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


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$![Image](https://cdntk.53zaixian.com/tq/1623875235415261186/img/compression/M_c1f42b9b1254ea0dacb12e3e3689d3e0.jpg)
$(1)求证：AC\parallel 平面PDB；$

Solution:
$证明：因为AD\perp DB，且DB=1，AB=2，所以AD=\sqrt{3}，所以\angle DBA=60^\circ .因为\triangle ABC为正三角形，所以\angle CAB=60^\circ ，易知四边形ACBD为平面四边形，所以DB\parallel AC.因为AC\not\subset 平面PDB，DB\subset 平面PDB，所以AC\parallel 平面PDB.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$<img_1809>
$(1)求证：AC\parallel 平面PDB；$

Solution:
$证明：因为AD\perp DB，且DB=1，AB=2，所以AD=\sqrt{3}，所以\angle DBA=60^\circ .因为\triangle ABC为正三角形，所以\angle CAB=60^\circ ，易知四边形ACBD为平面四边形，所以DB\parallel AC.因为AC\not\subset 平面PDB，DB\subset 平面PDB，所以AC\parallel 平面PDB.$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$<img_1809>
$(2)求二面角P-AB-C的余弦值；$

Solution:
$由点P在平面ABC上的射影为D可得PD\perp 平面ACBD，所以PD\perp DA，PD\perp DB.分别以\overrightarrow{DB},\overrightarrow{DA},\overrightarrow{DP}的方向为x,y,z轴的正方向建立空间直角坐标系，则B(1,0,0)，A(0,\sqrt{3},0)，P(0,0，1)，C(2,\sqrt{3},0).所以\overrightarrow{BA}=(-1，\sqrt{3}，0)，\overrightarrow{BP}=(-1，0，1).易知平面ABC的一个法向量为n=(0,0，1)，设平面PAB的法向量为m=(x,y,z)，由\left\{\begin{matrix}\overrightarrow{BA} \cdot \boldsymbol{m}=0,\\ \overrightarrow{BP} \cdot \boldsymbol{m}=0\end{matrix}\right.可得\left\{\begin{matrix}-x+\sqrt{3}y=0,\\ -x+z=0,\end{matrix}\right.令y=1，则x=\sqrt{3},z=\sqrt{3}，所以平面PAB的一个法向量为m=(\sqrt{3},1,\sqrt{3})，所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{\sqrt{3}}{\sqrt{7} \times 1}=\frac{\sqrt{21}}{7}，由题图可知二面角P-AB-C为钝二面角，所以二面角P-AB-C的余弦值为-\frac{\sqrt{21}}{7}.$

Final answer: $-\frac{\sqrt{21}}{7}$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$![Image](https://cdntk.53zaixian.com/tq/1623875235415261186/img/compression/M_c1f42b9b1254ea0dacb12e3e3689d3e0.jpg)
$(2)求二面角P-AB-C的余弦值；$

Solution:
$由点P在平面ABC上的射影为D可得PD\perp 平面ACBD，所以PD\perp DA，PD\perp DB.分别以\overrightarrow{DB},\overrightarrow{DA},\overrightarrow{DP}的方向为x,y,z轴的正方向建立空间直角坐标系，则B(1,0,0)，A(0,\sqrt{3},0)，P(0,0，1)，C(2,\sqrt{3},0).所以\overrightarrow{BA}=(-1，\sqrt{3}，0)，\overrightarrow{BP}=(-1，0，1).易知平面ABC的一个法向量为n=(0,0，1)，设平面PAB的法向量为m=(x,y,z)，由\left\{\begin{matrix}\overrightarrow{BA} \cdot \boldsymbol{m}=0,\\ \overrightarrow{BP} \cdot \boldsymbol{m}=0\end{matrix}\right.可得\left\{\begin{matrix}-x+\sqrt{3}y=0,\\ -x+z=0,\end{matrix}\right.令y=1，则x=\sqrt{3},z=\sqrt{3}，所以平面PAB的一个法向量为m=(\sqrt{3},1,\sqrt{3})，所以\\cos  \langle m,n\rangle =\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{\sqrt{3}}{\sqrt{7} \times 1}=\frac{\sqrt{21}}{7}，由题图可知二面角P-AB-C为钝二面角，所以二面角P-AB-C的余弦值为-\frac{\sqrt{21}}{7}.$

Final answer: $-\frac{\sqrt{21}}{7}$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$<img_1809>
$(3)线段PC上是否存在点E，使得PC\perp 平面ABE？如果存在，求\frac{CE}{CP}的值；如果不存在，请说明理由.$

