di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A'B'C'中，AC=BC=AA'，\angle ACB=90^\circ ，D，E分别为AB，BB'的中点.$<img_2660>
$(1)求证：\overrightarrow{CE}\perp \overrightarrow{A'D}；$

Solution:
$解法一：由题意得，CA,CB,CC'两两互相垂直.以\overrightarrow{CA}的方向为x轴正方向，\overrightarrow{CB}的方向为y轴正方向，\overrightarrow{CC'}的方向为z轴正方向建立空间直角坐标系(图略).设CA=CB=CC'=2，则C(0,0,0),E(0,2,1),A'(2,0,2),D(1,1,0),A(2,0,0),C'(0,0,2).证明：易得\overrightarrow{CE}=(0,2,1)，\overrightarrow{A'D}=(-1,1，-2).\because \overrightarrow{CE}\cdot \overrightarrow{A'D}=0，\therefore \overrightarrow{CE}\perp \overrightarrow{A'D}.
解法二：证明：设\overrightarrow{CA}=a，\overrightarrow{CB}=b，\overrightarrow{CC'}=c，根据题意知，|a|=|b|=|c|且a\cdot b=b\cdot c=c\cdot a=0.\because 向量a,b,c不共面，\therefore a，b，c可构成空间向量的一组基底，\therefore \overrightarrow{CE}=b＋\frac{1}{2}c，\overrightarrow{A'D}=-c＋\frac{1}{2}b-\frac{1}{2}a，\therefore \overrightarrow{CE}\cdot \overrightarrow{A'D}=-\frac{1}{2}c^{2}＋\frac{1}{2}b^{2}=0，\therefore \overrightarrow{CE}\perp \overrightarrow{A'D}.$
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A'B'C'中，AC=BC=AA'，\angle ACB=90^\circ ，D，E分别为AB，BB'的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094367140720643/img/compression/M_e5f1a7c9e21b62784ba7ff4790daae33.jpg)
$(2)求\overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值.$

Solution:
$解法一：由题意得，CA,CB,CC'两两互相垂直.以\overrightarrow{CA}的方向为x轴正方向，\overrightarrow{CB}的方向为y轴正方向，\overrightarrow{CC'}的方向为z轴正方向建立空间直角坐标系(图略).设CA=CB=CC'=2，则C(0,0,0),E(0,2,1),A'(2,0,2),D(1,1,0),A(2,0,0),C'(0,0,2).由(1)得\overrightarrow{CE}=(0,2,1).又\overrightarrow{AC'}=(-2,0,2)，\therefore \\cos  \langle \overrightarrow{AC'},\overrightarrow{CE}\rangle =\frac{2}{\sqrt{5} \times 2\sqrt{2}}=\frac{\sqrt{10}}{10}.\therefore \overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值为\frac{\sqrt{10}}{10}.解法二：易知\overrightarrow{AC'}=-a＋c，\overrightarrow{CE}=b+\frac{1}{2}c,|\overrightarrow{AC'}|=\sqrt{2}|a|，|\overrightarrow{CE}|=\frac{\sqrt{5}}{2}|a|，\therefore \overrightarrow{AC'}\cdot \overrightarrow{CE}=(-a＋c)\cdot \left(\boldsymbol{b}+\frac{1}{2}\boldsymbol{c}\right)=\frac{1}{2}c^{2}=\frac{1}{2}|a|^{2}，\therefore \\cos  \langle \overrightarrow{AC'}，\overrightarrow{CE}\rangle =\frac{\frac{1}{2}|\boldsymbol{a}|^2}{\sqrt{2} \times \frac{\sqrt{5}}{2}|\boldsymbol{a}|^2}=\frac{\sqrt{10}}{10}.\therefore \overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值为\frac{\sqrt{10}}{10}.$

Final answer: $\frac{\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $如图，已知直三棱柱ABC-A'B'C'中，AC=BC=AA'，\angle ACB=90^\circ ，D，E分别为AB，BB'的中点.$<img_2660>
$(2)求\overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值.$

