《2025年新课程同步导学高中数学选择性必修第二册湘教版》

例1 在平行六面体$ABCD - A_{1}B_{1}C_{1}D_{1}$中，$O$为$AC$的中点.
(1)化简：$A_{1}O-\frac{1}{2}\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}$；
(2)设$E$是棱$DD_{1}$上的点，且$\overrightarrow{DE}=\frac{2}{3}\overrightarrow{DD_{1}}$，若$\overrightarrow{EO}=x\overrightarrow{AB}+y\overrightarrow{AD}+z\overrightarrow{AA_{1}}$，试求实数$x,y,z$的值.

(1)$-\overrightarrow{AA_{1}}$；
(2)$x = \frac{1}{2}$，$y = -\frac{1}{2}$，$z = -\frac{2}{3}$.

答案：
(1) $\overrightarrow{A_{1}O} = \overrightarrow{A_{1}A} + \overrightarrow{AO}$，$O$为$AC$中点，$\overrightarrow{AO} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD})$，$\overrightarrow{A_{1}A} = -\overrightarrow{AA_{1}}$，则$\overrightarrow{A_{1}O} = -\overrightarrow{AA_{1}} + \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD})$。
$\overrightarrow{A_{1}O} - \frac{1}{2}\overrightarrow{AB} - \frac{1}{2}\overrightarrow{AD} = -\overrightarrow{AA_{1}} + \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD}) - \frac{1}{2}\overrightarrow{AB} - \frac{1}{2}\overrightarrow{AD} = -\overrightarrow{AA_{1}}$。
(2) $\overrightarrow{EO} = \overrightarrow{AO} - \overrightarrow{AE}$，$\overrightarrow{AO} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD})$，$\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AD} + \frac{2}{3}\overrightarrow{DD_{1}} = \overrightarrow{AD} + \frac{2}{3}\overrightarrow{AA_{1}}$。
$\overrightarrow{EO} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD}) - (\overrightarrow{AD} + \frac{2}{3}\overrightarrow{AA_{1}}) = \frac{1}{2}\overrightarrow{AB} - \frac{1}{2}\overrightarrow{AD} - \frac{2}{3}\overrightarrow{AA_{1}}$，故$x = \frac{1}{2}$，$y = -\frac{1}{2}$，$z = -\frac{2}{3}$。
(1) $-\overrightarrow{AA_{1}}$；
(2) $x = \frac{1}{2}$，$y = -\frac{1}{2}$，$z = -\frac{2}{3}$。