Solution:
$由(2)可得\overrightarrow{AB}=(1,-\sqrt{3},0)，\overrightarrow{PC}=(2,\sqrt{3},-1)，若PC\perp 平面ABE，则PC\perp AB，又\overrightarrow{PC}\cdot \overrightarrow{AB}=(2,\sqrt{3},-1)\cdot (1,-\sqrt{3},0)=-1\neq 0，所以PC与AB不垂直，所以在线段PC上不存在点E，使得PC\perp 平面ABE.$



Final answer: $在线段PC上不存在点E，使得PC\perp 平面ABE.$
Subfield: Solid Geometry


Question: $如图，在三棱锥P-ABC中，侧棱PA=2，底面三角形ABC为正三角形，边长为2，顶点P在平面ABC上的射影为D，有AD\perp DB，且DB=1.$![Image](https://cdntk.53zaixian.com/tq/1623875235415261186/img/compression/M_c1f42b9b1254ea0dacb12e3e3689d3e0.jpg)
$(3)线段PC上是否存在点E，使得PC\perp 平面ABE？如果存在，求\frac{CE}{CP}的值；如果不存在，请说明理由.$

Solution:
$由(2)可得\overrightarrow{AB}=(1,-\sqrt{3},0)，\overrightarrow{PC}=(2,\sqrt{3},-1)，若PC\perp 平面ABE，则PC\perp AB，又\overrightarrow{PC}\cdot \overrightarrow{AB}=(2,\sqrt{3},-1)\cdot (1,-\sqrt{3},0)=-1\neq 0，所以PC与AB不垂直，所以在线段PC上不存在点E，使得PC\perp 平面ABE.$

Final answer: $在线段PC上不存在点E，使得PC\perp 平面ABE.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$<img_1521>
$(1)求证：AA_{1}\perp 平面ABC；$

Solution:
$证明：因为四边形AA_{1}C_{1}C为正方形，所以AA_{1}\perp AC.因为平面ABC\perp 平面AA_{1}C_{1}C，且AA_{1}垂直于这两个平面的交线AC，所以AA_{1}\perp 平面ABC.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$![Image](https://cdntk.53zaixian.com/tq/1623875237856346114/img/compression/M_f510d7686a0562943289a3a35ff42b6d.jpg)
$(1)求证：AA_{1}\perp 平面ABC；$

Solution:
$证明：因为四边形AA_{1}C_{1}C为正方形，所以AA_{1}\perp AC.因为平面ABC\perp 平面AA_{1}C_{1}C，且AA_{1}垂直于这两个平面的交线AC，所以AA_{1}\perp 平面ABC.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$<img_1521>
$(2)求二面角A_{1}-BC_{1}-B_{1}的余弦值；$

Solution:
$由(1)知AA_{1}\perp AC，AA_{1}\perp AB.由题知AC=4，又AB=3，BC=5，所以AB^{2}+AC^{2}=BC^{2}，所以AB\perp AC.如图，以A为原点建立空间直角坐标系A-xyz，则B(0，3，0)，A_{1}(0，0，4)，B_{1}(0，3，4)，C_{1}(4，0，4).$<img_1536>$设平面A_{1}BC_{1}的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{A_1B}=0,\\ \boldsymbol{n} \cdot \overrightarrow{A_1C_1}=0,\end{matrix}\right.即\left\{\begin{matrix}3y-4z=0,\\ 4x=0.\end{matrix}\right.令z=3，则y=4，所以n=(0,4,3).同理可得，平面B_{1}BC_{1}的一个法向量为m=(3,4,0).所以\\cos  \langle n,m\rangle =\frac{\boldsymbol{n} \cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}=\frac{16}{25}.由图知二面角A_{1}-BC_{1}-B_{1}为锐二面角，所以二面角A_{1}-BC_{1}-B_{1}的余弦值为\frac{16}{25}.$

Final answer: $\frac{16}{25}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$![Image](https://cdntk.53zaixian.com/tq/1623875237856346114/img/compression/M_f510d7686a0562943289a3a35ff42b6d.jpg)
$(2)求二面角A_{1}-BC_{1}-B_{1}的余弦值；$

