《2025年教材帮高中数学选择性必修第二册苏教版》

5. (2025·河北省邢台一中月考)如图 6.2 -8,已知直三棱柱$ABC-A_{1}B_{1}C_{1}$中,$CA=CB=1,$$∠BCA=90^{\circ }$,棱$AA_{1}=2$,N 为$A_{1}A$的中点.
(1)求 BN 的长;
(2)求$BA_{1}$与$CB_{1}$所成角的余弦值.

答案：
5.如图D6.2 - 2，

以$C$为原点，以$\overrightarrow{CA}$,$\overrightarrow{CB}$,$\overrightarrow{CC_1}$为正交基底建立空间直角坐标系$Cxyz$。
(1)依题意得$B(0,1,0)$,$N(1,0,1)$。
$\therefore BN = |\overrightarrow{BN}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$。
(2)依题意得$A_1(1,0,2)$,$B(0,1,0)$，$C(0,0,0)$,$B_1(0,1,2)$。
$\therefore\overrightarrow{BA_1} = (1,-1,2)$,$\overrightarrow{CB_1} = (0,1,2)$，
$\therefore\overrightarrow{BA_1} · \overrightarrow{CB_1} = 3$，$|\overrightarrow{BA_1}| = \sqrt{6}$，$|\overrightarrow{CB_1}| = \sqrt{5}$。
$\therefore\cos\langle\overrightarrow{BA_1},\overrightarrow{CB_1}\rangle = \frac{\overrightarrow{BA_1} · \overrightarrow{CB_1}}{|\overrightarrow{BA_1}||\overrightarrow{CB_1}|} = \frac{\sqrt{30}}{10}$
故$BA_1$与$CB_1$所成角的余弦值为$\frac{\sqrt{30}}{10}$

例17 新定义 向量的叉乘 [多选题] (2025·北京市清华大学附属中学期中)向量的运算包含点乘和叉乘,其中点乘就是大家熟悉的向量的数量积. 现定义向量的叉乘:给定两个不共线的空间向量$\boldsymbol{a}$与$\boldsymbol{b},$规定①$\boldsymbol{a}×\boldsymbol{b}$为同时与$\boldsymbol{a},\boldsymbol{b}$垂直的向量;②$\boldsymbol{a},\boldsymbol{b},\boldsymbol{a}×\boldsymbol{b}$三个向量构成右手系(如图 6.2 -9);③$|\boldsymbol{a}×\boldsymbol{b}|=$$|\boldsymbol{a}||\boldsymbol{b}|· sin\langle \boldsymbol{a},\boldsymbol{b}\rangle $;④若$\boldsymbol{a}=(x_{1},y_{1},z_{1}),\boldsymbol{b}=(x_{2},y_{2},$$z_{2})$,则$\boldsymbol{a}×\boldsymbol{b}=(+$$\begin{vmatrix} y_{1}&z_{1}\\ y_{2}&z_{2}\end{vmatrix} ,-$$\begin{vmatrix} x_{1}&z_{1}\\ x_{2}&z_{2}\end{vmatrix} ,+$$\begin{vmatrix} x_{1}&y_{1}\\ x_{2}&y_{2}\end{vmatrix} )$,其中$\begin{vmatrix} a&b\\ c&d\end{vmatrix} =ad-bc$. 如图 6.2 -10,在长方体$ABCD-A_{1}B_{1}C_{1}D_{1}$中,$AB=AD=2,AA_{1}=3,$则下列结论正确的是( )



A.$|\overrightarrow {AB}×\overrightarrow {AD}|=|\overrightarrow {AA_{1}}|$
B.$\overrightarrow {AB}×\overrightarrow {AD}=\overrightarrow {AD}×\overrightarrow {AB}$
C.$(\overrightarrow {AB}-\overrightarrow {AD})×\overrightarrow {AA_{1}}=\overrightarrow {AB}×\overrightarrow {AA_{1}}-\overrightarrow {AD}×\overrightarrow {AA_{1}}$
D.长方体$ABCD-A_{1}B_{1}C_{1}D_{1}$的体积$V=(\overrightarrow {AB}×$$\overrightarrow {AD})· \overrightarrow {CC_{1}}$

