浙江 2013 · 数学 q17

题面

17．（4分）（2013•浙江）设$\overrightarrow{e_{1}}$、$\overrightarrow{e_{2}}$为单位向量，非零向量$\overrightarrow{b} = x \overrightarrow{e_{1}} + y \overrightarrow{e_{2}} , \ x , y \in R$．若$\overrightarrow{e_{1}}$、$\overrightarrow{e_{2}}$的夹角为$3 0^{\circ}$，则$\frac{\vert x \vert}{\vert \overrightarrow{\mathrm{b}} \vert}$的最大值等于

答案

2

解析

分析

由题意求得$\overrightarrow{e_{1}} \cdot \overrightarrow{e_{2}} = \frac{\sqrt{3}}{2} , \ \vert \overrightarrow{\mathrm{b}} \vert = \sqrt{\overrightarrow{\mathrm{b}}^{2}} = \sqrt{x^{2} + \sqrt{3} x y + y^{2}}$，从而可得$\frac{| x |}{| b |} = \frac{| x |}{\sqrt{x^{2} + \sqrt{3} x y + y^{2}}} = \sqrt{\frac{x^{2}}{x^{2} + \sqrt{3} x y + y^{2}}} = \sqrt{\frac{1}{( \frac{y}{x} + \frac{\sqrt{3}}{2} )^{2} + \frac{1}{4}}}$，再利用二次函数的性质求得$\frac{\vert x \vert}{\vert b \vert}$的最大值

解答

解：$\because \overrightarrow{e_{1}}$、$\overrightarrow{e_{2}}$为单位向量，$\overrightarrow{e_{1}}$和$\overrightarrow{e_{2}}$的夹角等于$3 0^{\circ}$，$\therefore \overrightarrow{e_{1}} \bullet \overrightarrow{e_{2}} = 1 \times 1 \times \cos 3 0^{\circ} = \frac{\sqrt{3}}{2}$

∵非零向量$\overrightarrow{b} = x \overrightarrow{e_{1}} + y \overrightarrow{e_{2}}$，$\therefore | \overrightarrow{b} | = \sqrt{\overrightarrow{b}^{2}} = \sqrt{x^{2} + 2 x y \overrightarrow{e_{1}} \bullet \overrightarrow{e_{2}} + y^{2}} = \sqrt{x^{2} + \sqrt{3} x y + y^{2}}$，

$\therefore \frac{| x |}{| b |} = \frac{| x |}{\sqrt{x^{2} + \sqrt{3} x y + y^{2}}} = \sqrt{\frac{x^{2}}{x^{2} + \sqrt{3} x y + y^{2}}} = \sqrt{\frac{1}{1 + \sqrt{3} \bullet \frac{y}{x} + ( \frac{y}{x} )^{2}}} = \sqrt{\frac{1}{( \frac{y}{x} + \frac{\sqrt{3}}{2} )^{2} + \frac{1}{4}}}$，

故当$\frac{y}{x} = - \frac{\sqrt{3}}{2}$时，$\frac{| x |}{| b |}$取得最大值为2，

故答案为 2