Solution:
$由(1)知AA_{1}\perp AC，AA_{1}\perp AB.由题知AC=4，又AB=3，BC=5，所以AB^{2}+AC^{2}=BC^{2}，所以AB\perp AC.如图，以A为原点建立空间直角坐标系A-xyz，则B(0，3，0)，A_{1}(0，0，4)，B_{1}(0，3，4)，C_{1}(4，0，4).$![Image](https://cdntk.53zaixian.com/tq/1623875238602932225/img/compression/M_a846298e18df3c896c9b090b525eceb9.jpg)$设平面A_{1}BC_{1}的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{A_1B}=0,\\ \boldsymbol{n} \cdot \overrightarrow{A_1C_1}=0,\end{matrix}\right.即\left\{\begin{matrix}3y-4z=0,\\ 4x=0.\end{matrix}\right.令z=3，则y=4，所以n=(0,4,3).同理可得，平面B_{1}BC_{1}的一个法向量为m=(3,4,0).所以\\cos  \langle n,m\rangle =\frac{\boldsymbol{n} \cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}=\frac{16}{25}.由图知二面角A_{1}-BC_{1}-B_{1}为锐二面角，所以二面角A_{1}-BC_{1}-B_{1}的余弦值为\frac{16}{25}.$

Final answer: $\frac{16}{25}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$<img_1521>
$(3)证明：在线段BC_{1}上存在点D，使得AD\perp A_{1}B.并求\frac{BD}{BC_1}的值.$

Solution:
$设D(x,y,z)是线段BC_{1}上一点，且\overrightarrow{BD}=\lambda \overrightarrow{BC_1}(0\leq \lambda \leq 1)，则(x,y-3,z)=\lambda (4,-3,4)，解得x=4\lambda ,y=3-3\lambda ,z=4\lambda .所以\overrightarrow{AD}=(4\lambda ,3-3\lambda ,4\lambda ).由\overrightarrow{AD}\cdot \overrightarrow{A_1B}=0，即9-25\lambda =0,解得\lambda =\frac{9}{25}.因为\frac{9}{25}\in [0,1]，所以在线段BC_{1}上存在点D，使得AD\perp A_{1}B.此时，\frac{BD}{BC_1}=\lambda =\frac{9}{25}.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，四边形AA_{1}C_{1}C是边长为4的正方形，平面ABC\perp 平面AA_{1}C_{1}C，AB=3，BC=5.$![Image](https://cdntk.53zaixian.com/tq/1623875237856346114/img/compression/M_f510d7686a0562943289a3a35ff42b6d.jpg)
$(3)证明：在线段BC_{1}上存在点D，使得AD\perp A_{1}B.并求\frac{BD}{BC_1}的值.$

Solution:
$设D(x,y,z)是线段BC_{1}上一点，且\overrightarrow{BD}=\lambda \overrightarrow{BC_1}(0\leq \lambda \leq 1)，则(x,y-3,z)=\lambda (4,-3,4)，解得x=4\lambda ,y=3-3\lambda ,z=4\lambda .所以\overrightarrow{AD}=(4\lambda ,3-3\lambda ,4\lambda ).由\overrightarrow{AD}\cdot \overrightarrow{A_1B}=0，即9-25\lambda =0,解得\lambda =\frac{9}{25}.因为\frac{9}{25}\in [0,1]，所以在线段BC_{1}上存在点D，使得AD\perp A_{1}B.此时，\frac{BD}{BC_1}=\lambda =\frac{9}{25}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，BB_{1}\perp 平面ABC，AB=BC=BB_{1}=1.$<img_3113>
$(1)求证：AC\parallel 平面BA_{1}C_{1}；$

Solution:
$证明：在三棱柱ABC－A_{1}B_{1}C_{1} 中，四边形 AA_{1}C_{1}C为平行四边形.所以AC\parallel A_{1}C_{1}.因为AC\not\subset 平面BA_{1}C_{1}，A_{1}C_{1}\subset 平面 BA_{1}C_{1}，所以AC\parallel 平面BA_{1}C_{1}.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，BB_{1}\perp 平面ABC，AB=BC=BB_{1}=1.$![Image](https://cdntk.53zaixian.com/tq/1623884770262609921/img/compression/M_afa62f1e1f6d8605ecceb3d27c8ab2a7.jpg)
$(1)求证：AC\parallel 平面BA_{1}C_{1}；$

Solution:
$证明：在三棱柱ABC－A_{1}B_{1}C_{1} 中，四边形 AA_{1}C_{1}C为平行四边形.所以AC\parallel A_{1}C_{1}.因为AC\not\subset 平面BA_{1}C_{1}，A_{1}C_{1}\subset 平面 BA_{1}C_{1}，所以AC\parallel 平面BA_{1}C_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，BB_{1}\perp 平面ABC，AB=BC=BB_{1}=1.$<img_3113>
$(2)若AB\perp BC，求：①AA_{1}与平面BA_{1}C_{1}所成角的正弦值；$

