《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

3. (多选)如图所示,平面内的两条相交直线$OP_{1}$和$OP_{2}$将该平面分割
成四个部分Ⅰ,Ⅱ,Ⅲ,Ⅳ(不包括
边界).若$\overrightarrow{OP}=a\overrightarrow{OP_{1}}+b\overrightarrow{OP_{2}}$,当
$ab＜0$时,点$P$可能落在第
部分.(AC)


A.Ⅰ
B.Ⅱ
C.Ⅲ
D.Ⅳ

答案： 3.AC 以$\overrightarrow{OP_1},\overrightarrow{OP_2}$为基.当$a ＞ 0,b ＜ 0$时,$P$点在第Ⅲ部分,当$a ＜ 0,b ＞ 0$时,$P$点在第Ⅰ部分,故选AC.

4. 如图所示的矩形$ABCD$中,$E$,$F$满足$\overrightarrow{BE}=\overrightarrow{EC}$,$\overrightarrow{CF}=$
$2\overrightarrow{FD}$,$G$为$EF$的中点,若$\overrightarrow{AG}=\lambda\overrightarrow{AB}+\mu\overrightarrow{AD}$,则$\lambda\mu$
的值为
(A)


A.$\frac{1}{2}$
B.3
C.$\frac{3}{4}$
D.2

答案： 4.A 因为$\overrightarrow{BE} =\overrightarrow{EC},\overrightarrow{CF} = 2\overrightarrow{FD},G$为$EF$的中点,所以$\overrightarrow{AG} =\frac{1}{2}\overrightarrow{AE} +\frac{1}{2}\overrightarrow{AF} =\frac{1}{2}(\overrightarrow{AB} + \overrightarrow{BE}) + \frac{1}{2}(\overrightarrow{AD} + \overrightarrow{DF}) =\frac{1}{2}(\overrightarrow{AB} + \frac{1}{2}\overrightarrow{BC}) +\frac{1}{2}(\overrightarrow{AD} + \frac{1}{3}\overrightarrow{DC}) =\frac{1}{2}(\overrightarrow{AB} + \frac{1}{2}\overrightarrow{AD}) + \frac{1}{2}(\overrightarrow{AD} + \frac{1}{3}\overrightarrow{AB}) =\frac{2}{3}\overrightarrow{AB} + \frac{3}{4}\overrightarrow{AD}$,所以$\lambda =\frac{2}{3},\mu =\frac{3}{4}$,所以$\lambda\mu =\frac{2}{3}×\frac{3}{4} =\frac{1}{2}$.故选A.

5. 已知$O$为$\triangle ABC$内一点,且$\overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{AO}$,且$\lambda\overrightarrow{AD}=$
$\overrightarrow{AC}$,若$B$,$O$,$D$三点共线,则实数$\lambda$的值为3
_.

答案： 5.3 设点$E$为边$BC$的中点,则
　$\frac{1}{2}(\overrightarrow{OB} + \overrightarrow{OC}) =\overrightarrow{OE}$,
　由题意,得$\overrightarrow{AO} =\overrightarrow{OE}$,
　所以$\overrightarrow{AO} =\frac{1}{2}\overrightarrow{AE} =\frac{1}{4}(\overrightarrow{AB} + \overrightarrow{AC}) =\frac{1}{4}\overrightarrow{AB} + \frac{1}{4}\overrightarrow{AD}$,因此若$B,O,D$三点共线,则$\frac{1}{4} + \frac{\lambda}{4} = 1$,即$\lambda = 3$.