Solution:
$解法一：由题意得，CA,CB,CC'两两互相垂直.以\overrightarrow{CA}的方向为x轴正方向，\overrightarrow{CB}的方向为y轴正方向，\overrightarrow{CC'}的方向为z轴正方向建立空间直角坐标系(图略).设CA=CB=CC'=2，则C(0,0,0),E(0,2,1),A'(2,0,2),D(1,1,0),A(2,0,0),C'(0,0,2).由(1)得\overrightarrow{CE}=(0,2,1).又\overrightarrow{AC'}=(-2,0,2)，\therefore \\cos  \langle \overrightarrow{AC'},\overrightarrow{CE}\rangle =\frac{2}{\sqrt{5} \times 2\sqrt{2}}=\frac{\sqrt{10}}{10}.\therefore \overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值为\frac{\sqrt{10}}{10}.解法二：易知\overrightarrow{AC'}=-a＋c，\overrightarrow{CE}=b+\frac{1}{2}c,|\overrightarrow{AC'}|=\sqrt{2}|a|，|\overrightarrow{CE}|=\frac{\sqrt{5}}{2}|a|，\therefore \overrightarrow{AC'}\cdot \overrightarrow{CE}=(-a＋c)\cdot \left(\boldsymbol{b}+\frac{1}{2}\boldsymbol{c}\right)=\frac{1}{2}c^{2}=\frac{1}{2}|a|^{2}，\therefore \\cos  \langle \overrightarrow{AC'}，\overrightarrow{CE}\rangle =\frac{\frac{1}{2}|\boldsymbol{a}|^2}{\sqrt{2} \times \frac{\sqrt{5}}{2}|\boldsymbol{a}|^2}=\frac{\sqrt{10}}{10}.\therefore \overrightarrow{CE}与\overrightarrow{AC'}夹角的余弦值为\frac{\sqrt{10}}{10}.$

Final answer: $\frac{\sqrt{10}}{10}$
Subfield: Solid Geometry


Question: $在如图所示的几何体中，正方形ABCD与梯形ABEF所在的平面相交，EB\parallel FA，FA=AB=\frac{1}{2}EB.$<img_1735>
$(1)证明：DF\parallel 平面BCE；$

Solution:
$证明：\because 四边形ABCD为正方形，\therefore AD\parallel BC，又BC\subset 平面BCE，AD\not\subset 平面BCE，\therefore AD\parallel 平面BCE.\because EB\parallel FA，EB\subset 平面BCE，FA\not\subset 平面BCE，\therefore FA\parallel 平面BCE，又AD\cap FA=A，AD,FA\subset 平面FAD，\therefore 平面FAD\parallel 平面BCE，\because DF\subset 平面FAD，\therefore DF\parallel 平面BCE.$
Subfield: Solid Geometry


Question: $在如图所示的几何体中，正方形ABCD与梯形ABEF所在的平面相交，EB\parallel FA，FA=AB=\frac{1}{2}EB.$![Image](https://cdntk.53zaixian.com/tq/1659094375307030529/img/compression/M_6072581924fa22d101fa2d956e9a717b.jpg)
$(1)证明：DF\parallel 平面BCE；$

Solution:
$证明：\because 四边形ABCD为正方形，\therefore AD\parallel BC，又BC\subset 平面BCE，AD\not\subset 平面BCE，\therefore AD\parallel 平面BCE.\because EB\parallel FA，EB\subset 平面BCE，FA\not\subset 平面BCE，\therefore FA\parallel 平面BCE，又AD\cap FA=A，AD,FA\subset 平面FAD，\therefore 平面FAD\parallel 平面BCE，\because DF\subset 平面FAD，\therefore DF\parallel 平面BCE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $在如图所示的几何体中，正方形ABCD与梯形ABEF所在的平面相交，EB\parallel FA，FA=AB=\frac{1}{2}EB.$<img_1735>
$(2)若BE\perp 平面ABCD，求异面直线DE与CF所成角的余弦值.$

Solution:
$易知BA,BC,BE两两互相垂直.以B为坐标原点，\overrightarrow{BE},\overrightarrow{BA},\overrightarrow{BC}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系(图略).设AB=2，则D(0,2,2)，E(4,0,0)，F(2,2,0)，C(0,0,2)，\therefore \overrightarrow{DE}=(4,-2,-2)，\overrightarrow{CF}=(2,2,-2)，\therefore |\\cos  \langle \overrightarrow{DE},\overrightarrow{CF}\rangle |=\frac{|\overrightarrow{DE} \cdot \overrightarrow{CF}|}{|\overrightarrow{DE}| \cdot |\overrightarrow{CF}|}=\frac{8}{2\sqrt{6} \times 2\sqrt{3}}=\frac{\sqrt{2}}{3}，\therefore 异面直线DE与CF所成角的余弦值为\frac{\sqrt{2}}{3}.$

Final answer: $\frac{\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $在如图所示的几何体中，正方形ABCD与梯形ABEF所在的平面相交，EB\parallel FA，FA=AB=\frac{1}{2}EB.$![Image](https://cdntk.53zaixian.com/tq/1659094375307030529/img/compression/M_6072581924fa22d101fa2d956e9a717b.jpg)
$(2)若BE\perp 平面ABCD，求异面直线DE与CF所成角的余弦值.$

