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


Question: $如图，三棱锥A-BCD中，DA=DB=DC,BD\perp CD，\angle ADB=\angle ADC=60^\circ ，E为BC的中点.$<img_2719>
$(2)点F满足\overrightarrow{EF}=\overrightarrow{DA}，求二面角D-AB-F的正弦值.$

Solution:
$设DA=DB=DC=2，则BC=2\sqrt{2}，DE=AE=\sqrt{2}，\therefore AE^{2}+DE^{2}=4=DA^{2},\therefore AE\perp DE.又\because AE\perp BC,DE\cap BC=E,DE\subset 平面BCD,BC\subset 平面BCD,\therefore AE\perp 平面BCD.以E为原点，\overrightarrow{ED}，\overrightarrow{EB}，\overrightarrow{EA}的方向分别为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系，$<img_2263>$则D(\sqrt{2}，0，0)，A(0，0，\sqrt{2})，B(0，\sqrt{2}，0)，E(0，0，0)，\therefore \overrightarrow{DA}=(-\sqrt{2}，0，\sqrt{2})，\overrightarrow{AB}=(0，\sqrt{2}，-\sqrt{2})，\because \overrightarrow{EF}=\overrightarrow{DA}，\therefore F(-\sqrt{2}，0，\sqrt{2})，\therefore \overrightarrow{AF}=(-\sqrt{2}，0，0).设平面DAB的法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\overrightarrow{DA}\cdot \boldsymbol{n}_1=0,\\ \overrightarrow{AB}\cdot \boldsymbol{n}_1=0,\end{matrix}\right.即\left\{\begin{matrix}-\sqrt{2}x_1+\sqrt{2}z_1=0,\\ \sqrt{2}y_1-\sqrt{2}z_1=0,\end{matrix}\right.令z_{1}=1,则n_{1}=(1,1,1).设平面ABF的法向量为n_{2}=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{AB}\cdot \boldsymbol{n}_2=0,\\ \overrightarrow{AF}\cdot \boldsymbol{n}_2=0,\end{matrix}\right.即\left\{\begin{matrix}\sqrt{2}y_2-\sqrt{2}z_2=0,\\ -\sqrt{2}x_2=0,\end{matrix}\right.令z_{2}=1,则n_{2}=(0,1,1).设二面角D-AB-F的平面角为\theta ,则|\\cos   \theta |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{2}{\sqrt{3}\times \sqrt{2}}=\frac{\sqrt{6}}{3}.又\because \theta \in [0,\pi ],\therefore \\sin   \theta =\sqrt{1-\\\cos   ^2\theta }=\sqrt{1-\left(\frac{\sqrt{6}}{3}\right)^2}=\frac{\sqrt{3}}{3},\therefore 二面角D-AB-F的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图，三棱锥A-BCD中，DA=DB=DC,BD\perp CD，\angle ADB=\angle ADC=60^\circ ，E为BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283941481594883/img/compression/M_de29da90feca03c43a87ef0366594e17.jpg)
$(2)点F满足\overrightarrow{EF}=\overrightarrow{DA}，求二面角D-AB-F的正弦值.$