答案：
解析▶方法1$|\overrightarrow {AB}×\overrightarrow {AD}|=|\overrightarrow {AB}||\overrightarrow {AD}|sin\langle \overrightarrow {AB},$$\overrightarrow {AD}\rangle =2×2×sin90^{\circ }=4≠|\overrightarrow {AA_{1}}|=3$,所以选项 A 错误.根据右手系知,$\overrightarrow {AB}×\overrightarrow {AD}$与$\overrightarrow {AD}×\overrightarrow {AB}$反向,所以$\overrightarrow {AB}×$$\overrightarrow {AD}≠\overrightarrow {AD}×\overrightarrow {AB}$,故选项 B 错误.因为$|(\overrightarrow {AB}-\overrightarrow {AD})×\overrightarrow {AA_{1}}|=$$|\overrightarrow {DB}×\overrightarrow {BB_{1}}|=2\sqrt {2}×3×sin90^{\circ }=6\sqrt {2}$,且$\overrightarrow {DB}×\overrightarrow {BB_{1}}=$$-\overrightarrow {BD}×\overrightarrow {BB_{1}}$与$\overrightarrow {CA}$同向共线;又$|\overrightarrow {AB}×\overrightarrow {AA_{1}}|=2×3×$$sin90^{\circ }=6$,且$\overrightarrow {AB}×\overrightarrow {AA_{1}}$与$\overrightarrow {DA}$同向共线,$|\overrightarrow {AD}×\overrightarrow {AA_{1}}|=$$2×3×sin90^{\circ }=6,\overrightarrow {AD}×\overrightarrow {AA_{1}}$与$\overrightarrow {DC}$同向共线,所以$|\overrightarrow {AB}×\overrightarrow {AA_{1}}-\overrightarrow {AD}×\overrightarrow {AA_{1}}|=6\sqrt {2}$,且$\overrightarrow {AB}×\overrightarrow {AA_{1}}-\overrightarrow {AD}×\overrightarrow {AA_{1}}$与$\overrightarrow {CA}$同向共线,故$(\overrightarrow {AB}-\overrightarrow {AD})×\overrightarrow {AA_{1}}=\overrightarrow {AB}×\overrightarrow {AA_{1}}-\overrightarrow {AD}×$$\overrightarrow {AA_{1}}$,故选项 C 正确.因为长方体$ABCD-A_{1}B_{1}C_{1}D_{1}$的体积为$2×2×3=12$,又由右手系知向量$\overrightarrow {AB}×\overrightarrow {AD}$的方向垂直底面向上,与$\overrightarrow {CC_{1}}$同向,所以$(\overrightarrow {AB}×\overrightarrow {AD})· \overrightarrow {CC_{1}}=(2×$$2×sin90^{\circ })×3×cos0^{\circ }=12$,故选项 D 正确.故选 CD.

方法2 建立空间直角坐标系，如图6.2-11所示.
易得$\overrightarrow {AB}=(0,2,0),\overrightarrow {AD}=(-2,$$0,0),\overrightarrow {AA_{1}}=(0,0,3),|\overrightarrow {AA_{1}}|=3,$则$\overrightarrow {AB}×\overrightarrow {AD}=(+\begin{vmatrix} 2&0\\ 0&0\end{vmatrix} ,-$$\begin{vmatrix} 0&0\\ -2&0\end{vmatrix} ,+\begin{vmatrix} 0&2\\ -2&0\end{vmatrix} )=(0,0,$4),$|\overrightarrow {AB}×\overrightarrow {AD}|=4≠|\overrightarrow {AA_{1}}|$,所以选

项

 A 错误；
$\overrightarrow {CC_{1}}=(0,0,3)$,则$(\overrightarrow {AB}×\overrightarrow {AD})· \overrightarrow {CC_{1}}=0+0+12=12$,故选项 D 正确；$\overrightarrow {AD}×\overrightarrow {AB}=(+\begin{vmatrix} 0&0\\ 2&0\end{vmatrix} ,-$$\begin{vmatrix} -2&0\\ 0&0\end{vmatrix} ,+\begin{vmatrix} -2&0\\ 0&2\end{vmatrix} )=(0,0,-4)$,故选项 B 错误；$\overrightarrow {AB}-\overrightarrow {AD}=$$\overrightarrow {DB}=(2,2,0)$,则$(\overrightarrow {AB}-\overrightarrow {AD})×\overrightarrow {AA_{1}}=(6,-6,0),\overrightarrow {AB}×$$\overrightarrow {AA_{1}}=(6,0,0),\overrightarrow {AD}×\overrightarrow {AA_{1}}=(0,6,0)$,则$\overrightarrow {AB}×\overrightarrow {AA_{1}}-$$\overrightarrow {AD}×\overrightarrow {AA_{1}}=(6,-6,0)$,所以$(\overrightarrow {AB}-\overrightarrow {AD})×\overrightarrow {AA_{1}}=\overrightarrow {AB}×\overrightarrow {AA_{1}}-\overrightarrow {AD}×\overrightarrow {AA_{1}}$,故选项 C 正确.故选 CD.
答案▶CD