《2025年小题狂做高中数学选择性必修第二册苏教版》

10. (多选题，2024 河北邢台期末)如图，在三棱柱$ABC - A_1B_1C_1$中，$M$，$N$分别是$A_1B$，$B_1C_1$上的点，且$BM = 2A_1M$，$C_1N = 2B_1N$。设$\overrightarrow{AB}=\boldsymbol{a}$，$\overrightarrow{AC}=\boldsymbol{b}$，$\overrightarrow{AA_1}=\boldsymbol{c}$，若$\angle BAC = 90°$，$\angle BAA_1 = \angle CAA_1 = 60°$，$AB = AC = AA_1 = 1$，则下列说法正确的是 (BD)

A.$\overrightarrow{MN}=\frac{1}{3}\boldsymbol{a}+\frac{1}{3}\boldsymbol{b}+\frac{2}{3}\boldsymbol{c}$
B.$|\overrightarrow{MN}|=\frac{\sqrt{5}}{3}$
C.$\overrightarrow{A_1B}\perp\overrightarrow{A_1C_1}$
D.$\cos\langle\overrightarrow{AB_1},\overrightarrow{BC_1}\rangle=\frac{1}{6}$

答案： 10. BD 因为$BM = 2A_1M$，$C_1N = 2B_1N$，所以$\overrightarrow{A_1M} = \frac{1}{3}\overrightarrow{A_1B_1} = \frac{1}{3}(\overrightarrow{AB} - \overrightarrow{AA_1})$，$\overrightarrow{A_1N} = \overrightarrow{A_1B_1} + \overrightarrow{B_1N} = \overrightarrow{AB} + \frac{1}{3}\overrightarrow{B_1C_1} = \overrightarrow{AB} + \frac{1}{3}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{2}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC}$，所以$\overrightarrow{MN} = \overrightarrow{A_1N} - \overrightarrow{A_1M} = \frac{2}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC} - \frac{1}{3}(\overrightarrow{AB} - \overrightarrow{AA_1}) = \frac{1}{3}\overrightarrow{a} + \frac{1}{3}\overrightarrow{b} + \frac{1}{3}\overrightarrow{c}$，故A错误。因为$|\boldsymbol{a}| = |\boldsymbol{b}| = |\boldsymbol{c}| = 1$，$\boldsymbol{a} · \boldsymbol{b} = 0$，$\boldsymbol{a} · \boldsymbol{c} = \boldsymbol{b} · \boldsymbol{c} = \frac{1}{2}$，所以$|\overrightarrow{MN}|^2 = \frac{1}{9}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^2 = \frac{1}{9}(\boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2 + 2\boldsymbol{a} · \boldsymbol{b} + 2\boldsymbol{a} · \boldsymbol{c} + 2\boldsymbol{b} · \boldsymbol{c}) = \frac{1}{9}(3 + 2) = \frac{5}{9}$，所以$|\overrightarrow{MN}| = \frac{\sqrt{5}}{3}$，故B正确。因为$\overrightarrow{A_1B} = \overrightarrow{AB} - \overrightarrow{AA_1} = \boldsymbol{a} - \boldsymbol{c}$，$\overrightarrow{BC_1} = \overrightarrow{BC} + \overrightarrow{BB_1} = \overrightarrow{AC} - \overrightarrow{AB} + \overrightarrow{AA_1} = \boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}$，所以$\overrightarrow{A_1B} · \overrightarrow{BC_1} = (\boldsymbol{a} + \boldsymbol{c}) · (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}) = \boldsymbol{a} · \boldsymbol{b} + \boldsymbol{b} · \boldsymbol{c} - \boldsymbol{a}^2 + \boldsymbol{c}^2 = \frac{1}{2}$。因为$\overrightarrow{A_1B}^2 = (\boldsymbol{a} + \boldsymbol{c})^2 = \boldsymbol{a}^2 + \boldsymbol{c}^2 + 2\boldsymbol{a} · \boldsymbol{c} = 3$，所以$|\overrightarrow{A_1B}| = \sqrt{3}$。又$\overrightarrow{BC_1}^2 = (-\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2 - 2\boldsymbol{a} · \boldsymbol{b} - 2\boldsymbol{a} · \boldsymbol{c} + 2\boldsymbol{b} · \boldsymbol{c} = 3$，所以$|\overrightarrow{BC_1}| = \sqrt{3}$，所以$\cos \langle\overrightarrow{A_1B},\overrightarrow{BC_1}\rangle = \frac{\overrightarrow{A_1B} · \overrightarrow{BC_1}}{|\overrightarrow{A_1B}||\overrightarrow{BC_1}|} = \frac{\frac{1}{2}}{\sqrt{3} × \sqrt{3}} = \frac{1}{6}$，故D正确。