Solution:
$设DA=DB=DC=2，则BC=2\sqrt{2}，DE=AE=\sqrt{2}，\therefore AE^{2}+DE^{2}=4=DA^{2},\therefore AE\perp DE.又\because AE\perp BC,DE\cap BC=E,DE\subset 平面BCD,BC\subset 平面BCD,\therefore AE\perp 平面BCD.以E为原点，\overrightarrow{ED}，\overrightarrow{EB}，\overrightarrow{EA}的方向分别为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1715283941921996802/img/compression/M_2ae0e961704f707a4acac4481f33ebf0.jpg)$则D(\sqrt{2}，0，0)，A(0，0，\sqrt{2})，B(0，\sqrt{2}，0)，E(0，0，0)，\therefore \overrightarrow{DA}=(-\sqrt{2}，0，\sqrt{2})，\overrightarrow{AB}=(0，\sqrt{2}，-\sqrt{2})，\because \overrightarrow{EF}=\overrightarrow{DA}，\therefore F(-\sqrt{2}，0，\sqrt{2})，\therefore \overrightarrow{AF}=(-\sqrt{2}，0，0).设平面DAB的法向量为n_{1}=(x_{1},y_{1},z_{1}),则\left\{\begin{matrix}\overrightarrow{DA}\cdot \boldsymbol{n}_1=0,\\ \overrightarrow{AB}\cdot \boldsymbol{n}_1=0,\end{matrix}\right.即\left\{\begin{matrix}-\sqrt{2}x_1+\sqrt{2}z_1=0,\\ \sqrt{2}y_1-\sqrt{2}z_1=0,\end{matrix}\right.令z_{1}=1,则n_{1}=(1,1,1).设平面ABF的法向量为n_{2}=(x_{2},y_{2},z_{2}),则\left\{\begin{matrix}\overrightarrow{AB}\cdot \boldsymbol{n}_2=0,\\ \overrightarrow{AF}\cdot \boldsymbol{n}_2=0,\end{matrix}\right.即\left\{\begin{matrix}\sqrt{2}y_2-\sqrt{2}z_2=0,\\ -\sqrt{2}x_2=0,\end{matrix}\right.令z_{2}=1,则n_{2}=(0,1,1).设二面角D-AB-F的平面角为\theta ,则|\\cos   \theta |=\frac{|\boldsymbol{n}_1\cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=\frac{2}{\sqrt{3}\times \sqrt{2}}=\frac{\sqrt{6}}{3}.又\because \theta \in [0,\pi ],\therefore \\sin   \theta =\sqrt{1-\\\cos   ^2\theta }=\sqrt{1-\left(\frac{\sqrt{6}}{3}\right)^2}=\frac{\sqrt{3}}{3},\therefore 二面角D-AB-F的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283942374981635/img/compression/M_07622067350e8856c0d52e7d8bfe96af.jpg)
$(1)求证：D_{1}F\parallel 平面A_{1}EC_{1}；$

Solution:
$如图，以A为坐标原点，AB,AD,AA_{1}所在直线分别为x轴,y轴,z轴建立空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1715283942781829121/img/compression/M_6c1a2ea84fda7eee4bbd368886bc914b.jpg)$则A(0,0,0),A_{1}(0,0,2),B(2,0,0),C(2,2,0),D(0,2,0),C_{1}(2,2,2),D_{1}(0,2,2)，则\overrightarrow{A_1C_1}=(2,2,0),因为E为BC的中点，F为CD的中点，所以E(2,1,0)，F(1,2,0)，所以\overrightarrow{D_1F}=(1,0,-2),\overrightarrow{A_1E}=(2,1,-2).设平面A_{1}EC_{1}的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1C_1}=2x+2y=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1E}=2x+y-2z=0,\end{matrix}\right.令x=2，则m=(2,-2,1)，因为\overrightarrow{D_1F}\cdot m=2-2=0，所以\overrightarrow{D_1F}\perp m，又因为D_{1}F\not\subset 平面A_{1}EC_{1}，所以D_{1}F\parallel 平面A_{1}EC_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$<img_1780>
$(1)求证：D_{1}F\parallel 平面A_{1}EC_{1}；$

Solution:
$如图，以A为坐标原点，AB,AD,AA_{1}所在直线分别为x轴,y轴,z轴建立空间直角坐标系，$<img_1659>$则A(0,0,0),A_{1}(0,0,2),B(2,0,0),C(2,2,0),D(0,2,0),C_{1}(2,2,2),D_{1}(0,2,2)，则\overrightarrow{A_1C_1}=(2,2,0),因为E为BC的中点，F为CD的中点，所以E(2,1,0)，F(1,2,0)，所以\overrightarrow{D_1F}=(1,0,-2),\overrightarrow{A_1E}=(2,1,-2).设平面A_{1}EC_{1}的法向量为m=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{m}\cdot \overrightarrow{A_1C_1}=2x+2y=0,\\ \boldsymbol{m}\cdot \overrightarrow{A_1E}=2x+y-2z=0,\end{matrix}\right.令x=2，则m=(2,-2,1)，因为\overrightarrow{D_1F}\cdot m=2-2=0，所以\overrightarrow{D_1F}\perp m，又因为D_{1}F\not\subset 平面A_{1}EC_{1}，所以D_{1}F\parallel 平面A_{1}EC_{1}.$
Subfield: Solid Geometry


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$<img_1780>
$(2)求直线AC_{1}与平面A_{1}EC_{1}所成角的正弦值；$

Solution:
$由(1)得\overrightarrow{AC_1}=(2,2,2)，设直线AC_{1}与平面A_{1}EC_{1}所成的角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{AC_1}\rangle |=\left|\frac{\boldsymbol{m}\cdot \overrightarrow{AC_1}}{|\boldsymbol{m}|\cdot |\overrightarrow{AC_1}|}\right|=\frac{\sqrt{3}}{9}.$

