《2025年教材帮高中数学选择性必修第二册苏教版》

例3-7 (2025·湖南省祁阳一中月考) 下列条件，能说明空间中不重合的$A$，$B$，$C$三点共线的是()

A.$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$
B.$\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$
C.$\overrightarrow{AB} = \overrightarrow{BC}$
D.$|\overrightarrow{AB}| = |\overrightarrow{BC}|$

答案： 解析 由$\overrightarrow{AB} = \overrightarrow{BC}$知$\overrightarrow{AB}$与$\overrightarrow{BC}$共线，又$\overrightarrow{AB}$与$\overrightarrow{BC}$有一共同的点$B$，故$A$，$B$，$C$三点共线.
答案 C

例4-8 在如图6.1-15所示的正方体$ABCD - A_1B_1C_1D_1$中，$E$，$F$分别为$AB$，$BC$的中点，求(1) $\langle \overrightarrow{AB}, \overrightarrow{EF} \rangle$; (2) $\langle \overrightarrow{B_1B}, \overrightarrow{EF} \rangle$; (3) $\langle \overrightarrow{C_1A_1}, \overrightarrow{EF} \rangle$; (4) $\langle \overrightarrow{B_1C}, \overrightarrow{EF} \rangle$; (5) $\langle \overrightarrow{D_1B_1}, \overrightarrow{EF} \rangle$; (6) $\langle \overrightarrow{AC}, \overrightarrow{CB_1} \rangle$.

答案： 解析
(1) $\because E$，$F$分别为$AB$，$BC$的中点，$\therefore EF // AC$.
$\because AB$与$AC$成$45°$角，$\therefore \langle \overrightarrow{AB}, \overrightarrow{EF} \rangle = 45°$.
(2) $\because B_1B \perp$平面$ABCD$，$\therefore B_1B \perp EF$，$\therefore \langle \overrightarrow{B_1B}, \overrightarrow{EF} \rangle = 90°$.
(3) $\because A_1C_1 // AC // EF$，$\overrightarrow{C_1A_1}$与$\overrightarrow{EF}$方向相反，$\therefore \langle \overrightarrow{C_1A_1}, \overrightarrow{EF} \rangle = 180°$.
(4) 连接$AB_1$，易知$B_1C = AB_1 = AC$，$\therefore \triangle AB_1C$为等边三角形.
$\therefore B_1C$与$AC$成$60°$角，又$EF // AC$，判定方向后可知$\langle \overrightarrow{B_1C}, \overrightarrow{EF} \rangle = 60°$.
(5) 连接$DB_1$，$\because D_1B_1 // DB_1$，$DB_1 \perp EF$，$\therefore D_1B_1 \perp EF$，$\therefore \langle \overrightarrow{D_1B_1}, \overrightarrow{EF} \rangle = 90°$.
(6) $\because AC$与$CB_1$成$60°$角，$\langle \overrightarrow{CA}, \overrightarrow{CB_1} \rangle = 60°$，$\langle \overrightarrow{AC}, \overrightarrow{CB_1} \rangle = 120°$.
(求向量夹角时，注意向量的方向)