Solution:
$因为 BB_{1}\perp 平面ABC，AB,BC\subset 平面ABC，所以BB_{1}\perp AB,BB_{1}\perp BC.又AB\perp BC，所以AB,BB_{1},BC两两垂直.如图,建立空间直角坐标系B－xyz，$<img_2582>$则A(1,0,0),B_{1}(0,1,0),C_{1}(0,1,1),A_{1}(1，1，0)，B(0,0,0).所以\overrightarrow{BA_1}=(1,1,0)，\overrightarrow{BC_1}=(0,1,1)，\overrightarrow{AA_1}=(0,1,0).设平面BA_{1}C_{1}的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{BA_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{BC_1}=0,\end{matrix}\right.即\left\{\begin{matrix}x+y=0,\\ y+z=0.\end{matrix}\right.令x＝1，则y＝－1，z＝1.于是n=(1,－1,1).①设直线 AA_{1}与平面 BA_{1}C_{1} 所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{AA_1},n\rangle |=\left|\frac{\overrightarrow{AA_1} \cdot \boldsymbol{n}}{|\overrightarrow{AA_1}||\boldsymbol{n}|}\right|=\frac{\sqrt{3}}{3}.所以AA_{1}与平面 BA_{1}C_{1}所成角的正弦值为\frac{\sqrt{3}}{3}.②因为 AC\parallel 平面 BA_{1}C_{1}，所以直线AC 与平面BA_{1}C_{1}的距离等于点A到平面BA_{1}C_{1}的距离.设点A到平面BA_{1}C_{1}的距离为h，则h=\frac{|\overrightarrow{AA_1} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{|(0,1,0) \cdot (1,-1,1)|}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$



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


Question: $如图，在三棱柱ABC－A_{1}B_{1}C_{1}中，BB_{1}\perp 平面ABC，AB=BC=BB_{1}=1.$![Image](https://cdntk.53zaixian.com/tq/1623884770262609921/img/compression/M_afa62f1e1f6d8605ecceb3d27c8ab2a7.jpg)
$(2)若AB\perp BC，求：①AA_{1}与平面BA_{1}C_{1}所成角的正弦值；②直线AC与平面BA_{1}C_{1}的距离.$

Solution:
$因为 BB_{1}\perp 平面ABC，AB,BC\subset 平面ABC，所以BB_{1}\perp AB,BB_{1}\perp BC.又AB\perp BC，所以AB,BB_{1},BC两两垂直.如图,建立空间直角坐标系B－xyz，$![Image](https://cdntk.53zaixian.com/tq/1623884771067916289/img/compression/M_3d5d41b1fa188ea36f03fc7e3715990e.jpg)$则A(1,0,0),B_{1}(0,1,0),C_{1}(0,1,1),A_{1}(1，1，0)，B(0,0,0).所以\overrightarrow{BA_1}=(1,1,0)，\overrightarrow{BC_1}=(0,1,1)，\overrightarrow{AA_1}=(0,1,0).设平面BA_{1}C_{1}的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{BA_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{BC_1}=0,\end{matrix}\right.即\left\{\begin{matrix}x+y=0,\\ y+z=0.\end{matrix}\right.令x＝1，则y＝－1，z＝1.于是n=(1,－1,1).①设直线 AA_{1}与平面 BA_{1}C_{1} 所成角为\theta ，则\\sin   \theta =|\\cos  \langle \overrightarrow{AA_1},n\rangle |=\left|\frac{\overrightarrow{AA_1} \cdot \boldsymbol{n}}{|\overrightarrow{AA_1}||\boldsymbol{n}|}\right|=\frac{\sqrt{3}}{3}.所以AA_{1}与平面 BA_{1}C_{1}所成角的正弦值为\frac{\sqrt{3}}{3}.②因为 AC\parallel 平面 BA_{1}C_{1}，所以直线AC 与平面BA_{1}C_{1}的距离等于点A到平面BA_{1}C_{1}的距离.设点A到平面BA_{1}C_{1}的距离为h，则h=\frac{|\overrightarrow{AA_1} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{|(0,1,0) \cdot (1,-1,1)|}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$

Final answer: $①AA_{1}与平面BA_{1}C_{1}所成角的正弦值为\frac{\sqrt{3}}{3}.
②直线AC与平面BA_{1}C_{1}的距离为\frac{\sqrt{3}}{3}.$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}=AB，D为BC的中点，平面A_{1}C_{1}D\cap 平面ABC=DP.$<img_2801>
$(1)求证：A_{1}C_{1}\parallel DP；$

Solution:
$证明：由题意得平面A_{1}B_{1}C_{1}\parallel 平面ABC.因为平面A_{1}C_{1}D\cap 平面ABC=DP，平面A_{1}C_{1}D\cap 平面A_{1}B_{1}C_{1}=A_{1}C_{1}，所以A_{1}C_{1}\parallel DP.$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}=AB，D为BC的中点，平面A_{1}C_{1}D\cap 平面ABC=DP.$![Image](https://cdntk.53zaixian.com/tq/1623884771457986563/img/compression/M_2741233e2102342ee95e9317f2885333.jpg)
$(1)求证：A_{1}C_{1}\parallel DP；$

