宝鸡中学2017届高三总复习同步微专题21:平面向量之“建系”

建立直角坐标系,把向量运算坐标化,是一个非常常见的思路。本微专题以4道宝鸡中学2017届高三同步复习题为例予以说明。


1.(理-考点规范练26-15)已知正方形ABCD的边长为2,\overrightarrow{DE}= 2\overrightarrow{EC}\overrightarrow{DF} = \dfrac12(\overrightarrow{DC} + \overrightarrow{DB}),则\overrightarrow{BE} \cdot \overrightarrow{DF} = \underline{\hbox to 10mm{}}

分析与解    以\overrightarrow{AB}方向为x轴正方形,A为原点建系:

    易知各点坐标为:B(2,0)E \left( \dfrac43,2 \right)D(0,2)F(2,1)

    故\overrightarrow{BE} = \left(-\dfrac23,2 \right)\overrightarrow{DF} = (2,-1)\therefore \overrightarrow{BE} \cdot \overrightarrow{DF} = -\dfrac43 - 2 = -\dfrac{10}{3}


2.(理-考点规范练26-16)已知直角梯形ABCD中,AD \parallel BC\angle ADC = 90^\circAD = 2BC = 1P是腰DC上的动点,则|\overrightarrow{PA} + 3\overrightarrow{PB}|的最小值为\underline{\hbox to 10mm{}}

分析与解    如图建立直角坐标系:

    易知A(2,0),设B(1,m)P(0,t),其中m>00 \le t \le m

    则\overrightarrow{PA} = (2,-t)\overrightarrow{PB} = (1,m-t)\therefore \overrightarrow{PA} + 3\overrightarrow{PB} =(5,3m-4t)

    令y = |\overrightarrow{PA} + 3\overrightarrow{PB}|^2 = 16t^2-24mt+9m^2+25, 0 \le t \le m,则y在对称轴t = \dfrac{3m}{4}处取到最小值y_{min} = 25\therefore |\overrightarrow{PA} + 3\overrightarrow{PB}|_{min} = 5


3.(理-考点规范练14,文-P252-1)已知\overrightarrow{AB} \bot \overrightarrow{AC}|\overrightarrow{AB}| = \dfrac{1}{t}|\overrightarrow{AC}| = t。若点P\triangle ABC所在平面内的一点,且\overrightarrow{AP} = \dfrac{\overrightarrow{AB}}{|\overrightarrow{AB}|} + \dfrac{4\overrightarrow{AC}}{|\overrightarrow{AC}|},则\overrightarrow{PB} \cdot \overrightarrow{PC}的最大值等于(   )

    A.13    B.15    C.19    D.21

分析与解    以\overrightarrow{AB}方向为x轴正方向,以A点为原点建系:

    则P(1,4)B\left( \dfrac{1}{t},0 \right)C(0,t)

    \therefore \overrightarrow{PB} = \left( \dfrac{1}{t}-1,-4 \right)\overrightarrow{PC} = (-1,t-4)

    \therefore \overrightarrow{PB} \cdot \overrightarrow{PC} = 17 - \left( 4t + \dfrac{1}{t} \right) \le 13。选A


4.(文-P251-11)已知三角形ABC中,AB = ACBC = 4\angle BAC = 120^\circ\overrightarrow{BE} = 3 \overrightarrow{EC},若PBC边上的动点,则\overrightarrow{AP} \cdot \overrightarrow{AE}的取值范围是\underline{\hbox to 10mm{}}

分析    以BCx轴,以BC的中点D为原点建系,很容易表示出各点的坐标。

      如图建系:

    容易求得A\left(0,\dfrac{2}{\sqrt3}\right)B(-2,0)C(2,0)E(1,0),设P(x,0)-2 \le x \le 2

    则\overrightarrow{AP} = \left(x, -\dfrac{2}{\sqrt3} \right)\overrightarrow{AE} = \left( 1, -\dfrac{2}{\sqrt3} \right)

    则\overrightarrow{AP} \cdot \overrightarrow{AE} = x + \dfrac43 \in \left[-\dfrac23,\dfrac{10}{3}\right]

说明    本题找“基底”的解法参见宝鸡中学2017届高三总复习同步微专题22:平面向量之“基底”

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