Final answer: $\frac{\sqrt{3}}{9}$
Subfield: Solid Geometry


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283942374981635/img/compression/M_07622067350e8856c0d52e7d8bfe96af.jpg)
$(2)求直线AC_{1}与平面A_{1}EC_{1}所成角的正弦值；$

Solution:
$由(1)得\overrightarrow{AC_1}=(2,2,2)，设直线AC_{1}与平面A_{1}EC_{1}所成的角为\theta ，则\\sin   \theta =|\\cos  \langle m,\overrightarrow{AC_1}\rangle |=\left|\frac{\boldsymbol{m}\cdot \overrightarrow{AC_1}}{|\boldsymbol{m}|\cdot |\overrightarrow{AC_1}|}\right|=\frac{\sqrt{3}}{9}.$

Final answer: $\frac{\sqrt{3}}{9}$
Subfield: Solid Geometry


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$<img_1780>
$(3)求二面角A-A_{1}C_{1}-E的正弦值.$

Solution:
$由正方体的特征可得平面AA_{1}C_{1}的一个法向量为\overrightarrow{DB}=(2,-2,0)，则\\cos  \langle \overrightarrow{DB},m\rangle =\frac{\overrightarrow{DB}\cdot \boldsymbol{m}}{|\overrightarrow{DB}|\cdot |\boldsymbol{m}|}=\frac{8}{3\times 2\sqrt{2}}=\frac{2\sqrt{2}}{3}，所以二面角A-A_{1}C_{1}-E的正弦值为\sqrt{1-\\\cos   ^2\langle \overrightarrow{DB},\boldsymbol{m}\rangle }=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在棱长为2的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，E为棱BC的中点，F为棱CD的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283942374981635/img/compression/M_07622067350e8856c0d52e7d8bfe96af.jpg)
$(3)求二面角A-A_{1}C_{1}-E的正弦值.$

Solution:
$由正方体的特征可得平面AA_{1}C_{1}的一个法向量为\overrightarrow{DB}=(2,-2,0)，则\\cos  \langle \overrightarrow{DB},m\rangle =\frac{\overrightarrow{DB}\cdot \boldsymbol{m}}{|\overrightarrow{DB}|\cdot |\boldsymbol{m}|}=\frac{8}{3\times 2\sqrt{2}}=\frac{2\sqrt{2}}{3}，所以二面角A-A_{1}C_{1}-E的正弦值为\sqrt{1-\\\cos   ^2\langle \overrightarrow{DB},\boldsymbol{m}\rangle }=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，PD=DC=1，M为BC的中点，且PB\perp AM.$<img_3088>
$(1)求BC；$

Solution:
$解法一：连接BD.因为PD\perp 底面ABCD，AM\subset 平面ABCD，所以PD\perp AM.又PB\perp AM，PD\cap PB=P，所以AM\perp 平面PBD，故AM\perp BD.于是\triangle ABD∽\triangle BMA，故\frac{AD}{AB}=\frac{AB}{BM}.因为BM=\frac{1}{2}BC，AD=BC，AB=1，所以\frac{1}{2}BC^{2}=1，即BC=\sqrt{2}.解法二：由题意知DA，DC，DP两两垂直，所以以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DP}的方向分别为x轴，y轴，z轴的正方向建立空间直角坐标系D-xyz，如图所示，设BC=t(t>0)，则A(t，0，0)，B(t，1，0)，M\left(\frac{t}{2},1,0\right)，P(0，0，1)，所以\overrightarrow{PB}=(t，1，-1)，\overrightarrow{AM}=\left(-\frac{t}{2},1,0\right)，因为PB\perp AM，所以\overrightarrow{PB}\cdot \overrightarrow{AM}=0，所以-\frac{t^2}{2}+1=0，即t^{2}=2.又t>0，所以t=\sqrt{2}，即BC=\sqrt{2}.$<img_2529>

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


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，PD=DC=1，M为BC的中点，且PB\perp AM.$![Image](https://cdntk.53zaixian.com/tq/1715283943247396866/img/compression/M_55a633b8173abc28ca1fd7038c6cc3c0.jpg)
$(1)求BC；$

