在平面上,\(\overrightarrow{AB_1}\perp\overrightarrow{AB_2}\),\(\left| \overrightarrow{OB_1}\right|=\left| \overrightarrow{OB_2} \right|=1\),\(\overrightarrow{AP}=\overrightarrow{AB_1}+\overrightarrow{AB_2}\).若\(\left|\overrightarrow{OP} \right|<\dfrac{1}{2}\),则\(\left| \overrightarrow{OA} \right|\)的取值范围是$(\qquad) \(
\)\mathrm{A}.\left(0,\dfrac{\sqrt{5}}{2} \right]$ \(\qquad\mathrm{B}.\left( \dfrac{\sqrt{5}}{2},\dfrac{\sqrt{7}}{2}\right]\) \(\qquad\mathrm{C}.\left(\dfrac{\sqrt{5}}{2},\sqrt{2}\right]\) \(\qquad\mathrm{D}.\left(\dfrac{\sqrt{7}}{2},\sqrt{2}\right]\)}
解析:
由于\(\angle B_1AB_2\)是直角,并且$$
\overrightarrow{AP}=\overrightarrow{AB_1}+\overrightarrow{AB_2}.$$所以四边形\(AB_1PB_2\)为矩形,如图所示.
结合矩形性质可知$$
|OA|2+|OP|2=|OB_1|2+|OB_2|2.$$所以\(|\overrightarrow{OA}|\)的取值范围为\(\left(\dfrac{\sqrt{7}}{2},\sqrt{2}\right]\).