Solution:
$证明：由题意得平面A_{1}B_{1}C_{1}\parallel 平面ABC.因为平面A_{1}C_{1}D\cap 平面ABC=DP，平面A_{1}C_{1}D\cap 平面A_{1}B_{1}C_{1}=A_{1}C_{1}，所以A_{1}C_{1}\parallel DP.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}=AB，D为BC的中点，平面A_{1}C_{1}D\cap 平面ABC=DP.$![Image](https://cdntk.53zaixian.com/tq/1623884771457986563/img/compression/M_2741233e2102342ee95e9317f2885333.jpg)
$(2)求平面A_{1}C_{1}D与平面AA_{1}D夹角的余弦值.$

Solution:
$取B_{1}C_{1}的中点D_{1}，连接DD_{1}，则DD_{1}\parallel CC_{1}.由题意得CC_{1}\perp 平面ABC，且底面ABC为等边三角形，所以DD_{1}\perp 平面ABC，AD\perp BC，所以DA、DB、DD_{1}两两垂直.以点D为坐标原点，DA，DB，DD_{1}所在直线分别为x轴、y轴、z轴，建立如图所示的空间直角坐标系D－xyz.$![Image](https://cdntk.53zaixian.com/tq/1623884772032606209/img/compression/M_745b0e156b6a671d82b7f5d0f70bb5bc.jpg)$设AA_{1}=2，则D(0,0,0),A_{1}(\sqrt{3},0,2),C_{1}(0,－1,2),B(0,1,0),所以\overrightarrow{DA_1}=(\sqrt{3},0,2),\overrightarrow{DC_1}=(0,－1,2).设平面A_{1}C_{1}D的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{DA_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{DC_1}=0,\end{matrix}\right.所以\left\{\begin{matrix}\sqrt{3}x+2z=0,\\ -y+2z=0.\end{matrix}\right.令z=－\sqrt{3}，则x=2，y=－2\sqrt{3}，所以n=(2,－2\sqrt{3},－\sqrt{3}).易知\overrightarrow{DB}=(0,1,0)是平面AA_{1}D的一个法向量.设平面A_{1}C_{1}D与平面AA_{1}D的夹角为\theta ,则\\cos   \theta =|\\cos  \langle n,\overrightarrow{DB}\rangle |=\frac{|\boldsymbol{n} \cdot \overrightarrow{DB}|}{|\boldsymbol{n}| \cdot |\overrightarrow{DB}|}=\frac{2\sqrt{57}}{19}，所以平面A_{1}C_{1}D与平面AA_{1}D夹角的余弦值为\frac{2\sqrt{57}}{19}.$

Final answer: $\frac{2\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图，在正三棱柱ABC－A_{1}B_{1}C_{1}中，AA_{1}=AB，D为BC的中点，平面A_{1}C_{1}D\cap 平面ABC=DP.$<img_2801>
$(2)求平面A_{1}C_{1}D与平面AA_{1}D夹角的余弦值.$

Solution:
$取B_{1}C_{1}的中点D_{1}，连接DD_{1}，则DD_{1}\parallel CC_{1}.由题意得CC_{1}\perp 平面ABC，且底面ABC为等边三角形，所以DD_{1}\perp 平面ABC，AD\perp BC，所以DA、DB、DD_{1}两两垂直.以点D为坐标原点，DA，DB，DD_{1}所在直线分别为x轴、y轴、z轴，建立如图所示的空间直角坐标系D－xyz.$<img_2880>$设AA_{1}=2，则D(0,0,0),A_{1}(\sqrt{3},0,2),C_{1}(0,－1,2),B(0,1,0),所以\overrightarrow{DA_1}=(\sqrt{3},0,2),\overrightarrow{DC_1}=(0,－1,2).设平面A_{1}C_{1}D的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{DA_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{DC_1}=0,\end{matrix}\right.所以\left\{\begin{matrix}\sqrt{3}x+2z=0,\\ -y+2z=0.\end{matrix}\right.令z=－\sqrt{3}，则x=2，y=－2\sqrt{3}，所以n=(2,－2\sqrt{3},－\sqrt{3}).易知\overrightarrow{DB}=(0,1,0)是平面AA_{1}D的一个法向量.设平面A_{1}C_{1}D与平面AA_{1}D的夹角为\theta ,则\\cos   \theta =|\\cos  \langle n,\overrightarrow{DB}\rangle |=\frac{|\boldsymbol{n} \cdot \overrightarrow{DB}|}{|\boldsymbol{n}| \cdot |\overrightarrow{DB}|}=\frac{2\sqrt{57}}{19}，所以平面A_{1}C_{1}D与平面AA_{1}D夹角的余弦值为\frac{2\sqrt{57}}{19}.$