Solution:
$解法一：连接BD.因为PD\perp 底面ABCD，AM\subset 平面ABCD，所以PD\perp AM.又PB\perp AM，PD\cap PB=P，所以AM\perp 平面PBD，故AM\perp BD.于是\triangle ABD∽\triangle BMA，故\frac{AD}{AB}=\frac{AB}{BM}.因为BM=\frac{1}{2}BC，AD=BC，AB=1，所以\frac{1}{2}BC^{2}=1，即BC=\sqrt{2}.解法二：由题意知DA，DC，DP两两垂直，所以以D为原点，\overrightarrow{DA}，\overrightarrow{DC}，\overrightarrow{DP}的方向分别为x轴，y轴，z轴的正方向建立空间直角坐标系D-xyz，如图所示，设BC=t(t>0)，则A(t，0，0)，B(t，1，0)，M\left(\frac{t}{2},1,0\right)，P(0，0，1)，所以\overrightarrow{PB}=(t，1，-1)，\overrightarrow{AM}=\left(-\frac{t}{2},1,0\right)，因为PB\perp AM，所以\overrightarrow{PB}\cdot \overrightarrow{AM}=0，所以-\frac{t^2}{2}+1=0，即t^{2}=2.又t>0，所以t=\sqrt{2}，即BC=\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1715283943570358274/img/compression/M_a2920142b13ea29a0a2da5ae0967d5cd.jpg)

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


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，PD=DC=1，M为BC的中点，且PB\perp AM.$<img_3088>
$(2)求二面角A-PM-B的正弦值.$

Solution:
$设平面PAM的法向量为n_{1}=(x_{1}，y_{1}，z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PA}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{AM}=0,\end{matrix}\right.由(1)知\overrightarrow{PA}=(\sqrt{2}，0，-1)，\overrightarrow{AM}=\left(-\frac{\sqrt{2}}{2},1,0\right)，所以\left\{\begin{matrix}\sqrt{2}x_1-z_1=0,\\ -\frac{\sqrt{2}}{2}x_1+y_1=0,\end{matrix}\right.令x_{1}=\sqrt{2}，则y_{1}=1，z_{1}=2，即n_{1}=(\sqrt{2}，1，2)，设平面PMB的法向量为n_{2}=(x_{2}，y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{PB}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{MB}=0,\end{matrix}\right.由(1)知\overrightarrow{PB}=(\sqrt{2}，1，-1)，\overrightarrow{MB}=\left(\frac{\sqrt{2}}{2},0,0\right)，所以\left\{\begin{matrix}\sqrt{2}x_2+y_2-z_2=0,\\ \frac{\sqrt{2}}{2}x_2=0,\end{matrix}\right.令y_{2}=1，则z_{2}=1，即n_{2}=(0，1，1)，所以\\cos  \langle n_{1}，n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1\times 1+2\times 1}{\sqrt{7}\times \sqrt{2}}=\frac{3}{\sqrt{14}}.由题图可知二面角A-PM-B的平面角为锐角，所以所求二面角的正弦值为\sqrt{1-\left(\frac{3}{\sqrt{14}}\right)^2}=\frac{\sqrt{70}}{14}.$

Final answer: $\frac{\sqrt{70}}{14}$
Subfield: Solid Geometry


Question: $如图，四棱锥P-ABCD的底面是矩形，PD\perp 底面ABCD，PD=DC=1，M为BC的中点，且PB\perp AM.$![Image](https://cdntk.53zaixian.com/tq/1715283943247396866/img/compression/M_55a633b8173abc28ca1fd7038c6cc3c0.jpg)
$(2)求二面角A-PM-B的正弦值.$

Solution:
$设平面PAM的法向量为n_{1}=(x_{1}，y_{1}，z_{1})，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot \overrightarrow{PA}=0,\\ \boldsymbol{n}_1\cdot \overrightarrow{AM}=0,\end{matrix}\right.由(1)知\overrightarrow{PA}=(\sqrt{2}，0，-1)，\overrightarrow{AM}=\left(-\frac{\sqrt{2}}{2},1,0\right)，所以\left\{\begin{matrix}\sqrt{2}x_1-z_1=0,\\ -\frac{\sqrt{2}}{2}x_1+y_1=0,\end{matrix}\right.令x_{1}=\sqrt{2}，则y_{1}=1，z_{1}=2，即n_{1}=(\sqrt{2}，1，2)，设平面PMB的法向量为n_{2}=(x_{2}，y_{2}，z_{2})，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{PB}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{MB}=0,\end{matrix}\right.由(1)知\overrightarrow{PB}=(\sqrt{2}，1，-1)，\overrightarrow{MB}=\left(\frac{\sqrt{2}}{2},0,0\right)，所以\left\{\begin{matrix}\sqrt{2}x_2+y_2-z_2=0,\\ \frac{\sqrt{2}}{2}x_2=0,\end{matrix}\right.令y_{2}=1，则z_{2}=1，即n_{2}=(0，1，1)，所以\\cos  \langle n_{1}，n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}=\frac{1\times 1+2\times 1}{\sqrt{7}\times \sqrt{2}}=\frac{3}{\sqrt{14}}.由题图可知二面角A-PM-B的平面角为锐角，所以所求二面角的正弦值为\sqrt{1-\left(\frac{3}{\sqrt{14}}\right)^2}=\frac{\sqrt{70}}{14}.$