11. (2024 浙江嘉兴期末)如图，在三棱锥$P - ABC$中，$AB\perp BC$，$PA\perp$平面$ABC$，$AE\perp PB$于点$E$，$M$是$AC$的中点，$PB = 1$，则$\overrightarrow{EP}·\overrightarrow{EM}$的最小值为$-\frac{1}{8}$。

答案：
11. $-\frac{1}{8}$ 连接$EC$，如图。因为$PA \perp$平面$ABC$，$BC \subset$平面$ABC$，所以$PA \perp BC$。又$AB \perp BC$，$PA \cap AB = A$，$PA,AB \subset$平面$PAB$，所以$BC \perp$平面$PAB$。又$PB \subset$平面$PAB$，所以$BC \perp PB$。因为$M$是$AC$的中点，所以$\overrightarrow{EM} = \frac{1}{2}(\overrightarrow{EA} + \overrightarrow{EC}) = \frac{1}{2}\overrightarrow{EA} + \frac{1}{2}(\overrightarrow{EB} + \overrightarrow{BC})$。又$AE \perp PB$，$\overrightarrow{EP} · \overrightarrow{EM} = \overrightarrow{EP} · [\frac{1}{2}\overrightarrow{EA} + \frac{1}{2}(\overrightarrow{EB} + \overrightarrow{BC})] = \frac{1}{2}\overrightarrow{EP} · \overrightarrow{EA} + \frac{1}{2}\overrightarrow{EP} · \overrightarrow{EB} + \frac{1}{2}\overrightarrow{EP} · \overrightarrow{BC} = \frac{1}{2}\overrightarrow{EP} · \overrightarrow{EB} = -\frac{1}{2}|\overrightarrow{EP}||\overrightarrow{EB}| \geq -\frac{1}{2}(\frac{|\overrightarrow{EP}| + |\overrightarrow{EB}|}{2})^2 = -\frac{1}{8}$，当且仅当$|\overrightarrow{EP}| = |\overrightarrow{EB}| = \frac{1}{2}$时，等号成立，所以$\overrightarrow{EP} · \overrightarrow{EM}$的最小值为$-\frac{1}{8}$。

12. (2024 湖北黄冈阶段练习)如图，在平行六面体$ABCD - A_1B_1C_1D_1$中，$AB = AD = 1$，$AA_1 = 2$，$\angle BAD = \angle BAA_1 = \angle DAA_1 = 60°$，求：
(1) $AC_1$的长；
(2) $BD_1$与$AC$所成角的余弦值。

