《2025年教材帮高中数学选择性必修第二册苏教版》

4. 如图6.1-30所示,在正方体$ABCD-A_1B_1C_1D_1$中,$\overrightarrow{AA_1}=\boldsymbol{a}$,$\overrightarrow{AB}=\boldsymbol{b}$,$\overrightarrow{AD}=\boldsymbol{c}$,点M,N分别是$A_1D$,$B_1D_1$的中点.

(1)试用$\boldsymbol{a}$,$\boldsymbol{b}$,$\boldsymbol{c}$表示$\overrightarrow{MN}$;
(2)求证:MN//平面$ABB_1A_1$.

答案： 4.
(1)因为$\overrightarrow{A_1D}=\overrightarrow{AD}-\overrightarrow{AA_1}=c - a$，所以$\overrightarrow{A_1M}=\frac{1}{2}\overrightarrow{A_1D}=\frac{1}{2}(c - a)$。连接$A_1N$，则$\overrightarrow{A_1N}=\frac{1}{2}(\overrightarrow{A_1B_1}+\overrightarrow{A_1D_1})=\frac{1}{2}(b + c)$，所以$\overrightarrow{MN}=\overrightarrow{A_1N}-\overrightarrow{A_1M}=\frac{1}{2}(b + c)-\frac{1}{2}(c - a)=\frac{1}{2}(b + a)=\frac{1}{2}a+\frac{1}{2}b$。\n
(2)连接$AB_1$，因为$\overrightarrow{AB_1}=\overrightarrow{AA_1}+\overrightarrow{AB}=a + b$，所以$\overrightarrow{MN}=\frac{1}{2}\overrightarrow{AB_1}$，即$MN// AB_1$。\n因为$AB_1\subset$平面$ABB_1A_1$，$MN\not\subset$平面$ABB_1A_1$，所以$MN//$平面$ABB_1A_1$。

例21 如图6.1-31,在三棱锥A-BCD中,DA,DB,DC两两垂直,且DB=DC=DA,E为BC的中点.证明:AE⊥BC.

答案： 解析 因为$\overrightarrow{AE}=\overrightarrow{DE}-\overrightarrow{DA}=\dfrac{1}{2}(\overrightarrow{DB}+\overrightarrow{DC})-\overrightarrow{DA}$,$\overrightarrow{BC}=\overrightarrow{DC}-\overrightarrow{DB}$,
所以$\overrightarrow{AE}·\overrightarrow{BC}=\left(\dfrac{1}{2}\overrightarrow{DB}+\dfrac{1}{2}\overrightarrow{DC}-\overrightarrow{DA}\right)·(\overrightarrow{DC}-\overrightarrow{DB})=\dfrac{1}{2}\overrightarrow{DB}·\overrightarrow{DC}-\dfrac{1}{2}\overrightarrow{DB}·\overrightarrow{DB}+\dfrac{1}{2}\overrightarrow{DC}·\overrightarrow{DC}-\dfrac{1}{2}\overrightarrow{DC}·\overrightarrow{DB}-\overrightarrow{DA}·\overrightarrow{DC}+\overrightarrow{DA}·\overrightarrow{DB}=0$,
所以$\overrightarrow{AE}\perp\overrightarrow{BC}$,即AE⊥BC.

例22 如图6.1-32所示,在正方体$ABCD-A_1B_1C_1D_1$中,O为AC与BD的交点,G为$CC_1$的中点,求证:$A_1O\perp$平面GBD.

答案： 解析 连接OG,设$\overrightarrow{A_1B_1}=\boldsymbol{a}$,$\overrightarrow{A_1D_1}=\boldsymbol{b}$,$\overrightarrow{A_1A}=\boldsymbol{c}$,
则$\boldsymbol{a}·\boldsymbol{b}=0$,$\boldsymbol{b}·\boldsymbol{c}=0$,$\boldsymbol{a}·\boldsymbol{c}=0$.
∵ $\overrightarrow{A_1O}=\overrightarrow{A_1A}+\overrightarrow{AO}=\overrightarrow{A_1A}+\dfrac{1}{2}(\overrightarrow{AB}+\overrightarrow{AD})=\boldsymbol{c}+\dfrac{1}{2}(\boldsymbol{a}+\boldsymbol{b})$,$\overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}=\boldsymbol{b}-\boldsymbol{a}$,$\overrightarrow{OG}=\overrightarrow{OC}+\overrightarrow{CG}=\dfrac{1}{2}(\overrightarrow{AB}+\overrightarrow{AD})+\dfrac{1}{2}\overrightarrow{CC_1}=\dfrac{1}{2}(\boldsymbol{a}+\boldsymbol{b})-\dfrac{1}{2}\boldsymbol{c}$,
∴ $\overrightarrow{A_1O}·\overrightarrow{BD}=\left(\boldsymbol{c}+\dfrac{1}{2}\boldsymbol{a}+\dfrac{1}{2}\boldsymbol{b}\right)·(\boldsymbol{b}-\boldsymbol{a})=\boldsymbol{c}·(\boldsymbol{b}-\boldsymbol{a})+\dfrac{1}{2}(\boldsymbol{a}+\boldsymbol{b})·(\boldsymbol{b}-\boldsymbol{a})=\boldsymbol{c}·\boldsymbol{b}-\boldsymbol{c}·\boldsymbol{a}+\dfrac{1}{2}(\boldsymbol{b}^2-\boldsymbol{a}^2)=\dfrac{1}{2}(|\boldsymbol{b}|^2-|\boldsymbol{a}|^2)=0$,([另解]也可用三垂线定理求解,$A_1O$在平面ABCD上的射影为AO,AO⊥BD,即$A_1O\perp BD$)
$\overrightarrow{A_1O}·\overrightarrow{OG}=\left(\boldsymbol{c}+\dfrac{1}{2}\boldsymbol{a}+\dfrac{1}{2}\boldsymbol{b}\right)·\left(\dfrac{1}{2}\boldsymbol{a}+\dfrac{1}{2}\boldsymbol{b}-\dfrac{1}{2}\boldsymbol{c}\right)=\dfrac{1}{4}(\boldsymbol{a}+\boldsymbol{b})^2+\dfrac{1}{4}\boldsymbol{c}·(\boldsymbol{a}+\boldsymbol{b})-\dfrac{1}{2}\boldsymbol{c}^2=\dfrac{1}{4}(\boldsymbol{a}^2+\boldsymbol{b}^2)-\dfrac{1}{2}\boldsymbol{c}^2=\dfrac{1}{4}(|\boldsymbol{a}|^2+|\boldsymbol{b}|^2)-\dfrac{1}{2}|\boldsymbol{c}|^2=0$.
∴ $A_1O\perp BD$,$A_1O\perp OG$.
又BD∩OG=O,
∴ $A_1O\perp$平面GBD.