Final answer: $\frac{\sqrt{70}}{14}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$<img_3083>
$(1)求证：C_{1}M\perp B_{1}D；$

Solution:
$依题意，以C为原点，分别以\overrightarrow{CA}，\overrightarrow{CB}，\overrightarrow{CC_1}的方向为x轴，y轴，z轴的正方向建立空间直角坐标系(如图)，可得C(0，0，0)，A(2，0，0)，B(0，2，0)，C_{1}(0，0，3)，A_{1}(2，0，3)，B_{1}(0，2，3)，D(2，0，1)，E(0，0，2)，M(1，1，3).$<img_3122>$证明：依题意，\overrightarrow{C_1M}=(1，1，0)，\overrightarrow{B_1D}=(2，-2，-2)，从而\overrightarrow{C_1M}\cdot \overrightarrow{B_1D}=2-2+0=0，所以C_{1}M\perp B_{1}D.$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283944124006402/img/compression/M_cf86dfafa7705cfa0144781d38ed0e9c.jpg)
$(1)求证：C_{1}M\perp B_{1}D；$

Solution:
$依题意，以C为原点，分别以\overrightarrow{CA}，\overrightarrow{CB}，\overrightarrow{CC_1}的方向为x轴，y轴，z轴的正方向建立空间直角坐标系(如图)，可得C(0，0，0)，A(2，0，0)，B(0，2，0)，C_{1}(0，0，3)，A_{1}(2，0，3)，B_{1}(0，2，3)，D(2，0，1)，E(0，0，2)，M(1，1，3).$![Image](https://cdntk.53zaixian.com/tq/1715283944124006402/img/compression/M_67d3d874bed05109cacf47cc6c5c2d1b.jpg)$证明：依题意，\overrightarrow{C_1M}=(1，1，0)，\overrightarrow{B_1D}=(2，-2，-2)，从而\overrightarrow{C_1M}\cdot \overrightarrow{B_1D}=2-2+0=0，所以C_{1}M\perp B_{1}D.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283944124006402/img/compression/M_cf86dfafa7705cfa0144781d38ed0e9c.jpg)
$(2)求二面角B-B_{1}E-D的正弦值；$

Solution:
$依题意，\overrightarrow{CA}=(2，0，0)是平面BB_{1}E的一个法向量，\overrightarrow{EB_1}=(0，2，1)，\overrightarrow{ED}=(2，0，-1).设n=(x，y，z)为平面DB_{1}E的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{EB_1}=0,\\ \boldsymbol{n}\cdot \overrightarrow{ED}=0,\end{matrix}\right.即\left\{\begin{matrix}2y+z=0,\\ 2x-z=0.\end{matrix}\right.不妨设x=1，可得n=(1，-1，2).因此有\\cos  \langle \overrightarrow{CA}，n\rangle =\frac{\overrightarrow{CA}\cdot \boldsymbol{n}}{|\overrightarrow{CA}||\boldsymbol{n}|}=\frac{\sqrt{6}}{6}，于是\\sin  \langle \overrightarrow{CA}，n\rangle =\frac{\sqrt{30}}{6}.所以，二面角B-B_{1}E-D的正弦值为\frac{\sqrt{30}}{6}.$

Final answer: $\frac{\sqrt{30}}{6}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$<img_3083>
$(2)求二面角B-B_{1}E-D的正弦值；$

Solution:
$依题意，\overrightarrow{CA}=(2，0，0)是平面BB_{1}E的一个法向量，\overrightarrow{EB_1}=(0，2，1)，\overrightarrow{ED}=(2，0，-1).设n=(x，y，z)为平面DB_{1}E的法向量，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{EB_1}=0,\\ \boldsymbol{n}\cdot \overrightarrow{ED}=0,\end{matrix}\right.即\left\{\begin{matrix}2y+z=0,\\ 2x-z=0.\end{matrix}\right.不妨设x=1，可得n=(1，-1，2).因此有\\cos  \langle \overrightarrow{CA}，n\rangle =\frac{\overrightarrow{CA}\cdot \boldsymbol{n}}{|\overrightarrow{CA}||\boldsymbol{n}|}=\frac{\sqrt{6}}{6}，于是\\sin  \langle \overrightarrow{CA}，n\rangle =\frac{\sqrt{30}}{6}.所以，二面角B-B_{1}E-D的正弦值为\frac{\sqrt{30}}{6}.$

