题目内容
3.分析 由已知可得$\overrightarrow{BF}$=$\overrightarrow{BD}$+$\overrightarrow{DF}$,$\overrightarrow{CF}$=-$\overrightarrow{BD}$+$\overrightarrow{DF}$,$\overrightarrow{BA}$=$\overrightarrow{BD}$+3$\overrightarrow{DF}$,$\overrightarrow{CA}$=-$\overrightarrow{BD}$+3$\overrightarrow{DF}$,$\overrightarrow{BE}$=$\overrightarrow{BD}$+2$\overrightarrow{DF}$,$\overrightarrow{CE}$=-$\overrightarrow{BD}$+2$\overrightarrow{DF}$,结合已知求出$\overrightarrow{DF}$2=$\frac{5}{8}$,$\overrightarrow{BD}$2=$\frac{13}{8}$,可得答案.
解答 解:∵D是BC的中点,E,F是AD上的两个三等分点,
∴$\overrightarrow{BF}$=$\overrightarrow{BD}$+$\overrightarrow{DF}$,$\overrightarrow{CF}$=-$\overrightarrow{BD}$+$\overrightarrow{DF}$,
$\overrightarrow{BA}$=$\overrightarrow{BD}$+3$\overrightarrow{DF}$,$\overrightarrow{CA}$=-$\overrightarrow{BD}$+3$\overrightarrow{DF}$,
∴$\overrightarrow{BF}$•$\overrightarrow{CF}$=$\overrightarrow{DF}$2-$\overrightarrow{BD}$2=-1,
$\overrightarrow{BA}$•$\overrightarrow{CA}$=9$\overrightarrow{DF}$2-$\overrightarrow{BD}$2=4,
∴$\overrightarrow{DF}$2=$\frac{5}{8}$,$\overrightarrow{BD}$2=$\frac{13}{8}$,
又∵$\overrightarrow{BE}$=$\overrightarrow{BD}$+2$\overrightarrow{DF}$,$\overrightarrow{CE}$=-$\overrightarrow{BD}$+2$\overrightarrow{DF}$,
∴$\overrightarrow{BE}$•$\overrightarrow{CE}$=4$\overrightarrow{DF}$2-$\overrightarrow{BD}$2=$\frac{7}{8}$,
故答案为:$\frac{7}{8}$
点评 本题考查的知识是平面向量的数量积运算,平面向量的线性运算,难度中档.
| A. | $\frac{1}{3}$ | B. | $\frac{1}{2}$ | C. | $\frac{2}{3}$ | D. | $\frac{3}{4}$ |
| A. | 0 | B. | m | C. | 2m | D. | 4m |
| A. | 1 | B. | 2 | C. | $\sqrt{2}$ | D. | 2$\sqrt{2}$ |