Question

15．在邊長(zhǎng)為1的正△ABC中，點(diǎn)P₁，P₂滿足$\overrightarrow{B{P}_{1}}$=$\overrightarrow{{P}_{1}{P}_{2}}$=$\overrightarrow{{P}_{2}C}$，則$\overrightarrow{AB}$$•\overrightarrow{A{P}_{1}}$$+\overrightarrow{A{P}_{1}}$$•\overrightarrow{A{P}_{2}}$+$\overrightarrow{A{P}_{2}}$$•\overrightarrow{AC}$的值為$\frac{43}{18}$．

Answer 1

分析 利用已知將$\overrightarrow{AB}$$•\overrightarrow{A{P}_{1}}$$+\overrightarrow{A{P}_{1}}$$•\overrightarrow{A{P}_{2}}$+$\overrightarrow{A{P}_{2}}$$•\overrightarrow{AC}$表示為$\overrightarrow{AB}•（\overrightarrow{AB}+\overrightarrow{B{P}_{1}}）+（\overrightarrow{AB}+\overrightarrow{B{P}_{1}}）（\overrightarrow{AB}+\overrightarrow{B{P}_{2}}$）+（$\overrightarrow{AB}+\overrightarrow{B{P}_{2}}）\overrightarrow{AC}$$•\overrightarrow{AC}$，利用等邊三角形的性質(zhì)解答．

解答 解：因?yàn)檫呴L(zhǎng)為1的正△ABC中，點(diǎn)P₁，P₂滿足$\overrightarrow{B{P}_{1}}$=$\overrightarrow{{P}_{1}{P}_{2}}$=$\overrightarrow{{P}_{2}C}$，
則$\overrightarrow{AB}$$•\overrightarrow{A{P}_{1}}$$+\overrightarrow{A{P}_{1}}$$•\overrightarrow{A{P}_{2}}$+$\overrightarrow{A{P}_{2}}$$•\overrightarrow{AC}$=$\overrightarrow{AB}•（\overrightarrow{AB}+\overrightarrow{B{P}_{1}}）+（\overrightarrow{AB}+\overrightarrow{B{P}_{1}}）（\overrightarrow{AB}+\overrightarrow{B{P}_{2}}$）+（$\overrightarrow{AB}+\overrightarrow{B{P}_{2}}）\overrightarrow{AC}$$•\overrightarrow{AC}$
=${\overrightarrow{AB}}^{2}$$+\frac{1}{3}\overrightarrow{AB}•\overrightarrow{BC}$$+{\overrightarrow{AB}}^{2}$$+\frac{2}{3}\overrightarrow{AB}•\overrightarrow{BC}$$+\frac{1}{3}\overrightarrow{BC}•\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}•\frac{2}{3}\overrightarrow{BC}$$+\overrightarrow{AB}•\overrightarrow{AC}+\frac{2}{3}\overrightarrow{BC}•\overrightarrow{AC}$
=1-$\frac{1}{6}$+1-$\frac{1}{3}$-$\frac{1}{6}$+$\frac{2}{9}$+$\frac{1}{2}$+$\frac{1}{3}$
=$\frac{43}{18}$．

點(diǎn)評(píng) 本題考查了向量加法的三角形法則以及向量的數(shù)量積公式的運(yùn)用；屬于基礎(chǔ)題．