答案： 12. 解：
(1) 设$\overrightarrow{AB} = \boldsymbol{a}$，$\overrightarrow{AD} = \boldsymbol{b}$，$\overrightarrow{AA_1} = \boldsymbol{c}$，则$\boldsymbol{a} · \boldsymbol{b} = |\boldsymbol{a}||\boldsymbol{b}|\cos 60^{\circ} = \frac{1}{2}$，$\boldsymbol{a} · \boldsymbol{c} = |\boldsymbol{a}||\boldsymbol{c}|\cos 60^{\circ} = 1$，$\boldsymbol{b} · \boldsymbol{c} = |\boldsymbol{b}||\boldsymbol{c}|\cos 60^{\circ} = 1$。
因为$\overrightarrow{AC_1} = \boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}$，所以在平行四边形$AA_1C_1C$中，$\overrightarrow{AC_1} = \overrightarrow{AC} + \overrightarrow{AA_1} = \boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}$，则$|\overrightarrow{AC_1}|^2 = \overrightarrow{AC_1}^2 = (\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2 + 2\boldsymbol{a} · \boldsymbol{b} + 2\boldsymbol{a} · \boldsymbol{c} + 2\boldsymbol{b} · \boldsymbol{c} = 1 + 1 + 4 + 2 × \frac{1}{2} + 2 × 1 + 2 × 1 = 11$，所以$|\overrightarrow{AC_1}| = \sqrt{11}$，所以$AC_1$的长为$\sqrt{11}$。
(2) 由$\overrightarrow{AC} = \boldsymbol{a} + \boldsymbol{b}$，可得$|\overrightarrow{AC}|^2 = (\boldsymbol{a} + \boldsymbol{b})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + 2\boldsymbol{a} · \boldsymbol{b} = 1 + 1 + 1 = 3$，所以$|\overrightarrow{AC}| = \sqrt{3}$。
由$\overrightarrow{BD_1} = \overrightarrow{AD_1} - \overrightarrow{AB} = \boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}$，可得$|\overrightarrow{BD_1}|^2 = (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{c}^2 - 2\boldsymbol{a} · \boldsymbol{b} - 2\boldsymbol{a} · \boldsymbol{c} + 2\boldsymbol{b} · \boldsymbol{c} = 1 + 1 + 4 - 1 - 2 + 2 = 5$，所以$|\overrightarrow{BD_1}| = \sqrt{5}$，$\cos \langle\overrightarrow{AC},\overrightarrow{BD_1}\rangle = \frac{\overrightarrow{AC} · \overrightarrow{BD_1}}{|\overrightarrow{AC}||\overrightarrow{BD_1}|} = \frac{(\boldsymbol{a} + \boldsymbol{b}) · (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a})}{\sqrt{3} × \sqrt{5}} = \frac{\boldsymbol{a} · \boldsymbol{c} - \boldsymbol{a}^2 + \boldsymbol{b}^2 + \boldsymbol{b} · \boldsymbol{c}}{\sqrt{15}} = \frac{1 - 1 + 1 + 1}{\sqrt{15}} = \frac{2\sqrt{15}}{15}$。
所以$BD_1$与$AC$所成角的余弦值为$\frac{2\sqrt{15}}{15}$。

13. (2024 福建三明阶段检测)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$\langle\overrightarrow{DM},\overrightarrow{AO}\rangle$的大小。