Final answer: $\frac{\sqrt{30}}{6}$
Subfield: Solid Geometry


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283944124006402/img/compression/M_cf86dfafa7705cfa0144781d38ed0e9c.jpg)
$(3)求直线AB与平面DB_{1}E所成角的正弦值.$

Solution:
$依题意，\overrightarrow{AB}=(-2，2，0).由(2)知n=(1，-1，2)为平面DB_{1}E的一个法向量，故\\cos  \langle \overrightarrow{AB}，n\rangle =\frac{\overrightarrow{AB}\cdot \boldsymbol{n}}{|\overrightarrow{AB}||\boldsymbol{n}|}=-\frac{\sqrt{3}}{3}.所以，直线AB与平面DB_{1}E所成角的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图，在三棱柱ABC-A_{1}B_{1}C_{1}中，CC_{1}\perp 平面ABC，AC\perp BC，AC=BC=2，CC_{1}=3，点D，E分别在棱AA_{1}和棱CC_{1}上，且AD=1，CE=2，M为棱A_{1}B_{1}的中点.$<img_3083>
$(3)求直线AB与平面DB_{1}E所成角的正弦值.$

Solution:
$依题意，\overrightarrow{AB}=(-2，2，0).由(2)知n=(1，-1，2)为平面DB_{1}E的一个法向量，故\\cos  \langle \overrightarrow{AB}，n\rangle =\frac{\overrightarrow{AB}\cdot \boldsymbol{n}}{|\overrightarrow{AB}||\boldsymbol{n}|}=-\frac{\sqrt{3}}{3}.所以，直线AB与平面DB_{1}E所成角的正弦值为\frac{\sqrt{3}}{3}.$

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


Question: $如图,已知四边形ABCD和四边形CDEF都是直角梯形,AB\parallel DC,DC\parallel EF,AB=5,DC=3,EF=1,\angle BAD=\angle CDE=60^\circ ,二面角F-DC-B的平面角为60^\circ .设M,N分别为AE,BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283944862203907/img/compression/M_7878b48b4bd971b18485325842d32d31.jpg)
$(1)证明:FN\perp AD;$

Solution:
$证明：由题意知CD\perp CB,CD\perp CF,CB\cap CF=C,\therefore CD\perp 平面CBF,又FN\subset 平面CBF，\therefore CD\perp FN.易知\angle FCB即为二面角F-DC-B的平面角，故\angle FCB=60^\circ ,又CF=(CD-EF)\\tan  \angle CDE=2\sqrt{3},CB=(AB-CD)\cdot \\tan  \angle BAD=2\sqrt{3},\therefore \triangle BCF是等边三角形，又N是BC的中点，\therefore CB\perp FN,\because CB\cap CD=C，CB，CD\subset 平面ABCD，\therefore FN\perp 平面ABCD，\because AD\subset 平面ABCD，\therefore FN\perp AD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,已知四边形ABCD和四边形CDEF都是直角梯形,AB\parallel DC,DC\parallel EF,AB=5,DC=3,EF=1,\angle BAD=\angle CDE=60^\circ ,二面角F-DC-B的平面角为60^\circ .设M,N分别为AE,BC的中点.$<img_2274>
$(1)证明:FN\perp AD;$

Solution:
$证明：由题意知CD\perp CB,CD\perp CF,CB\cap CF=C,\therefore CD\perp 平面CBF,又FN\subset 平面CBF，\therefore CD\perp FN.易知\angle FCB即为二面角F-DC-B的平面角，故\angle FCB=60^\circ ,又CF=(CD-EF)\\tan  \angle CDE=2\sqrt{3},CB=(AB-CD)\cdot \\tan  \angle BAD=2\sqrt{3},\therefore \triangle BCF是等边三角形，又N是BC的中点，\therefore CB\perp FN,\because CB\cap CD=C，CB，CD\subset 平面ABCD，\therefore FN\perp 平面ABCD，\because AD\subset 平面ABCD，\therefore FN\perp AD.$
Subfield: Solid Geometry


Question: $如图,已知四边形ABCD和四边形CDEF都是直角梯形,AB\parallel DC,DC\parallel EF,AB=5,DC=3,EF=1,\angle BAD=\angle CDE=60^\circ ,二面角F-DC-B的平面角为60^\circ .设M,N分别为AE,BC的中点.$![Image](https://cdntk.53zaixian.com/tq/1715283944862203907/img/compression/M_7878b48b4bd971b18485325842d32d31.jpg)
$(2)求直线BM与平面ADE所成角的正弦值.$

