《2025年资源库高中数学人教版》

例539 如图6-3-31，在平行六面体$ABCD-A_{1}B_{1}C_{1}D_{1}$中，$AB=2$，$AA_{1}=3$，$AD=1$，$B_{1}D=\sqrt{15}$，且$\angle DAB=\angle BAA_{1}=\angle DAA_{1}=\frac{\pi}{3}$，求$\overrightarrow {CD_{1}}$与$\overrightarrow {B_{1}D}$夹角的余弦值.

答案： $-\frac{3\sqrt{105}}{70}$
解析：连接$BA_{1}$，由题意，$\overrightarrow {CD_{1}}=\overrightarrow {BA_{1}}=\overrightarrow {AA_{1}}-\overrightarrow {AB}$.
则$|\overrightarrow {CD_{1}}|=\sqrt{(\overrightarrow {AA_{1}}-\overrightarrow {AB})^{2}}=\sqrt{|\overrightarrow {AA_{1}}|^{2}+|\overrightarrow {AB}|^{2}-2\overrightarrow {AA_{1}}\cdot\overrightarrow {AB}}=\sqrt{3^{2}+2^{2}-2×3×2×\frac{1}{2}}=\sqrt{7}$.
$\overrightarrow {CD_{1}}\cdot\overrightarrow {B_{1}D}=(\overrightarrow {AA_{1}}-\overrightarrow {AB})\cdot(\overrightarrow {AD}-\overrightarrow {AB}-\overrightarrow {AA_{1}})=\overrightarrow {AD}\cdot\overrightarrow {AA_{1}}-\overrightarrow {AD}\cdot\overrightarrow {AB}+\overrightarrow {AB}^{2}-\overrightarrow {AA_{1}}^{2}=1×3×\frac{1}{2}-1×2×\frac{1}{2}+2^{2}-3^{2}=-\frac{9}{2}$.
所以$\cos\langle\overrightarrow {CD_{1}},\overrightarrow {B_{1}D}\rangle=\frac{\overrightarrow {CD_{1}}\cdot\overrightarrow {B_{1}D}}{|\overrightarrow {CD_{1}}||\overrightarrow {B_{1}D}|}=\frac{-\frac{9}{2}}{\sqrt{7}×\sqrt{15}}=-\frac{3\sqrt{105}}{70}$.

例540 如图6-3-32，已知线段$AB$在平面$\alpha$内，线段$AC\perp\alpha$于$A$，线段$BD\perp AB$，线段$DD'\perp\alpha$于$D'$，如果$\angle DBD'=30^{\circ}$，$AB=a$，$AC=BD=b$，求$CD$的长.

答案： $\sqrt{a^{2}+3b^{2}}$
解析：由$AC\perp\alpha$，$AB\subset\alpha$，可知$AC\perp AB$.
由$\angle DBD'=30^{\circ}$，可知$\langle\overrightarrow {CA},\overrightarrow {BD}\rangle=60^{\circ}$.
因为$|\overrightarrow {CD}|^{2}=\overrightarrow {CD}\cdot\overrightarrow {CD}=(\overrightarrow {CA}+\overrightarrow {AB}+\overrightarrow {BD})^{2}=|\overrightarrow {CA}|^{2}+|\overrightarrow {AB}|^{2}+|\overrightarrow {BD}|^{2}+2(\overrightarrow {CA}\cdot\overrightarrow {AB}+\overrightarrow {CA}\cdot\overrightarrow {BD}+\overrightarrow {AB}\cdot\overrightarrow {BD})=b^{2}+a^{2}+b^{2}+2(0 + b^{2}\cos60^{\circ}+0)=a^{2}+3b^{2}$.
所以$|\overrightarrow {CD}|=\sqrt{a^{2}+3b^{2}}$，即$CD=\sqrt{a^{2}+3b^{2}}$.

例541 已知矩形$ABCD$，$P$为平面$ABCD$外一点，$M,N$分别为$PC,PD$上的点，且$M$是$PC$上靠近$C$的三等分点，$N$为$PD$的中点，求满足$\overrightarrow {MN}=x\overrightarrow {AB}+y\overrightarrow {AD}+z\overrightarrow {AP}$的实数$x,y,z$的值.

答案： $x=-\frac{2}{3}$，$y=-\frac{1}{6}$，$z=\frac{1}{6}$
解析：方法一：取$PC$的中点$E$，连接$NE$，则$\overrightarrow {MN}=\overrightarrow {EN}-\overrightarrow {EM}$.
$\overrightarrow {EN}=\frac{1}{2}\overrightarrow {CD}=\frac{1}{2}\overrightarrow {BA}=-\frac{1}{2}\overrightarrow {AB}$.
$\overrightarrow {EM}=\overrightarrow {PM}-\overrightarrow {PE}=\frac{2}{3}\overrightarrow {PC}-\frac{1}{2}\overrightarrow {PC}=\frac{1}{6}\overrightarrow {PC}$.
连接$AC$，则$\overrightarrow {PC}=\overrightarrow {AC}-\overrightarrow {AP}=\overrightarrow {AB}+\overrightarrow {AD}-\overrightarrow {AP}$.
所以$\overrightarrow {MN}=-\frac{1}{2}\overrightarrow {AB}-\frac{1}{6}(\overrightarrow {AB}+\overrightarrow {AD}-\overrightarrow {AP})=-\frac{2}{3}\overrightarrow {AB}-\frac{1}{6}\overrightarrow {AD}+\frac{1}{6}\overrightarrow {AP}$.
因为$\overrightarrow {AB},\overrightarrow {AD},\overrightarrow {AP}$不共面，所以$x=-\frac{2}{3}$，$y=-\frac{1}{6}$，$z=\frac{1}{6}$.
方法二：$\overrightarrow {MN}=\overrightarrow {PN}-\overrightarrow {PM}=\frac{1}{2}\overrightarrow {PD}-\frac{2}{3}\overrightarrow {PC}=\frac{1}{2}(\overrightarrow {PA}+\overrightarrow {AD})-\frac{2}{3}(\overrightarrow {PA}+\overrightarrow {AC})=-\frac{1}{2}\overrightarrow {AP}+\frac{1}{2}\overrightarrow {AD}-\frac{2}{3}(-\overrightarrow {AP}+\overrightarrow {AB}+\overrightarrow {AD})=-\frac{2}{3}\overrightarrow {AB}-\frac{1}{6}\overrightarrow {AD}+\frac{1}{6}\overrightarrow {AP}$.
因为$\overrightarrow {AB},\overrightarrow {AD},\overrightarrow {AP}$不共面，所以$x=-\frac{2}{3}$，$y=-\frac{1}{6}$，$z=\frac{1}{6}$.