Question

8．在边长为1的正方体ABCD-A′B′C′D′中，E，F，G分别在BB′，BC，BA上，并且满足$\overrightarrow{BE}=\frac{3}{4}\overrightarrow{BB'}$，$\overrightarrow{BF}=\frac{1}{2}\overrightarrow{BC}$，$\overrightarrow{BG}=\frac{1}{2}\overrightarrow{BA}$．若平面AB′F，平面ACE，平面B′CG交于一点O，$\overrightarrow{BO}=x\overrightarrow{BG}+y\overrightarrow{BF}+z\overrightarrow{BE}$，则x+y+z=$\frac{4}{3}$，$|\overrightarrow{OD}|$=$\frac{\sqrt{59}}{6}$．

Answer 1

分析 根据四点共面列出方程组解出x，y，z，用两两垂直的向量$\overrightarrow{BA}，\overrightarrow{BC}，\overrightarrow{BB′}$表示出$\overrightarrow{OD}$，计算${\overrightarrow{OD}}^{2}$开方即为|$\overrightarrow{OD}$|．

解答 解$\overrightarrow{BO}=x\overrightarrow{BG}+y\overrightarrow{BF}+z\overrightarrow{BE}$=$\frac{x}{2}\overrightarrow{BA}+y\overrightarrow{BF}+\frac{3z}{4}\overrightarrow{B{B}^{'}}$=$\frac{x}{2}\overrightarrow{BA}+\frac{y}{2}\overrightarrow{BC}+z\overrightarrow{BE}$=x$\overrightarrow{BG}+\frac{y}{2}\overrightarrow{BC}+\frac{3z}{4}\overrightarrow{B{B}^{'}}$．
∵O，A，B′，F四点共面，O，A，C，E四点共面，O，B′，C，G四点共面，
∴$\left\{\begin{array}{l}{\frac{x}{2}+y+\frac{3z}{4}=1}\\{\frac{x}{2}+\frac{y}{2}+z=1}\\{x+\frac{y}{2}+\frac{3z}{4}=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=\frac{1}{3}}\\{y=\frac{1}{3}}\\{z=\frac{2}{3}}\end{array}\right.$，
∴x+y+z=$\frac{4}{3}$．
∴$\overrightarrow{BO}$=$\frac{1}{6}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{B{B}^{'}}$．
∵$\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{BC}$，
∴$\overrightarrow{OD}=\overrightarrow{BD}-\overrightarrow{BO}$=$\frac{5}{6}\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}-\frac{1}{2}\overrightarrow{B{B}^{'}}$，
∴${\overrightarrow{OD}}^{2}$=（$\frac{5}{6}\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}-\frac{1}{2}\overrightarrow{B{B}^{'}}$）²=$\frac{25}{36}{\overrightarrow{BA}}^{2}+\frac{25}{36}{\overrightarrow{BC}}^{2}+\frac{1}{4}{\overrightarrow{BB′}}^{2}$=$\frac{25}{36}+\frac{25}{36}+\frac{1}{4}$=$\frac{59}{36}$．
∴|$\overrightarrow{OD}$|=$\frac{\sqrt{59}}{6}$．
故答案为：$\frac{4}{3}$，$\frac{\sqrt{59}}{6}$．

点评 本题考查了平面向量线性运算的几何意义，平面向量的基本定理，属于中档题．