Solution:
$建立如图所示的空间直角坐标系.$![Image](https://cdntk.53zaixian.com/tq/1715283945193553922/img/compression/M_833855238bf1f0aa0d01ffaeeb7663d9.jpg)$由题意得B(0,\sqrt{3},0),A(5,\sqrt{3},0),E(1,0,3),D(3,-\sqrt{3},0),则M\left(3,\frac{\sqrt{3}}{2},\frac{3}{2}\right)，\therefore \overrightarrow{BM}=\left(3,-\frac{\sqrt{3}}{2},\frac{3}{2}\right),\overrightarrow{DA}=(2,2\sqrt{3},0),\overrightarrow{DE}=(-2,\sqrt{3},3).设平面ADE的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DA}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DE}=0,\end{matrix}\right.即\left\{\begin{matrix}2x+2\sqrt{3}y=0,\\ -2x+\sqrt{3}y+3z=0,\end{matrix}\right.令y=-1，则n=(\sqrt{3},-1,\sqrt{3}).设直线BM与平面ADE所成角为\theta ,则\\sin   \theta =|\\cos  \langle n,\overrightarrow{BM}\rangle |=\frac{|\overrightarrow{BM}\cdot \boldsymbol{n}|}{|\overrightarrow{BM}||\boldsymbol{n}|}=\frac{5\sqrt{7}}{14}.\therefore 直线BM与平面ADE所成角的正弦值为\frac{5\sqrt{7}}{14}.$

Final answer: $\frac{5\sqrt{7}}{14}$
Subfield: Solid Geometry


Question: $如图,已知四边形ABCD和四边形CDEF都是直角梯形,AB\parallel DC,DC\parallel EF,AB=5,DC=3,EF=1,\angle BAD=\angle CDE=60^\circ ,二面角F-DC-B的平面角为60^\circ .设M,N分别为AE,BC的中点.$<img_2274>
$(2)求直线BM与平面ADE所成角的正弦值.$

Solution:
$建立如图所示的空间直角坐标系.$<img_2439>$由题意得B(0,\sqrt{3},0),A(5,\sqrt{3},0),E(1,0,3),D(3,-\sqrt{3},0),则M\left(3,\frac{\sqrt{3}}{2},\frac{3}{2}\right)，\therefore \overrightarrow{BM}=\left(3,-\frac{\sqrt{3}}{2},\frac{3}{2}\right),\overrightarrow{DA}=(2,2\sqrt{3},0),\overrightarrow{DE}=(-2,\sqrt{3},3).设平面ADE的法向量为n=(x,y,z)，则\left\{\begin{matrix}\boldsymbol{n}\cdot \overrightarrow{DA}=0,\\ \boldsymbol{n}\cdot \overrightarrow{DE}=0,\end{matrix}\right.即\left\{\begin{matrix}2x+2\sqrt{3}y=0,\\ -2x+\sqrt{3}y+3z=0,\end{matrix}\right.令y=-1，则n=(\sqrt{3},-1,\sqrt{3}).设直线BM与平面ADE所成角为\theta ,则\\sin   \theta =|\\cos  \langle n,\overrightarrow{BM}\rangle |=\frac{|\overrightarrow{BM}\cdot \boldsymbol{n}|}{|\overrightarrow{BM}||\boldsymbol{n}|}=\frac{5\sqrt{7}}{14}.\therefore 直线BM与平面ADE所成角的正弦值为\frac{5\sqrt{7}}{14}.$

Final answer: $\frac{5\sqrt{7}}{14}$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E，F分别在棱DD_{1}，BB_{1}上，且2DE=ED_{1}，BF=2FB_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283945571041282/img/compression/M_96ff668e6ba196b909bfa7be8108807e.jpg)
$(1)证明：点C_{1}在平面AEF内；$