6. 如图,经过$\triangle OAB$的重心$G$的直线与$OA$,$OB$分别交于
点$P$,$Q$,设$\overrightarrow{OP}=m\overrightarrow{OA}$,$\overrightarrow{OQ}=n\overrightarrow{OB}$,$m,n\in\mathbf{R}$,则$\frac{1}{m}+\frac{1}{n}$
的值为
_.3

答案： 6.3 方法一:设$\overrightarrow{OA} =\boldsymbol{a},\overrightarrow{OB} =\boldsymbol{b}$,由题意知$\overrightarrow{OG} =\frac{2}{3}×\frac{1}{2}(\overrightarrow{OA} +\overrightarrow{OB}) =\frac{1}{3}(\boldsymbol{a} + \boldsymbol{b})$,$\overrightarrow{PQ} =\overrightarrow{OQ} - \overrightarrow{OP} =\boldsymbol{n}\boldsymbol{b} - \boldsymbol{m}\boldsymbol{a}$,$\overrightarrow{PG} =\overrightarrow{OG} - \overrightarrow{OP} =(\frac{1}{3} - \boldsymbol{m})\boldsymbol{a} + \frac{1}{3}\boldsymbol{b}$,
　由$P,G,Q$三点共线得,存在实数$\lambda$,使得$\overrightarrow{PQ} =\lambda\overrightarrow{PG}$,即$\boldsymbol{n}\boldsymbol{b} -\boldsymbol{m}\boldsymbol{a} =\lambda[(\frac{1}{3} - \boldsymbol{m})\boldsymbol{a} + \frac{1}{3}\boldsymbol{b}]$,
　从而$\begin{cases} -\boldsymbol{m} =\lambda(\frac{1}{3} - \boldsymbol{m}),\\\boldsymbol{n} =\frac{1}{3}\lambda\end{cases}$
　消去$\lambda$,得$\frac{1}{\boldsymbol{m}} + \frac{1}{\boldsymbol{n}} = 3$.
　方法二:由题意知$\overrightarrow{OG} =\frac{2}{3}×\frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) =\frac{1}{3}(\overrightarrow{OP} + \overrightarrow{OQ}) =\frac{1}{3\boldsymbol{m}}\overrightarrow{OP} + \frac{1}{3\boldsymbol{n}}\overrightarrow{OQ}$,
　又$P,G,Q$三点共线,由三点共线性质定理可知$\frac{1}{3\boldsymbol{m}} + \frac{1}{3\boldsymbol{n}} = 1$,
　即$\frac{1}{\boldsymbol{m}} + \frac{1}{\boldsymbol{n}} = 3$.
　方法三:(特例)当$PQ// AB$时,$m = n =\frac{2}{3}$,$\therefore\frac{1}{\boldsymbol{m}} + \frac{1}{\boldsymbol{n}} = 3$.

7. 设$\boldsymbol{e}_{1}$,$\boldsymbol{e}_{2}$是不共线的非零向量,且$\boldsymbol{a}=\boldsymbol{e}_{1}-2\boldsymbol{e}_{2}$,$\boldsymbol{b}=$
$\boldsymbol{e}_{1}+3\boldsymbol{e}_{2}$.
(1)证明:$\boldsymbol{a}$,$\boldsymbol{b}$可以作为一组基底;
(2)以$\boldsymbol{a}$,$\boldsymbol{b}$为基底,求向量$\boldsymbol{c}=3\boldsymbol{e}_{1}-\boldsymbol{e}_{2}$的分解式;
(3)若$4\boldsymbol{e}_{1}-3\boldsymbol{e}_{2}=\lambda\boldsymbol{a}+\mu\boldsymbol{b}$,求$\lambda$,$\mu$的值.