Solution:
$易知BA,BC,BE两两互相垂直.以B为坐标原点，\overrightarrow{BE},\overrightarrow{BA},\overrightarrow{BC}的方向分别为x轴,y轴,z轴的正方向建立空间直角坐标系(图略).设AB=2，则D(0,2,2)，E(4,0,0)，F(2,2,0)，C(0,0,2)，\therefore \overrightarrow{DE}=(4,-2,-2)，\overrightarrow{CF}=(2,2,-2)，\therefore |\\cos  \langle \overrightarrow{DE},\overrightarrow{CF}\rangle |=\frac{|\overrightarrow{DE} \cdot \overrightarrow{CF}|}{|\overrightarrow{DE}| \cdot |\overrightarrow{CF}|}=\frac{8}{2\sqrt{6} \times 2\sqrt{3}}=\frac{\sqrt{2}}{3}，\therefore 异面直线DE与CF所成角的余弦值为\frac{\sqrt{2}}{3}.$

Final answer: $\frac{\sqrt{2}}{3}$
Subfield: Solid Geometry


Question: $如图，已知平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AD=AA_{1}=AB=1，\angle A_{1}AB=\angle DAB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=4\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1659094334818426883/img/compression/M_5d93fcef597db08746a103d2674a38f0.jpg)
$(1)试用a，b，c表示\overrightarrow{MN}；$

Solution:
$连接BD，\overrightarrow{MN}=\overrightarrow{MD_1}+\overrightarrow{D_1A_1}+\overrightarrow{A_1N}=-\frac{3}{4}\overrightarrow{D_1B}-\overrightarrow{AD}+\frac{2}{3}\overrightarrow{A_1C_1}=-\frac{3}{4}(\overrightarrow{D_1D}+\overrightarrow{DB})-\overrightarrow{AD}+\frac{2}{3}(\overrightarrow{AB}+\overrightarrow{AD})=\frac{3}{4}c-\frac{3}{4}(a-b)-b+\frac{2}{3}(a+b)=-\frac{1}{12}a+\frac{5}{12}b+\frac{3}{4}c.$

Final answer: $\overrightarrow{MN}=-\frac{1}{12}a+\frac{5}{12}b+\frac{3}{4}c.$
Subfield: Solid Geometry


Question: $如图，已知平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AD=AA_{1}=AB=1，\angle A_{1}AB=\angle DAB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=4\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_2313>
$(1)试用a，b，c表示\overrightarrow{MN}；$

Solution:
$连接BD，\overrightarrow{MN}=\overrightarrow{MD_1}+\overrightarrow{D_1A_1}+\overrightarrow{A_1N}=-\frac{3}{4}\overrightarrow{D_1B}-\overrightarrow{AD}+\frac{2}{3}\overrightarrow{A_1C_1}=-\frac{3}{4}(\overrightarrow{D_1D}+\overrightarrow{DB})-\overrightarrow{AD}+\frac{2}{3}(\overrightarrow{AB}+\overrightarrow{AD})=\frac{3}{4}c-\frac{3}{4}(a-b)-b+\frac{2}{3}(a+b)=-\frac{1}{12}a+\frac{5}{12}b+\frac{3}{4}c.$

Final answer: $\overrightarrow{MN}=-\frac{1}{12}a+\frac{5}{12}b+\frac{3}{4}c.$
Subfield: Solid Geometry


Question: $如图，已知平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AD=AA_{1}=AB=1，\angle A_{1}AB=\angle DAB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=4\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$<img_2313>
$(2)求|\overrightarrow{MN}|.$

Solution:
$由(1)得\overrightarrow{MN}^2=\left(-\frac{1}{12}\boldsymbol{a}+\frac{5}{12}\boldsymbol{b}+\frac{3}{4}\boldsymbol{c}\right)^2=\frac{1}{144}a^{2}+\frac{25}{144}b^{2}+\frac{9}{16}c^{2}-2\times \frac{1}{12}\times \frac{5}{12}a\cdot b-2\times \frac{1}{12}\times \frac{3}{4}a\cdot c+2\times \frac{5}{12}\times \frac{3}{4}b\cdot c=\frac{1}{144}+\frac{25}{144}+\frac{9}{16}-2\times \frac{1}{12}\times \frac{5}{12}\times \\cos   60^\circ -2\times \frac{1}{12}\times \frac{3}{4}\times \\cos   60^\circ +2\times \frac{5}{12}\times \frac{3}{4}\times \\cos   60^\circ =\frac{23}{24}，\therefore |\overrightarrow{MN}|=\frac{\sqrt{138}}{12}.$

