浙江 2017 · 数学 q10

题面

10．（5分）如图，已知平面四边形 ABCD，A B \bot B C$$A B = B C = A D = 2，CD=3，AC 与BD 交于点O，记$1_{1} = \overrightarrow{O A} \bullet \overrightarrow{O B} , \vert_{2} = \overrightarrow{O B} \bullet \overrightarrow{O C} , \vert_{3} = \overrightarrow{O C} \bullet \overrightarrow{O D} ,$，则（ ）

A$1_{1} < 1_{2} < 1_{3}$ B$1_{1} < 1_{3} < 1_{2}$ C$\vert{\scriptscriptstyle 3} < \vert{\scriptscriptstyle 1} < \vert{\scriptscriptstyle 2}$ D$1_{2} < 1_{1} < 1_{3}$

A$1_{1} < 1_{2} < 1_{3}$B$1_{1} < 1_{3} < 1_{2}$C$\vert{\scriptscriptstyle 3} < \vert{\scriptscriptstyle 1} < \vert{\scriptscriptstyle 2}$D$1_{2} < 1_{1} < 1_{3}$

答案

C

解析

分析

根据向量数量积的定义结合图象边角关系进行判断即可

解答

解：\because A B \bot B C$$A B = B C = A D = 2$$C D = 3$$\therefore A C = 2 \sqrt{2}

$\therefore \angle A O B = \angle C O D > 9 0^{\circ}$

由图象知$0 A < O C , O B < O D$

$\therefore 0 > \overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}} > \overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{OD}}, \quad \overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} > 0,$

即$\vert{_{3}} < \vert{_{1}} < \vert{_{2}}$

故选：C