答案： 13.
(1) 证明：设$\overrightarrow{VA} = \boldsymbol{a}$，$\overrightarrow{VB} = \boldsymbol{b}$，$\overrightarrow{VC} = \boldsymbol{c}$，正四面体的棱长为1。因为$\overrightarrow{VD} = \overrightarrow{VB} + \overrightarrow{BD} = \overrightarrow{VB} + \frac{2}{3} × \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{BC}) = \overrightarrow{VB} + \frac{1}{3}(\overrightarrow{VA} - \overrightarrow{VB} + \overrightarrow{VC} - \overrightarrow{VB}) = \frac{1}{3}(\overrightarrow{VA} + \overrightarrow{VB} + \overrightarrow{VC}) = \frac{1}{3}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})$，
$\overrightarrow{AO} = \overrightarrow{VO} - \overrightarrow{VA} = \frac{1}{2}\overrightarrow{VD} - \overrightarrow{VA} = \frac{1}{6}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}) - \boldsymbol{a} = \frac{1}{6}(\boldsymbol{b} + \boldsymbol{c} - 5\boldsymbol{a})$，
$\overrightarrow{BO} = \overrightarrow{VO} - \overrightarrow{VB} = \frac{1}{2}\overrightarrow{VD} - \overrightarrow{VB} = \frac{1}{6}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}) - \boldsymbol{b} = \frac{1}{6}(\boldsymbol{a} + \boldsymbol{c} - 5\boldsymbol{b})$，
$\overrightarrow{CO} = \overrightarrow{VO} - \overrightarrow{VC} = \frac{1}{2}\overrightarrow{VD} - \overrightarrow{VC} = \frac{1}{6}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}) - \boldsymbol{c} = \frac{1}{6}(\boldsymbol{a} + \boldsymbol{b} - 5\boldsymbol{c})$，
所以$\overrightarrow{AO} · \overrightarrow{BO} = \frac{1}{36}(\boldsymbol{b} + \boldsymbol{c} - 5\boldsymbol{a}) · (\boldsymbol{a} + \boldsymbol{c} - 5\boldsymbol{b}) = \frac{1}{36}(18\boldsymbol{a} · \boldsymbol{b} - 9|\boldsymbol{a}|^2) = \frac{1}{36}(18 × 1 × 1 × \cos \frac{\pi}{3} - 9) = 0$，
所以$\overrightarrow{AO} \perp \overrightarrow{BO}$，即$AO \perp BO$。
同理，$AO \perp CO$，$BO \perp CO$，所以$AO$，$BO$，$CO$两两垂直。
(2) 解：$\overrightarrow{DM} = \overrightarrow{DV} + \overrightarrow{VM} = -\frac{1}{3}(\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}) + \frac{1}{2}\boldsymbol{c} = \frac{1}{6}(-2\boldsymbol{a} - 2\boldsymbol{b} + \boldsymbol{c})$，
所以$|\overrightarrow{DM}| = \sqrt{[\frac{1}{6}(-2\boldsymbol{a} - 2\boldsymbol{b} + \boldsymbol{c})]^2} = \sqrt{\frac{1}{36}(9\boldsymbol{a}^2 + 8\boldsymbol{a} · \boldsymbol{b} - 4\boldsymbol{a} · \boldsymbol{c} - 4\boldsymbol{b} · \boldsymbol{c})} = \sqrt{\frac{1}{36} × 9} = \frac{1}{2}$。
又$|\overrightarrow{AO}| = \sqrt{[\frac{1}{6}(\boldsymbol{b} + \boldsymbol{c} - 5\boldsymbol{a})]^2} = \sqrt{\frac{1}{36}(27\boldsymbol{a}^2 - 18\boldsymbol{a} · \boldsymbol{b})} = \sqrt{\frac{1}{36} × (27 - 9)} = \frac{\sqrt{2}}{2}$，
$\overrightarrow{DM} · \overrightarrow{AO} = \frac{1}{6}(-2\boldsymbol{a} - 2\boldsymbol{b} + \boldsymbol{c}) · \frac{1}{6}(\boldsymbol{b} + \boldsymbol{c} - 5\boldsymbol{a}) = \frac{1}{36} × 9\boldsymbol{a}^2 = \frac{1}{36} × 9 = \frac{1}{4}$，
所以$\cos \langle\overrightarrow{DM},\overrightarrow{AO}\rangle = \frac{\overrightarrow{DM} · \overrightarrow{AO}}{|\overrightarrow{DM}||\overrightarrow{AO}|} = \frac{\frac{1}{4}}{\frac{1}{2} × \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{2}$。
又$\langle\overrightarrow{DM},\overrightarrow{AO}\rangle \in [0,\pi]$，所以$\langle\overrightarrow{DM},\overrightarrow{AO}\rangle = \frac{\pi}{4}$。