答案： 7.
(1)证明:若$\boldsymbol{a},\boldsymbol{b}$共线,则存在$\lambda\in\mathbf{R}$,使$\boldsymbol{a} =\lambda\boldsymbol{b}$,则$\boldsymbol{e}_1 - 2\boldsymbol{e}_2 =\lambda(\boldsymbol{e}_1 + 3\boldsymbol{e}_2)$.
　由$\boldsymbol{e}_1,\boldsymbol{e}_2$不共线,
　得$\begin{cases}\lambda = 1,\\3\lambda = - 2\end{cases}\Rightarrow\begin{cases}\lambda = 1,\\\lambda = -\frac{2}{3}\end{cases}$
　$\therefore\lambda$不存在,故$\boldsymbol{a}$与$\boldsymbol{b}$不共线,可以作为一组基底.
(2)设$\boldsymbol{c} = m\boldsymbol{a} + n\boldsymbol{b}(m,n\in\mathbf{R})$,
　则$3\boldsymbol{e}_1 - \boldsymbol{e}_2 = m(\boldsymbol{e}_1 - 2\boldsymbol{e}_2) + n(\boldsymbol{e}_1 + 3\boldsymbol{e}_2) = (m + n)\boldsymbol{e}_1 + ( - 2m + 3n)\boldsymbol{e}_2$.
　$\begin{cases}m + n = 3,\\ - 2m + 3n = - 1\end{cases}\Rightarrow\begin{cases}m = 2,\\n = 1\end{cases}$
　$\therefore\boldsymbol{c} = 2\boldsymbol{a} + \boldsymbol{b}$.
(3)由$4\boldsymbol{e}_1 - 3\boldsymbol{e}_2 =\lambda\boldsymbol{a} + \mu\boldsymbol{b}$,得$4\boldsymbol{e}_1 - 3\boldsymbol{e}_2 =\lambda(\boldsymbol{e}_1 - 2\boldsymbol{e}_2) + \mu(\boldsymbol{e}_1 +3\boldsymbol{e}_2) = (\lambda + \mu)\boldsymbol{e}_1 + ( - 2\lambda + 3\mu)\boldsymbol{e}_2$.
　$\begin{cases}\lambda + \mu = 4,\\ - 2\lambda + 3\mu = - 3\end{cases}\Rightarrow\begin{cases}\lambda = 3,\\ \mu = 1\end{cases}$
　故所求$\lambda,\mu$的值分别为$3$和$1$.

8. 如图所示,在$\triangle ABC$中,$M$是$AB$的中点,且$\overrightarrow{AN}=$
$\frac{1}{3}\overrightarrow{AC}$,$BN$与$CM$相交于点$E$,设$\overrightarrow{AB}=\boldsymbol{a}$,$\overrightarrow{AC}=\boldsymbol{b}$,试用基底
$\{\boldsymbol{a},\boldsymbol{b}\}$表示向量$\overrightarrow{AE}$.

答案： 8.易得$\overrightarrow{AN} =\frac{1}{3}\overrightarrow{AC} =\frac{1}{3}\boldsymbol{b}$,$\overrightarrow{AM} =\frac{1}{2}\overrightarrow{AB} =\frac{1}{2}\boldsymbol{a}$,
　由$N,E,B$三点共线可知,存在实数$m$使$\overrightarrow{AE} = m\overrightarrow{AN} + (1 - m)\overrightarrow{AB} =\frac{1}{3}m\boldsymbol{b} + (1 - m)\boldsymbol{a}$.
　由$C,E,M$三点共线可知,存在实数$n$使$\overrightarrow{AE} = n\overrightarrow{AM} + (1 - n)\overrightarrow{AC} =\frac{1}{2}n\boldsymbol{a} + (1 - n)\boldsymbol{b}$.
　所以$\frac{1}{3}m\boldsymbol{b} + (1 - m)\boldsymbol{a} =\frac{1}{2}n\boldsymbol{a} + (1 - n)\boldsymbol{b}$,由于$\{\boldsymbol{a},\boldsymbol{b}\}$为基底,
　所以$\begin{cases}1 - m =\frac{1}{2}n,\\ \frac{1}{3}m = 1 - n,\end{cases}$
　解得$\begin{cases}m =\frac{3}{5},\\n =\frac{4}{5}\end{cases}$ 所以$\overrightarrow{AE} =\frac{2}{5}\boldsymbol{a} + \frac{1}{5}\boldsymbol{b}$.