《2025年世纪金榜新视野暑假作业高一数学》

13. 已知$M$,$G分别是空间四边形ABCD的两边BC$,$CD$的中点,化简下列各式.
(1)$\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{BD}+\overrightarrow{BC})$.
(2)$\overrightarrow{AG}-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$.

答案：
(1) 解：因为M是BC的中点，所以$\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BC}$。又因为$\frac{1}{2}\overrightarrow{BD}=\overrightarrow{MD}$（D为BD中点表述不准确，应为$\overrightarrow{BD}=2\overrightarrow{MD}$，M是BC中点，此处应为$\overrightarrow{BM}+\overrightarrow{MD}=\overrightarrow{BD}$，但更简便的是：$\overrightarrow{BD}+\overrightarrow{BC}=2\overrightarrow{BM}+\overrightarrow{BC}$错误，正确应为：$\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{BD}+\overrightarrow{BC})=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BD}+\frac{1}{2}\overrightarrow{BC}$，因为M是BC中点，所以$\frac{1}{2}\overrightarrow{BC}=\overrightarrow{BM}$，$\frac{1}{2}\overrightarrow{BD}=\overrightarrow{BH}$（H为BD中点），但空间四边形中，连接BH、HM，可得$\overrightarrow{BH}+\overrightarrow{BM}=\overrightarrow{BM}+\overrightarrow{BH}=\overrightarrow{BA}+\overrightarrow{AH}+\overrightarrow{BM}$复杂，正确方法：$\overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}$，$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$，所以原式$=\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AB})=\overrightarrow{AB}+\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{AC}-2\overrightarrow{AB})=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{AC})$，又因为G是CD中点，$\overrightarrow{AG}=\frac{1}{2}(\overrightarrow{AC}+\overrightarrow{AD})$，所以原式$=\overrightarrow{AG}$。
(2) 解：因为G是CD的中点，所以$\overrightarrow{AG}=\frac{1}{2}(\overrightarrow{AC}+\overrightarrow{AD})$。则$\overrightarrow{AG}-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{2}(\overrightarrow{AC}+\overrightarrow{AD})-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})=\frac{1}{2}\overrightarrow{AD}-\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}(\overrightarrow{AD}-\overrightarrow{AB})=\frac{1}{2}\overrightarrow{BD}$。
答案：
(1)$\overrightarrow{AG}$；
(2)$\frac{1}{2}\overrightarrow{BD}$

14. 如图,正四面体$V - ABC的高VD的中点为O$,$VC的中点为M$.
(1)求证：$AO$,$BO$,$CO$两两垂直.
(2)求$\langle \overrightarrow{DM},\overrightarrow{AO}\rangle$.

答案：
(1)证明：设正四面体棱长为$2$，以$D$为原点，$DA$为$x$轴，$DC$为$y$轴，$DV$为$z$轴建立坐标系。则$D(0,0,0)$，$A\left(\dfrac{2\sqrt{3}}{3},0,0\right)$，$B\left(-\dfrac{\sqrt{3}}{3},1,0\right)$，$C\left(-\dfrac{\sqrt{3}}{3},-1,0\right)$，$V(0,0,h)$，由$|VA|=2$得$h=\dfrac{2\sqrt{6}}{3}$，$O\left(0,0,\dfrac{\sqrt{6}}{3}\right)$。
$\overrightarrow{AO}=\left(-\dfrac{2\sqrt{3}}{3},0,\dfrac{\sqrt{6}}{3}\right)$，$\overrightarrow{BO}=\left(\dfrac{\sqrt{3}}{3},-1,\dfrac{\sqrt{6}}{3}\right)$，$\overrightarrow{CO}=\left(\dfrac{\sqrt{3}}{3},1,\dfrac{\sqrt{6}}{3}\right)$。
$\overrightarrow{AO}\cdot\overrightarrow{BO}=-\dfrac{2}{3}+0+\dfrac{2}{3}=0$，同理$\overrightarrow{AO}\cdot\overrightarrow{CO}=0$，$\overrightarrow{BO}\cdot\overrightarrow{CO}=0$，故$AO$，$BO$，$CO$两两垂直。
(2)解：$M\left(-\dfrac{\sqrt{3}}{6},-\dfrac{1}{2},\dfrac{\sqrt{6}}{3}\right)$，$\overrightarrow{DM}=\left(-\dfrac{\sqrt{3}}{6},-\dfrac{1}{2},\dfrac{\sqrt{6}}{3}\right)$，$\overrightarrow{AO}=\left(-\dfrac{2\sqrt{3}}{3},0,\dfrac{\sqrt{6}}{3}\right)$。
$|\overrightarrow{DM}|=\sqrt{\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{2}{3}}=\sqrt{1}=1$，$|\overrightarrow{AO}|=\sqrt{\dfrac{4}{3}+0+\dfrac{2}{3}}=\sqrt{2}$。
$\overrightarrow{DM}\cdot\overrightarrow{AO}=\dfrac{1}{3}+0+\dfrac{2}{3}=1$，$\cos\theta=\dfrac{1}{1×\sqrt{2}}=\dfrac{\sqrt{2}}{2}$，故$\langle\overrightarrow{DM},\overrightarrow{AO}\rangle=\dfrac{\pi}{4}$。
答案：
(1)见证明；
(2)$\dfrac{\pi}{4}$