Solution:
$证法一：在棱CC_{1}上取点G，使得C_{1}G=\frac{1}{2}CG，连接DG、FG、C_{1}E、C_{1}F，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1715283945902391298/img/compression/M_83e618c78c5b4771d55c8c66c8fd5004.jpg)$在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，BF\parallel CG,BF=CG，所以四边形BCGF为平行四边形，则BC\parallel FG,BC=FG，而BC=AD,BC\parallel AD，所以AD\parallel FG,AD=FG，所以四边形ADGF为平行四边形，即有AF\parallel DG，同理可证四边形DEC_{1}G为平行四边形，所以C_{1}E\parallel DG，所以C_{1}E\parallel AF,因此点C_{1}在平面AEF内.证法二：设AB=a,AD=b，AA_{1}=c，如图，$![Image](https://cdntk.53zaixian.com/tq/1715283945902391298/img/compression/M_dced72775c4255528f4f3f1c5c2e6f4a.jpg)$以C_{1}为坐标原点，\overrightarrow{C_1D_1}的方向为x轴正方向，建立空间直角坐标系C_{1}-xyz.连接C_{1}F，则C_{1}(0，0，0)，A(a,b,c)，E\left(a,0,\frac{2}{3}c\right)，F\left(0,b,\frac{1}{3}c\right)，\overrightarrow{EA}=\left(0,b,\frac{1}{3}c\right)，\overrightarrow{C_1F}=\left(0,b,\frac{1}{3}c\right)，得\overrightarrow{EA}=\overrightarrow{C_1F}，因此EA\parallel C_{1}F，即A，E，F，C_{1}四点共面，所以点C_{1}在平面AEF内.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E，F分别在棱DD_{1}，BB_{1}上，且2DE=ED_{1}，BF=2FB_{1}.$<img_2011>
$(1)证明：点C_{1}在平面AEF内；$

Solution:
$证法一：在棱CC_{1}上取点G，使得C_{1}G=\frac{1}{2}CG，连接DG、FG、C_{1}E、C_{1}F，如图所示.$<img_1726>$在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，BF\parallel CG,BF=CG，所以四边形BCGF为平行四边形，则BC\parallel FG,BC=FG，而BC=AD,BC\parallel AD，所以AD\parallel FG,AD=FG，所以四边形ADGF为平行四边形，即有AF\parallel DG，同理可证四边形DEC_{1}G为平行四边形，所以C_{1}E\parallel DG，所以C_{1}E\parallel AF,因此点C_{1}在平面AEF内.证法二：设AB=a,AD=b，AA_{1}=c，如图，$<img_2769>$以C_{1}为坐标原点，\overrightarrow{C_1D_1}的方向为x轴正方向，建立空间直角坐标系C_{1}-xyz.连接C_{1}F，则C_{1}(0，0，0)，A(a,b,c)，E\left(a,0,\frac{2}{3}c\right)，F\left(0,b,\frac{1}{3}c\right)，\overrightarrow{EA}=\left(0,b,\frac{1}{3}c\right)，\overrightarrow{C_1F}=\left(0,b,\frac{1}{3}c\right)，得\overrightarrow{EA}=\overrightarrow{C_1F}，因此EA\parallel C_{1}F，即A，E，F，C_{1}四点共面，所以点C_{1}在平面AEF内.$
Subfield: Solid Geometry


Question: $如图，在长方体ABCD-A_{1}B_{1}C_{1}D_{1}中，点E，F分别在棱DD_{1}，BB_{1}上，且2DE=ED_{1}，BF=2FB_{1}.$![Image](https://cdntk.53zaixian.com/tq/1715283945571041282/img/compression/M_96ff668e6ba196b909bfa7be8108807e.jpg)
$(2)若AB=2，AD=1，AA_{1}=3，求二面角A-EF-A_{1}的正弦值.$

Solution:
$建系同(1).由已知得A(2，1，3)，E(2，0，2)，F(0，1，1)，A_{1}(2，1，0)，\overrightarrow{AE}=(0，-1， -1)，\overrightarrow{AF}=(-2，0，-2)，\overrightarrow{A_1E}=(0，-1，2)，\overrightarrow{A_1F}=(-2，0，1).设n_{1}=(x,y,z)为平面AEF的法向量，则\left\{\begin{matrix}\boldsymbol{n}_1\cdot A\overrightarrow{E}=0,\\ \boldsymbol{n}_1\cdot A\overrightarrow{F}=0,\end{matrix}\right.即\left\{\begin{matrix}-y-z=0,\\ -2x-2z=0,\end{matrix}\right.可取n_{1}=(-1,-1,1).设n_{2}为平面A_{1}EF的法向量，则\left\{\begin{matrix}\boldsymbol{n}_2\cdot \overrightarrow{A_1E}=0,\\ \boldsymbol{n}_2\cdot \overrightarrow{A_1F}=0,\end{matrix}\right.同理可取n_{2}=\left(\frac{1}{2},2,1\right).因为\\cos  \langle n_{1},n_{2}\rangle =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1|\cdot |\boldsymbol{n}_2|}=-\frac{\sqrt{7}}{7}，所以二面角A-EF-A_{1}的正弦值为\frac{\sqrt{42}}{7}.$

Final answer: $\frac{\sqrt{42}}{7}$