Final answer: $\frac{2\sqrt{57}}{19}$
Subfield: Solid Geometry


Question: $如图，已知长方体ABCD－A_{1}B_{1}C_{1}D_{1}中，AB=AD=2，AA_{1}=1.E为A_{1}D_{1}的中点，平面CB_{1}E交棱DD_{1}于点F.$![Image](https://cdntk.53zaixian.com/tq/1623884775782313986/img/compression/M_a6f2282eb5a0a2be8d0620b07682713a.jpg)
$(1)求证：B_{1}C\parallel EF；$

Solution:
$证明：由长方体的性质知：平面BCC_{1}B_{1}\parallel 平面ADD_{1}A_{1}，又B_{1}C\subset 平面BCC_{1}B_{1}，\therefore B_{1}C\parallel 平面ADD_{1}A_{1}，又平面CB_{1}E\cap 平面ADD_{1}A_{1}=EF，且B_{1}C\subset 平面CB_{1}E，\therefore B_{1}C\parallel EF.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，已知长方体ABCD－A_{1}B_{1}C_{1}D_{1}中，AB=AD=2，AA_{1}=1.E为A_{1}D_{1}的中点，平面CB_{1}E交棱DD_{1}于点F.$<img_3095>
$(1)求证：B_{1}C\parallel EF；$

Solution:
$证明：由长方体的性质知：平面BCC_{1}B_{1}\parallel 平面ADD_{1}A_{1}，又B_{1}C\subset 平面BCC_{1}B_{1}，\therefore B_{1}C\parallel 平面ADD_{1}A_{1}，又平面CB_{1}E\cap 平面ADD_{1}A_{1}=EF，且B_{1}C\subset 平面CB_{1}E，\therefore B_{1}C\parallel EF.$
Subfield: Solid Geometry


Question: $如图，已知长方体ABCD－A_{1}B_{1}C_{1}D_{1}中，AB=AD=2，AA_{1}=1.E为A_{1}D_{1}的中点，平面CB_{1}E交棱DD_{1}于点F.$<img_3095>
$(2)求二面角C－B_{1}E－C_{1}的余弦值，并求点A到平面CB_{1}E的距离.$

Solution:
$以A为坐标原点，AB，AD，AA_{1}所在直线分别为x轴，y轴，z轴建立空间直角坐标系，则A(0,0,0)，B_{1}(2,0,1)，E(0,1,1)，C(2,2,0)，C_{1}(2,2,1)，$<img_3045>$\therefore \overrightarrow{EB_1}=(2,－1,0)，\overrightarrow{CB_1}=(0,－2,1)，设平面CB_{1}E的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{EB_1}=0,\\ \boldsymbol{m} \cdot \overrightarrow{CB_1}=0,\end{matrix}\right.即\left\{\begin{matrix}2x-y=0,\\ -2y+z=0,\end{matrix}\right.令y=2，则m=(1,2,4)，易知平面C_{1}B_{1}E的一个法向量为n=(0,0,1)，\therefore \\cos  \langle m,n\rangle =\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{4}{\sqrt{1+4+16}}=\frac{4\sqrt{21}}{21}，由图可知二面角C-B_{1}E-C_{1}是锐二面角，所以二面角C-B_{1}E-C_{1}的余弦值是\frac{4\sqrt{21}}{21}.\overrightarrow{AC}=(2,2,0)，\therefore A到平面CB_{1}E的距离为\frac{|\boldsymbol{m} \cdot \overrightarrow{AC}|}{|\boldsymbol{m}|}=\frac{6}{\sqrt{21}}=\frac{2\sqrt{21}}{7}.$

Final answer: $\frac{4\sqrt{21}}{21}$,$\frac{2\sqrt{21}}{7}$
Subfield: Solid Geometry


Question: $如图，已知长方体ABCD－A_{1}B_{1}C_{1}D_{1}中，AB=AD=2，AA_{1}=1.E为A_{1}D_{1}的中点，平面CB_{1}E交棱DD_{1}于点F.$![Image](https://cdntk.53zaixian.com/tq/1623884775782313986/img/compression/M_a6f2282eb5a0a2be8d0620b07682713a.jpg)
$(2)求二面角C－B_{1}E－C_{1}的余弦值，并求点A到平面CB_{1}E的距离.$