14. (多选题，2024 辽宁大连阶段练习)在三维空间中，定义向量的外积：$\boldsymbol{a}×\boldsymbol{b}$叫作向量$\boldsymbol{a}$与$\boldsymbol{b}$的外积。它是一个向量，满足下列两个条件：①$\boldsymbol{a}\perp(\boldsymbol{a}×\boldsymbol{b})$，$\boldsymbol{b}\perp(\boldsymbol{a}×\boldsymbol{b})$，且$\boldsymbol{a}$，$\boldsymbol{b}$和$\boldsymbol{a}×\boldsymbol{b}$构成右手系(三个向量的方向依次与右手的拇指、食指、中指的指向一致，如图所示)；②$\boldsymbol{a}×\boldsymbol{b}$的模$|\boldsymbol{a}×\boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\sin\langle\boldsymbol{a},\boldsymbol{b}\rangle$($\langle\boldsymbol{a},\boldsymbol{b}\rangle$表示向量$\boldsymbol{a}$，$\boldsymbol{b}$的夹角)。在正方体$ABCD - A_1B_1C_1D_1$中，有以下四个结论，正确的有 (ACD)

A.$|\overrightarrow{AB_1}×\overrightarrow{AC}| = |\overrightarrow{AD_1}×\overrightarrow{D_1B_1}|$
B.$\overrightarrow{AB}×\overrightarrow{AD}=\overrightarrow{AD}×\overrightarrow{AB}$
C.$\overrightarrow{A_1C_1}×\overrightarrow{A_1D}$与$\overrightarrow{BD_1}$共线
D.$(\overrightarrow{BC}×\overrightarrow{AC})·\overrightarrow{A_1A}$与正方体体积数值相等

答案：
14. ACD 如图，设正方体的棱长为1。对于A，易知$\langle\overrightarrow{AB_1},\overrightarrow{AC_1}\rangle = \langle\overrightarrow{D_1A_1},\overrightarrow{D_1C_1}\rangle = \frac{\pi}{3}$，$|\overrightarrow{AB_1} × \overrightarrow{AC_1}| = |\overrightarrow{AB_1}| · |\overrightarrow{AC_1}|\sin\frac{\pi}{3} = (\sqrt{2})^2 × \frac{\sqrt{3}}{2} = \sqrt{3}$，
$|\overrightarrow{AD_1} × \overrightarrow{D_1B_1}| = |\overrightarrow{AD_1}| · |\overrightarrow{D_1B_1}| · \sin\frac{2\pi}{3} = \sqrt{3}$，所以$|\overrightarrow{AB_1} × \overrightarrow{AC_1}| = |\overrightarrow{AD_1} × \overrightarrow{D_1B_1}|$，所以A正确；对于B，由$\boldsymbol{a}$，$\boldsymbol{b}$和$\boldsymbol{a} × \boldsymbol{b}$构成右手系知，$\boldsymbol{a} × \boldsymbol{b}$与$\boldsymbol{b} × \boldsymbol{a}$方向相反，即$\boldsymbol{a} × \boldsymbol{b} = -\boldsymbol{b} × \boldsymbol{a}$，所以B错误；对于C，$A_1C_1 \perp B_1D_1$，$A_1C_1 \perp BB_1$，$B_1D_1 \cap BB_1 = B_1$，则$A_1C_1 \perp$平面$BB_1D_1D$，又$BD_1 \subset$平面$BB_1D_1D$，所以$BD_1 \perp A_1C_1$，同理可得$BD_1 \perp A_1D$，再由右手系知，$A_1C_1 × A_1D_1$与$BD_1$共线，所以C正确；对于D，$|\overrightarrow{BC} × \overrightarrow{AC}| = |\overrightarrow{BC}| · |\overrightarrow{AC}| · \sin\frac{\pi}{4} = 1 × \sqrt{2} × \frac{\sqrt{2}}{2} = 1$，$\overrightarrow{BC} × \overrightarrow{AC}$与$\overrightarrow{AA_1}$共线，$(\overrightarrow{BC} × \overrightarrow{AC}) · \overrightarrow{AA_1} = |\overrightarrow{BC} × \overrightarrow{AC}||\overrightarrow{AA_1}|\cos \langle\overrightarrow{BC} × \overrightarrow{AC},\overrightarrow{AA_1}\rangle = \cos \langle\overrightarrow{BC} × \overrightarrow{AC},\overrightarrow{AA_1}\rangle = \cos 0 = 1$，正方体体积为1，所以D正确。