15. 如图,在三棱柱$ABC - A_{1}B_{1}C_{1}$中,$M$,$N分别是A_{1}B$,$B_{1}C_{1}$上的点,且$BM = 2A_{1}M$,$C_{1}N = 2B_{1}N$.设$\overrightarrow{AB}= \boldsymbol{a}$,$\overrightarrow{AC}= \boldsymbol{b}$,$\overrightarrow{A{A}_{1}}= \boldsymbol{c}$.
(1)试用$\boldsymbol{a}$,$\boldsymbol{b}$,$\boldsymbol{c}表示向量\overrightarrow{MN}$.
(2)若$\angle BAC = 90^{\circ}$,$\angle BAA_{1} = \angle CAA_{1} = 60^{\circ}$,$AB = AC = AA_{1} = 1$,求$MN$的长.

答案：
(1) 解：
$\overrightarrow{A_{1}B} = \overrightarrow{AB} - \overrightarrow{AA_{1}} = \boldsymbol{a} - \boldsymbol{c}$，
$\because BM = 2A_{1}M$，$\therefore \overrightarrow{A_{1}M} = \frac{1}{3}\overrightarrow{A_{1}B} = \frac{1}{3}(\boldsymbol{a} - \boldsymbol{c})$，
$\overrightarrow{A_{1}M} = \frac{1}{3}\boldsymbol{a} - \frac{1}{3}\boldsymbol{c}$，
$\overrightarrow{A_{1}N} = \overrightarrow{A_{1}B_{1}} + \overrightarrow{B_{1}N} = \boldsymbol{a} + \frac{1}{3}\overrightarrow{B_{1}C_{1}} = \boldsymbol{a} + \frac{1}{3}(\boldsymbol{b} - \boldsymbol{a}) = \frac{2}{3}\boldsymbol{a} + \frac{1}{3}\boldsymbol{b}$，
$\overrightarrow{MN} = \overrightarrow{A_{1}N} - \overrightarrow{A_{1}M} = \left(\frac{2}{3}\boldsymbol{a} + \frac{1}{3}\boldsymbol{b}\right) - \left(\frac{1}{3}\boldsymbol{a} - \frac{1}{3}\boldsymbol{c}\right) = \frac{1}{3}\boldsymbol{a} + \frac{1}{3}\boldsymbol{b} + \frac{1}{3}\boldsymbol{c}$。
(2) 解：
$\because \angle BAC = 90^\circ$，$\angle BAA_{1} = \angle CAA_{1} = 60^\circ$，$AB = AC = AA_{1} = 1$，
$\therefore |\boldsymbol{a}| = |\boldsymbol{b}| = |\boldsymbol{c}| = 1$，$\boldsymbol{a} \cdot \boldsymbol{b} = 0$，$\boldsymbol{a} \cdot \boldsymbol{c} = |\boldsymbol{a}||\boldsymbol{c}|\cos 60^\circ = \frac{1}{2}$，$\boldsymbol{b} \cdot \boldsymbol{c} = |\boldsymbol{b}||\boldsymbol{c}|\cos 60^\circ = \frac{1}{2}$，
$|\overrightarrow{MN}|^2 = \left(\frac{1}{3}\boldsymbol{a} + \frac{1}{3}\boldsymbol{b} + \frac{1}{3}\boldsymbol{c}\right)^2 = \frac{1}{9}(|\boldsymbol{a}|^2 + |\boldsymbol{b}|^2 + |\boldsymbol{c}|^2 + 2\boldsymbol{a} \cdot \boldsymbol{b} + 2\boldsymbol{a} \cdot \boldsymbol{c} + 2\boldsymbol{b} \cdot \boldsymbol{c})$，
$= \frac{1}{9}(1 + 1 + 1 + 0 + 2 × \frac{1}{2} + 2 × \frac{1}{2}) = \frac{1}{9}(3 + 1 + 1) = \frac{5}{9}$，
$\therefore |\overrightarrow{MN}| = \frac{\sqrt{5}}{3}$。
答案：
(1) $\overrightarrow{MN} = \frac{1}{3}\boldsymbol{a} + \frac{1}{3}\boldsymbol{b} + \frac{1}{3}\boldsymbol{c}$；
(2) $\frac{\sqrt{5}}{3}$。