Solution:
$以A为坐标原点，AB，AD，AA_{1}所在直线分别为x轴，y轴，z轴建立空间直角坐标系，则A(0,0,0)，B_{1}(2,0,1)，E(0,1,1)，C(2,2,0)，C_{1}(2,2,1)，$![Image](https://cdntk.53zaixian.com/tq/1623884776046555138/img/compression/M_4245fd755b25d8701bd16ca9031263ef.jpg)$\therefore \overrightarrow{EB_1}=(2,－1,0)，\overrightarrow{CB_1}=(0,－2,1)，设平面CB_{1}E的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m} \cdot \overrightarrow{EB_1}=0,\\ \boldsymbol{m} \cdot \overrightarrow{CB_1}=0,\end{matrix}\right.即\left\{\begin{matrix}2x-y=0,\\ -2y+z=0,\end{matrix}\right.令y=2，则m=(1,2,4)，易知平面C_{1}B_{1}E的一个法向量为n=(0,0,1)，\therefore \\cos  \langle m,n\rangle =\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{4}{\sqrt{1+4+16}}=\frac{4\sqrt{21}}{21}，由图可知二面角C-B_{1}E-C_{1}是锐二面角，所以二面角C-B_{1}E-C_{1}的余弦值是\frac{4\sqrt{21}}{21}.\overrightarrow{AC}=(2,2,0)，\therefore A到平面CB_{1}E的距离为\frac{|\boldsymbol{m} \cdot \overrightarrow{AC}|}{|\boldsymbol{m}|}=\frac{6}{\sqrt{21}}=\frac{2\sqrt{21}}{7}.$

Final answer: $\frac{4\sqrt{21}}{21}$,$\frac{2\sqrt{21}}{7}$
Subfield: Solid Geometry


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$<img_2663>
$(1)求证：AB_{1}\perp 平面A_{1}BM；$

Solution:
$证明：在正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,因为BC\perp 平面AA_{1}B_{1}B,AB_{1}\subset 平面AA_{1}B_{1}B,所以BC\perp AB_{1},即BM\perp AB_{1}.因为四边形AA_{1}B_{1}B是正方形,所以A_{1}B\perp AB_{1}.因为A_{1}B\cap BM=B,A_{1}B,BM\subset 平面A_{1}BM,所以AB_{1}\perp 平面A_{1}BM.$
Subfield: Solid Geometry


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1623884776419848195/img/compression/M_e4ee4af25511c26f9740eab900a8e9f4.jpg)
$(1)求证：AB_{1}\perp 平面A_{1}BM；$

Solution:
$证明：在正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,因为BC\perp 平面AA_{1}B_{1}B,AB_{1}\subset 平面AA_{1}B_{1}B,所以BC\perp AB_{1},即BM\perp AB_{1}.因为四边形AA_{1}B_{1}B是正方形,所以A_{1}B\perp AB_{1}.因为A_{1}B\cap BM=B,A_{1}B,BM\subset 平面A_{1}BM,所以AB_{1}\perp 平面A_{1}BM.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1623884776419848195/img/compression/M_e4ee4af25511c26f9740eab900a8e9f4.jpg)
$(2)求二面角B－A_{1}M－C_{1}的大小；$

Solution:
$如图,建立空间直角坐标系D－xyz,则A(2,0,0),B_{1}(2,2,2),A_{1}(2,0,2),M(1,2,0),C_{1}(0,2,2).所以\overrightarrow{AB_1}=(0,2,2),\overrightarrow{A_1C_1}=(－2,2,0),\overrightarrow{C_1M}=(1,0,－2).$![Image](https://cdntk.53zaixian.com/tq/1623884776877027330/img/compression/M_c51489bcbc31abfcb094a49f08cd8f59.jpg)$由(1)知,平面A_{1}BM的一个法向量为\overrightarrow{AB_1}=(0,2,2),设平面A_{1}MC_{1}的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{A_1C_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{C_1M}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+2y=0,\\ x-2z=0,\end{matrix}\right.令x=2,则y=2,z=1,所以n=(2,2,1).所以\\cos  \langle n,\overrightarrow{AB_1}\rangle =\frac{\boldsymbol{n} \cdot \overrightarrow{AB_1}}{|\boldsymbol{n}| \cdot |\overrightarrow{AB_1}|}=\frac{\sqrt{2}}{2}.由图可知,二面角B－A_{1}M－C_{1}为钝二面角,所以二面角B－A_{1}M－C_{1}的大小为135^\circ .$

Final answer: $135^\circ $
Subfield: Solid Geometry


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$<img_2663>
$(2)求二面角B－A_{1}M－C_{1}的大小；$