Final answer: $\frac{\sqrt{138}}{12}$
Subfield: Solid Geometry


Question: $如图，已知平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AD=AA_{1}=AB=1，\angle A_{1}AB=\angle DAB=\angle DAA_{1}=60^\circ ，\overrightarrow{A_1C_1}=3\overrightarrow{NC_1}，\overrightarrow{D_1B}=4\overrightarrow{MB}，设\overrightarrow{AB}=a，\overrightarrow{AD}=b，\overrightarrow{AA_1}=c.$![Image](https://cdntk.53zaixian.com/tq/1659094334818426883/img/compression/M_5d93fcef597db08746a103d2674a38f0.jpg)
$(2)求|\overrightarrow{MN}|.$

Solution:
$由(1)得\overrightarrow{MN}^2=\left(-\frac{1}{12}\boldsymbol{a}+\frac{5}{12}\boldsymbol{b}+\frac{3}{4}\boldsymbol{c}\right)^2=\frac{1}{144}a^{2}+\frac{25}{144}b^{2}+\frac{9}{16}c^{2}-2\times \frac{1}{12}\times \frac{5}{12}a\cdot b-2\times \frac{1}{12}\times \frac{3}{4}a\cdot c+2\times \frac{5}{12}\times \frac{3}{4}b\cdot c=\frac{1}{144}+\frac{25}{144}+\frac{9}{16}-2\times \frac{1}{12}\times \frac{5}{12}\times \\cos   60^\circ -2\times \frac{1}{12}\times \frac{3}{4}\times \\cos   60^\circ +2\times \frac{5}{12}\times \frac{3}{4}\times \\cos   60^\circ =\frac{23}{24}，\therefore |\overrightarrow{MN}|=\frac{\sqrt{138}}{12}.$

Final answer: $\frac{\sqrt{138}}{12}$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD－A_{1}B_{1}C_{1}D_{1}中，\angle A_{1}AD=\angle A_{1}AB=\angle BAD=60^\circ ，AB=AD=2，AA_{1}=1，P为线段BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094316326858754/img/compression/M_70c8892b22775f46bf6e534e1138d598.jpg)
$(1)求|\overrightarrow{D_1P}|；$

Solution:
$连接AP,AD_{1},\because \overrightarrow{D_1P}=\overrightarrow{AP}-\overrightarrow{AD_1}=(\overrightarrow{AB}+\overrightarrow{BP})-(\overrightarrow{AD}+\overrightarrow{AA_1})=\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}，\therefore |\overrightarrow{D_1P}|^{2}=\left(\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}\right)^2=|\overrightarrow{AB}|^{2}+\frac{1}{4}|\overrightarrow{AD}|^{2}+|\overrightarrow{AA_1}|^{2}-\overrightarrow{AB}\cdot \overrightarrow{AD}-2\overrightarrow{AB}\cdot \overrightarrow{AA_1}+\overrightarrow{AD}\cdot \overrightarrow{AA_1}=4+1+1-2-2+1=3，\therefore |\overrightarrow{D_1P}|=\sqrt{3}.$

Final answer: $\sqrt{3}$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD－A_{1}B_{1}C_{1}D_{1}中，\angle A_{1}AD=\angle A_{1}AB=\angle BAD=60^\circ ，AB=AD=2，AA_{1}=1，P为线段BC的中点.$<img_2335>
$(1)求|\overrightarrow{D_1P}|；$

Solution:
$连接AP,AD_{1},\because \overrightarrow{D_1P}=\overrightarrow{AP}-\overrightarrow{AD_1}=(\overrightarrow{AB}+\overrightarrow{BP})-(\overrightarrow{AD}+\overrightarrow{AA_1})=\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}，\therefore |\overrightarrow{D_1P}|^{2}=\left(\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}\right)^2=|\overrightarrow{AB}|^{2}+\frac{1}{4}|\overrightarrow{AD}|^{2}+|\overrightarrow{AA_1}|^{2}-\overrightarrow{AB}\cdot \overrightarrow{AD}-2\overrightarrow{AB}\cdot \overrightarrow{AA_1}+\overrightarrow{AD}\cdot \overrightarrow{AA_1}=4+1+1-2-2+1=3，\therefore |\overrightarrow{D_1P}|=\sqrt{3}.$

Final answer: $\sqrt{3}$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD－A_{1}B_{1}C_{1}D_{1}中，\angle A_{1}AD=\angle A_{1}AB=\angle BAD=60^\circ ，AB=AD=2，AA_{1}=1，P为线段BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094316326858754/img/compression/M_70c8892b22775f46bf6e534e1138d598.jpg)
$(2)求\overrightarrow{AB_1}与\overrightarrow{D_1P}夹角的余弦值.$

