设$$(oldsymbol{a},oldsymbol{b},oldsymbol{e})=left( overrightarrow{OA} , overrightarrow{OB} , overrightarrow{OE} ight)$$如图所示,


易知$EAperp EB$,构造矩形$FAEB$,不妨固定$E$点,则由矩形性质可知$$ |OF|^2+|OE|^2=|OA|^2+|OB|^2.$$所以$F$点的轨迹方程为$$x^2+y^2=12.$$从而$|oldsymbol{a}-oldsymbol{b}|$,即$|AB|$,也即$|EF|$的最小值为$2sqrt{3}-1$.

法二 由题$$
|oldsymbol{a}cdot oldsymbol{b}+1|=|oldsymbol{e}cdotleft(oldsymbol{a}+oldsymbol{b} ight)|leqslant |oldsymbol{a}+oldsymbol{b}|.$$两边平方可得$$
left(oldsymbol{a}cdotoldsymbol{b} ight)^2+2cdotoldsymbol{a}cdotoldsymbol{b}+1leqslant oldsymbol{a}^{2+oldsymbol{b}}2+2cdot oldsymbol{a}cdotoldsymbol{b}. $$所以$$-2sqrt3leqslant oldsymbol{a}cdot oldsymbol{b}leqslant 2sqrt{3}.$$因此$$
|oldsymbol{a}-oldsymbol{b}|=sqrt{oldsymbol{a}^{2+oldsymbol{b}}2-2cdot oldsymbol{a}cdot oldsymbol{b}}geqslant sqrt{13-4sqrt{3}}=2sqrt{3}-1.$$
当(oldsymbol{e})与(oldsymbol{a}+oldsymbol{b})同向时取等.因此所求表达式的最小值为(2sqrt{3}-1).