Solution:
$如图,建立空间直角坐标系D－xyz,则A(2,0,0),B_{1}(2,2,2),A_{1}(2,0,2),M(1,2,0),C_{1}(0,2,2).所以\overrightarrow{AB_1}=(0,2,2),\overrightarrow{A_1C_1}=(－2,2,0),\overrightarrow{C_1M}=(1,0,－2).$<img_2284>$由(1)知,平面A_{1}BM的一个法向量为\overrightarrow{AB_1}=(0,2,2),设平面A_{1}MC_{1}的法向量为n=(x,y,z),则\left\{\begin{matrix}\boldsymbol{n} \cdot \overrightarrow{A_1C_1}=0,\\ \boldsymbol{n} \cdot \overrightarrow{C_1M}=0,\end{matrix}\right.即\left\{\begin{matrix}-2x+2y=0,\\ x-2z=0,\end{matrix}\right.令x=2,则y=2,z=1,所以n=(2,2,1).所以\\cos  \langle n,\overrightarrow{AB_1}\rangle =\frac{\boldsymbol{n} \cdot \overrightarrow{AB_1}}{|\boldsymbol{n}| \cdot |\overrightarrow{AB_1}|}=\frac{\sqrt{2}}{2}.由图可知,二面角B－A_{1}M－C_{1}为钝二面角,所以二面角B－A_{1}M－C_{1}的大小为135^\circ .$

Final answer: $135^\circ $
Subfield: Solid Geometry


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$<img_2663>
$(3)求点A到平面A_{1}MC_{1}的距离.$

Solution:
$设点A到平面A_{1}MC_{1}的距离为d,因为\overrightarrow{AA_1}=(0,0,2),n=(2,2,1),所以d=\frac{|\overrightarrow{AA_1} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{2}{3}.所以点A到平面A_{1}MC_{1}的距离为\frac{2}{3}.$

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


Question: $如图,在棱长为2的正方体ABCD－A_{1}B_{1}C_{1}D_{1}中,点M是BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1623884776419848195/img/compression/M_e4ee4af25511c26f9740eab900a8e9f4.jpg)
$(3)求点A到平面A_{1}MC_{1}的距离.$

Solution:
$设点A到平面A_{1}MC_{1}的距离为d,因为\overrightarrow{AA_1}=(0,0,2),n=(2,2,1),所以d=\frac{|\overrightarrow{AA_1} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{2}{3}.所以点A到平面A_{1}MC_{1}的距离为\frac{2}{3}.$

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


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，侧面ABB_{1}A_{1}为矩形，平面ABB_{1}A_{1}\perp 平面ACC_{1}A_{1}，AB=2，AA_{1}=4，D，E分别是BC，A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1623884777413898242/img/compression/M_9f4508f0a2185a75460dfce6843f9e18.jpg)
$(1)求证：DE\parallel 平面ACC_{1}A_{1}；$

Solution:
$证明：取AC中点F，连接DF,A_{1}F，如图所示，$![Image](https://cdntk.53zaixian.com/tq/1623884777887854593/img/compression/M_8462f14fe9b129f5183fe07b728d00ef.jpg)$因为点D,F分别为BC,AC的中点，故DF\parallel AB，DF=\frac{1}{2}AB，又点E为A_{1}B_{1}的中点，且四边形ABB_{1}A_{1}为矩形，故A_{1}E\parallel AB，A_{1}E=\frac{1}{2}AB，所以A_{1}E\parallel DF，且A_{1}E=DF，所以四边形A_{1}EDF是平行四边形，则DE\parallel A_{1}F，又DE\not\subset 平面ACC_{1}A_{1},A_{1}F\subset 平面ACC_{1}A_{1}，所以DE\parallel 平面ACC_{1}A_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，侧面ABB_{1}A_{1}为矩形，平面ABB_{1}A_{1}\perp 平面ACC_{1}A_{1}，AB=2，AA_{1}=4，D，E分别是BC，A_{1}B_{1}的中点.$<img_2691>
$(1)求证：DE\parallel 平面ACC_{1}A_{1}；$

Solution:
$证明：取AC中点F，连接DF,A_{1}F，如图所示，$<img_1664>$因为点D,F分别为BC,AC的中点，故DF\parallel AB，DF=\frac{1}{2}AB，又点E为A_{1}B_{1}的中点，且四边形ABB_{1}A_{1}为矩形，故A_{1}E\parallel AB，A_{1}E=\frac{1}{2}AB，所以A_{1}E\parallel DF，且A_{1}E=DF，所以四边形A_{1}EDF是平行四边形，则DE\parallel A_{1}F，又DE\not\subset 平面ACC_{1}A_{1},A_{1}F\subset 平面ACC_{1}A_{1}，所以DE\parallel 平面ACC_{1}A_{1}.$