Solution:
$由(1)知，\overrightarrow{D_1P}=\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}，|\overrightarrow{D_1P}|=\sqrt{3}.\because \overrightarrow{AB_1}=\overrightarrow{AB}+\overrightarrow{AA_1}，\therefore |\overrightarrow{AB_1}|=\sqrt{(\overrightarrow{AB}+\overrightarrow{AA_1})^2}=\sqrt{|\overrightarrow{AB}|^2+|\overrightarrow{AA_1}|^2+2\overrightarrow{AB} \cdot \overrightarrow{AA_1}}=\sqrt{4+1+2}=\sqrt{7}，\overrightarrow{AB_1}\cdot \overrightarrow{D_1P}=(\overrightarrow{AB}+\overrightarrow{AA_1})\cdot \left(\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}\right)=|\overrightarrow{AB}|^{2}-\frac{1}{2}\overrightarrow{AB}\cdot \overrightarrow{AD}-\overrightarrow{AB}\cdot \overrightarrow{AA_1}+\overrightarrow{AB}\cdot \overrightarrow{AA_1}-\frac{1}{2}\overrightarrow{AD}\cdot \overrightarrow{AA_1}-|\overrightarrow{AA_1}|^{2}=|\overrightarrow{AB}|^{2}-\frac{1}{2}\overrightarrow{AB}\cdot \overrightarrow{AD}-\frac{1}{2}\overrightarrow{AD}\cdot \overrightarrow{AA_1}-|\overrightarrow{AA_1}|^{2}=4-1-\frac{1}{2}-1=\frac{3}{2}，\therefore \\cos  \langle \overrightarrow{AB_1}，\overrightarrow{D_1P}\rangle =\frac{\overrightarrow{AB_1} \cdot \overrightarrow{D_1P}}{|\overrightarrow{AB_1}||\overrightarrow{D_1P}|}=\frac{\frac{3}{2}}{\sqrt{7} \times \sqrt{3}}=\frac{\sqrt{21}}{14}.$

Final answer: $\frac{\sqrt{21}}{14}$
Subfield: Solid Geometry


Question: $如图，在平行六面体ABCD－A_{1}B_{1}C_{1}D_{1}中，\angle A_{1}AD=\angle A_{1}AB=\angle BAD=60^\circ ，AB=AD=2，AA_{1}=1，P为线段BC的中点.$<img_2335>
$(2)求\overrightarrow{AB_1}与\overrightarrow{D_1P}夹角的余弦值.$

Solution:
$由(1)知，\overrightarrow{D_1P}=\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}，|\overrightarrow{D_1P}|=\sqrt{3}.\because \overrightarrow{AB_1}=\overrightarrow{AB}+\overrightarrow{AA_1}，\therefore |\overrightarrow{AB_1}|=\sqrt{(\overrightarrow{AB}+\overrightarrow{AA_1})^2}=\sqrt{|\overrightarrow{AB}|^2+|\overrightarrow{AA_1}|^2+2\overrightarrow{AB} \cdot \overrightarrow{AA_1}}=\sqrt{4+1+2}=\sqrt{7}，\overrightarrow{AB_1}\cdot \overrightarrow{D_1P}=(\overrightarrow{AB}+\overrightarrow{AA_1})\cdot \left(\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AD}-\overrightarrow{AA_1}\right)=|\overrightarrow{AB}|^{2}-\frac{1}{2}\overrightarrow{AB}\cdot \overrightarrow{AD}-\overrightarrow{AB}\cdot \overrightarrow{AA_1}+\overrightarrow{AB}\cdot \overrightarrow{AA_1}-\frac{1}{2}\overrightarrow{AD}\cdot \overrightarrow{AA_1}-|\overrightarrow{AA_1}|^{2}=|\overrightarrow{AB}|^{2}-\frac{1}{2}\overrightarrow{AB}\cdot \overrightarrow{AD}-\frac{1}{2}\overrightarrow{AD}\cdot \overrightarrow{AA_1}-|\overrightarrow{AA_1}|^{2}=4-1-\frac{1}{2}-1=\frac{3}{2}，\therefore \\cos  \langle \overrightarrow{AB_1}，\overrightarrow{D_1P}\rangle =\frac{\overrightarrow{AB_1} \cdot \overrightarrow{D_1P}}{|\overrightarrow{AB_1}||\overrightarrow{D_1P}|}=\frac{\frac{3}{2}}{\sqrt{7} \times \sqrt{3}}=\frac{\sqrt{21}}{14}.$

Final answer: $\frac{\sqrt{21}}{14}$
Subfield: Solid Geometry


Question: $如图，在底面是正方形的四棱锥P-ABCD中，PA\perp 平面ABCD，AP=AB=2，E,F,G分别是BC,PC,CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094387180118020/img/compression/M_50bc6f3349afc9ea7fe43078bb6de8eb.jpg)
$(1)求证：BG\perp 平面PAE；$

Solution:
$因为四棱锥P-ABCD的底面是正方形，且PA\perp 平面ABCD，所以PA，AB，AD两两互相垂直.以A为坐标原点，AB,AD,AP所在直线分别为x轴,y轴,z轴建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1659094387180118020/img/compression/M_719a167f8e1539304ab491d5b5449bbf.jpg)$则A(0,0,0),B(2,0,0),P(0,0,2),C(2,2,0),D(0,2,0).因为E,F,G分别是BC,PC,CD的中点，所以E(2,1,0),F(1,1,1),G(1,2,0).证明：\overrightarrow{BG}=(-1,2,0)，\overrightarrow{AP}=(0,0,2),\overrightarrow{AE}=(2,1,0),因为\overrightarrow{BG}\cdot \overrightarrow{AP}=0，\overrightarrow{BG}\cdot \overrightarrow{AE}=0，所以BG\perp AP，BG\perp AE，又AE\cap AP=A，AE，AP\subset 平面PAE，所以BG\perp 平面PAE.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在底面是正方形的四棱锥P-ABCD中，PA\perp 平面ABCD，AP=AB=2，E,F,G分别是BC,PC,CD的中点.$<img_1585>
$(1)求证：BG\perp 平面PAE；$

Solution:
$因为四棱锥P-ABCD的底面是正方形，且PA\perp 平面ABCD，所以PA，AB，AD两两互相垂直.以A为坐标原点，AB,AD,AP所在直线分别为x轴,y轴,z轴建立如图所示的空间直角坐标系，$<img_2408>$则A(0,0,0),B(2,0,0),P(0,0,2),C(2,2,0),D(0,2,0).因为E,F,G分别是BC,PC,CD的中点，所以E(2,1,0),F(1,1,1),G(1,2,0).证明：\overrightarrow{BG}=(-1,2,0)，\overrightarrow{AP}=(0,0,2),\overrightarrow{AE}=(2,1,0),因为\overrightarrow{BG}\cdot \overrightarrow{AP}=0，\overrightarrow{BG}\cdot \overrightarrow{AE}=0，所以BG\perp AP，BG\perp AE，又AE\cap AP=A，AE，AP\subset 平面PAE，所以BG\perp 平面PAE.$
Subfield: Solid Geometry


Question: $如图，在底面是正方形的四棱锥P-ABCD中，PA\perp 平面ABCD，AP=AB=2，E,F,G分别是BC,PC,CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094387180118020/img/compression/M_50bc6f3349afc9ea7fe43078bb6de8eb.jpg)
$(2)在线段BG上是否存在点H，使得FH\parallel 平面PAE？若存在，求出\frac{BH}{BG}的值；若不存在，请说明理由.$

Solution:
$假设在线段BG上存在点H，使得FH\parallel 平面PAE.连接FB.设\overrightarrow{BH}=\lambda \overrightarrow{BG}(0\leq \lambda \leq 1)，则\overrightarrow{FH}=\overrightarrow{FB}+\overrightarrow{BH}=\overrightarrow{FB}+\lambda \overrightarrow{BG}=(1-\lambda ,2\lambda -1,-1).因为FH\parallel 平面PAE，BG\perp 平面PAE，所以\overrightarrow{BG}\cdot \overrightarrow{FH}=-1\times (1-\lambda )+2\times (2\lambda -1)+0\times (-1)=5\lambda -3=0，所以\lambda =\frac{3}{5}，所以在线段BG上存在点H，使得FH\parallel 平面PAE，且\frac{BH}{BG}=\frac{3}{5}.$

Final answer: $在线段BG上存在点H，使得FH\parallel 平面PAE，且\frac{BH}{BG}=\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，在底面是正方形的四棱锥P-ABCD中，PA\perp 平面ABCD，AP=AB=2，E,F,G分别是BC,PC,CD的中点.$<img_1585>
$(2)在线段BG上是否存在点H，使得FH\parallel 平面PAE？若存在，求出\frac{BH}{BG}的值；若不存在，请说明理由.$

Solution:
$假设在线段BG上存在点H，使得FH\parallel 平面PAE.连接FB.设\overrightarrow{BH}=\lambda \overrightarrow{BG}(0\leq \lambda \leq 1)，则\overrightarrow{FH}=\overrightarrow{FB}+\overrightarrow{BH}=\overrightarrow{FB}+\lambda \overrightarrow{BG}=(1-\lambda ,2\lambda -1,-1).因为FH\parallel 平面PAE，BG\perp 平面PAE，所以\overrightarrow{BG}\cdot \overrightarrow{FH}=-1\times (1-\lambda )+2\times (2\lambda -1)+0\times (-1)=5\lambda -3=0，所以\lambda =\frac{3}{5}，所以在线段BG上存在点H，使得FH\parallel 平面PAE，且\frac{BH}{BG}=\frac{3}{5}.$



Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 平面ABC，AC=BC=AA_{1}=1，AC\perp BC，且D，E，F分别为棱AB，BC，AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094388195139586/img/compression/M_844ecbbf4045e4241c204183239e6b9d.jpg)
$(1)证明直线A_{1}F与B_{1}E共面，并求其所成角的余弦值；$

Solution:
$\because E，F分别是棱BC，AC的中点，\therefore EF\parallel AB.由棱柱的性质易得A_{1}B_{1}\parallel AB，\therefore EF\parallel A_{1}B_{1}，\therefore E，F，A_{1}，B_{1}四点共面，即直线A_{1}F与B_{1}E共面.取A_{1}B_{1}的中点H，连接EH(图略)，易知四边形EFA_{1}H为平行四边形，故A_{1}F\parallel HE，则\angle HEB_{1}(或其补角)为直线A_{1}F与B_{1}E所成的角.\because AC=BC=1，AC\perp BC，\therefore AB=A_{1}B_{1}=\sqrt{2}，在\triangle HEB_{1}中，HB_{1}=\frac{1}{2}A_{1}B_{1}=\frac{\sqrt{2}}{2}，HE=A_{1}F=B_{1}E=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}，\therefore \\cos  \angle HEB_{1}=\frac{HE^2+B_1E^2-HB^2_1}{2 \cdot HE \cdot B_1E}=\frac{\frac{5}{4}+\frac{5}{4}-\frac{1}{2}}{2 \times \frac{\sqrt{5}}{2} \times \frac{\sqrt{5}}{2}}=\frac{4}{5}，即直线A_{1}F与B_{1}E所成角的余弦值为\frac{4}{5}.$

Final answer: $证明题，\frac{4}{5}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 平面ABC，AC=BC=AA_{1}=1，AC\perp BC，且D，E，F分别为棱AB，BC，AC的中点.$<img_2819>
$(1)证明直线A_{1}F与B_{1}E共面，并求其所成角的余弦值；$

Solution:
$\because E，F分别是棱BC，AC的中点，\therefore EF\parallel AB.由棱柱的性质易得A_{1}B_{1}\parallel AB，\therefore EF\parallel A_{1}B_{1}，\therefore E，F，A_{1}，B_{1}四点共面，即直线A_{1}F与B_{1}E共面.取A_{1}B_{1}的中点H，连接EH(图略)，易知四边形EFA_{1}H为平行四边形，故A_{1}F\parallel HE，则\angle HEB_{1}(或其补角)为直线A_{1}F与B_{1}E所成的角.\because AC=BC=1，AC\perp BC，\therefore AB=A_{1}B_{1}=\sqrt{2}，在\triangle HEB_{1}中，HB_{1}=\frac{1}{2}A_{1}B_{1}=\frac{\sqrt{2}}{2}，HE=A_{1}F=B_{1}E=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}，\therefore \\cos  \angle HEB_{1}=\frac{HE^2+B_1E^2-HB^2_1}{2 \cdot HE \cdot B_1E}=\frac{\frac{5}{4}+\frac{5}{4}-\frac{1}{2}}{2 \times \frac{\sqrt{5}}{2} \times \frac{\sqrt{5}}{2}}=\frac{4}{5}，即直线A_{1}F与B_{1}E所成角的余弦值为\frac{4}{5}.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 平面ABC，AC=BC=AA_{1}=1，AC\perp BC，且D，E，F分别为棱AB，BC，AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094388195139586/img/compression/M_844ecbbf4045e4241c204183239e6b9d.jpg)
$(2)在棱CC_{1}上是否存在点M，使得DM\perp 平面A_{1}B_{1}EF？若存在，求出\frac{CM}{CC_1}的值；若不存在，请说明理由.$

Solution:
$以C为坐标原点，建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1659094388459380737/img/compression/M_e49ebe072e3bd7d09a11763c634af8be.jpg)$则A_{1}(0,1,1)，B_{1}(1,0,1)，F\left(0,\frac{1}{2},0\right)，D\left(\frac{1}{2},\frac{1}{2},0\right).所以\overrightarrow{A_1B_1}=(1,-1,0)，\overrightarrow{A_1F}=\left(0,-\frac{1}{2},-1\right)，设M(0,0,m)(0\leq m\leq 1)，则\overrightarrow{DM}=\left(-\frac{1}{2},-\frac{1}{2},m\right)，要使DM\perp 平面A_{1}B_{1}EF，则\left\{\begin{matrix}\overrightarrow{DM} \cdot \overrightarrow{A_1B_1}=0,\\ \overrightarrow{DM} \cdot \overrightarrow{A_1F}=0,\end{matrix}\right.即\left\{\begin{matrix}-\frac{1}{2} \times 1+\left(-\frac{1}{2}\right) \times (-1)+m \times 0=0,\\ -\frac{1}{2} \times 0+\left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right)+m \times (-1)=0,\end{matrix}\right.解得m=\frac{1}{4}\in [0,1]，即\frac{CM}{CC_1}=\frac{1}{4}.故在棱CC_{1}上存在点M，使得DM\perp 平面A_{1}B_{1}EF，且\frac{CM}{CC_1}=\frac{1}{4}.$

Final answer: $1/4$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，AA_{1}\perp 平面ABC，AC=BC=AA_{1}=1，AC\perp BC，且D，E，F分别为棱AB，BC，AC的中点.$<img_2819>
$(2)在棱CC_{1}上是否存在点M，使得DM\perp 平面A_{1}B_{1}EF？若存在，求出\frac{CM}{CC_1}的值；若不存在，请说明理由.$

Solution:
$以C为坐标原点，建立如图所示的空间直角坐标系，$<img_2279>$则A_{1}(0,1,1)，B_{1}(1,0,1)，F\left(0,\frac{1}{2},0\right)，D\left(\frac{1}{2},\frac{1}{2},0\right).所以\overrightarrow{A_1B_1}=(1,-1,0)，\overrightarrow{A_1F}=\left(0,-\frac{1}{2},-1\right)，设M(0,0,m)(0\leq m\leq 1)，则\overrightarrow{DM}=\left(-\frac{1}{2},-\frac{1}{2},m\right)，要使DM\perp 平面A_{1}B_{1}EF，则\left\{\begin{matrix}\overrightarrow{DM} \cdot \overrightarrow{A_1B_1}=0,\\ \overrightarrow{DM} \cdot \overrightarrow{A_1F}=0,\end{matrix}\right.即\left\{\begin{matrix}-\frac{1}{2} \times 1+\left(-\frac{1}{2}\right) \times (-1)+m \times 0=0,\\ -\frac{1}{2} \times 0+\left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right)+m \times (-1)=0,\end{matrix}\right.解得m=\frac{1}{4}\in [0,1]，即\frac{CM}{CC_1}=\frac{1}{4}.故在棱CC_{1}上存在点M，使得DM\perp 平面A_{1}B_{1}EF，且\frac{CM}{CC_1}=\frac{1}{4}.$

Final answer: $1/4$
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=AB=AA_{1}，\angle CAB=90^\circ ，M是B_{1}C_{1}的中点，N是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1659094399700115457/img/compression/M_c322e081ae7cdc690532344a74b47e57.jpg)
$(1)求证：MN\parallel 平面ABB_{1}A_{1}；$

Solution:
$证明：取AB的中点H，连接HN，B_{1}H，则HN是\triangle ABC的中位线，所以HN\parallel BC，且HN=\frac{1}{2}BC，又因为B_{1}M\parallel BC，且B_{1}M=\frac{1}{2}BC，所以HN\parallel B_{1}M，且HN=B_{1}M，所以四边形HNMB_{1}是平行四边形，所以MN\parallel B_{1}H，又因为MN\not\subset 平面ABB_{1}A_{1}，B_{1}H\subset 平面ABB_{1}A_{1}，所以MN\parallel 平面ABB_{1}A_{1}.$![Image](https://cdntk.53zaixian.com/tq/1659094399964356610/img/compression/M_9077e0e1c3f88a3356eae140ed595baa.jpg)

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在直三棱柱ABC-A_{1}B_{1}C_{1}中，AC=AB=AA_{1}，\angle CAB=90^\circ ，M是B_{1}C_{1}的中点，N是AC的中点.$<img_2190>
$(1)求证：MN\parallel 平面ABB_{1}A_{1}；$

Solution:
$证明：取AB的中点H，连接HN，B_{1}H，则HN是\triangle ABC的中位线，所以HN\parallel BC，且HN=\frac{1}{2}BC，又因为B_{1}M\parallel BC，且B_{1}M=\frac{1}{2}BC，所以HN\parallel B_{1}M，且HN=B_{1}M，所以四边形HNMB_{1}是平行四边形，所以MN\parallel B_{1}H，又因为MN\not\subset 平面ABB_{1}A_{1}，B_{1}H\subset 平面ABB_{1}A_{1}，所以MN\parallel 平面ABB_{1}A_{1}.$<img_2392>