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


Question: $下图是一个半圆柱，E为半圆弧\overset{\LARGE{\frown}}{CD}上一点，|CD|=\sqrt{5}.$![Image](https://cdntk.53zaixian.com/tq/1658378886166679553/img/compression/M_3e22e5090c317680f2d1e527825456a9.jpg)
$(2)有三个条件：①4\overrightarrow{DE}\cdot \overrightarrow{DC}=\overrightarrow{EC}\cdot \overrightarrow{DC}；②直线AD与BE所成角的正弦值为\frac{2}{3}；③\frac{\\\sin   \angle EAB}{\\\sin   \angle EBA}=\frac{\sqrt{6}}{2}.请你从中选择两个作为已知条件，求直线AD与平面EAB所成角的余弦值.$

Solution:
$由条件①，得4|\overrightarrow{DE}||\overrightarrow{DC}|\\cos  \angle CDE=|\overrightarrow{CE}||\overrightarrow{CD}|\\cos  \angle DCE,即4|DE|^{2}=|CE|^{2},所以2|DE|=|CE|,又因为|DE|^{2}+|CE|^{2}=5,所以|DE|=1，|CE|=2.因为AD\parallel BC,BC\perp 平面DCE,所以\angle CBE为异面直线AD与BE所成的角，BC\perp CE,所以Rt\triangle CBE中，\\sin  \angle CBE=\frac{|CE|}{|BE|},\\tan  \angle CBE=\frac{|CE|}{|BC|}.由条件②，得\\sin  \angle CBE=\frac{2}{3},所以 \\tan  \angle CBE=\frac{2\sqrt{5}}{5}.由条件③，得\frac{\\\sin   \angle EAB}{\\\sin   \angle EBA}=\frac{|EB|}{|EA|}=\frac{\sqrt{6}}{2},设|AD|=x,则\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2}.(6分)若选条件①②，则|DE|=1,|CE|=2,且\frac{|CE|}{|BC|}=\\tan  \angle CBE=\frac{2\sqrt{5}}{5},所以|AD|=|BC|=\sqrt{5}.若选条件①③，则|DE|=1，|CE|=2，且\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2},所以|AD|=|BC|=\sqrt{5}.若选条件②③，则\frac{|CE|}{x}=\\tan  \angle CBE=\frac{2\sqrt{5}}{5},且\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2},|DE|^{2}+|CE|^{2}=5,所以|DE|=1，|CE|=2，|AD|=|BC|=\sqrt{5}.即从①②③中任选两个作为条件，都可以得到|DE|=1，|CE|=2，|AD|=|BC|=\sqrt{5}.(9分)下面求AD与平面EAB所成角的正弦值.以A为原点，AB所在直线为x轴，AD所在直线为z轴，过点A且与平面ABCD垂直的直线为y轴,建立如图所示的空间直角坐标系，则A(0,0,0),B(\sqrt{5},0,0),D(0,0,\sqrt{5}),E\left(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5},\sqrt{5}\right),所以\overrightarrow{AE}=\left(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5},\sqrt{5}\right)，\overrightarrow{AB}=(\sqrt{5},0,0),\overrightarrow{AD}=(0,0,\sqrt{5}),设平面EAB的一个法向量为m=(a,b,c),则\left\{\begin{matrix}\overrightarrow{AE} \cdot \boldsymbol{m}=0,\\ \overrightarrow{AB} \cdot \boldsymbol{m}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{\sqrt{5}}{5}a+\frac{2\sqrt{5}}{5}b+\sqrt{5}c=0,\\ \sqrt{5}a=0,\end{matrix}\right.令c=1,则a=0,b=-\frac{5}{2}，所以m=\left(0,-\frac{5}{2},1\right).(10分)设直线AD与平面EAB所成的角为\theta ，\theta \in \left[0,\frac{\pi }{2}\right],则\\sin   \theta =|\\cos  \langle \overrightarrow{AD},m\rangle |=\frac{\sqrt{5}}{\sqrt{5} \times \frac{\sqrt{29}}{2}}=\frac{2\sqrt{29}}{29},(11分)所以\\cos   \theta =\sqrt{1-\\\sin   ^2\theta }=\frac{5\sqrt{29}}{29},所以AD与平面EAB所成角的余弦值为\frac{5\sqrt{29}}{29}.(12分)$![Image](https://cdntk.53zaixian.com/tq/1658378886166679553/img/compression/M_444c597a27611108886c57676f0f050c.jpg)

Final answer: $|AD|=|BC|=\sqrt 5
AD与平面EAB所成角的余弦值为5\sqrt 29/29$
Subfield: Solid Geometry


Question: $下图是一个半圆柱，E为半圆弧\overset{\LARGE{\frown}}{CD}上一点，|CD|=\sqrt{5}.$<img_2679>
$(2)有三个条件：①4\overrightarrow{DE}\cdot \overrightarrow{DC}=\overrightarrow{EC}\cdot \overrightarrow{DC}；②直线AD与BE所成角的正弦值为\frac{2}{3}；③\frac{\\\sin   \angle EAB}{\\\sin   \angle EBA}=\frac{\sqrt{6}}{2}.请你从中选择两个作为已知条件，求直线AD与平面EAB所成角的余弦值.$

Solution:
$由条件①，得4|\overrightarrow{DE}||\overrightarrow{DC}|\\cos  \angle CDE=|\overrightarrow{CE}||\overrightarrow{CD}|\\cos  \angle DCE,即4|DE|^{2}=|CE|^{2},所以2|DE|=|CE|,又因为|DE|^{2}+|CE|^{2}=5,所以|DE|=1，|CE|=2.因为AD\parallel BC,BC\perp 平面DCE,所以\angle CBE为异面直线AD与BE所成的角，BC\perp CE,所以Rt\triangle CBE中，\\sin  \angle CBE=\frac{|CE|}{|BE|},\\tan  \angle CBE=\frac{|CE|}{|BC|}.由条件②，得\\sin  \angle CBE=\frac{2}{3},所以 \\tan  \angle CBE=\frac{2\sqrt{5}}{5}.由条件③，得\frac{\\\sin   \angle EAB}{\\\sin   \angle EBA}=\frac{|EB|}{|EA|}=\frac{\sqrt{6}}{2},设|AD|=x,则\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2}.(6分)若选条件①②，则|DE|=1,|CE|=2,且\frac{|CE|}{|BC|}=\\tan  \angle CBE=\frac{2\sqrt{5}}{5},所以|AD|=|BC|=\sqrt{5}.若选条件①③，则|DE|=1，|CE|=2，且\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2},所以|AD|=|BC|=\sqrt{5}.若选条件②③，则\frac{|CE|}{x}=\\tan  \angle CBE=\frac{2\sqrt{5}}{5},且\frac{x^2+|CE|^2}{x^2+|DE|^2}=\frac{3}{2},|DE|^{2}+|CE|^{2}=5,所以|DE|=1，|CE|=2，|AD|=|BC|=\sqrt{5}.即从①②③中任选两个作为条件，都可以得到|DE|=1，|CE|=2，|AD|=|BC|=\sqrt{5}.(9分)下面求AD与平面EAB所成角的正弦值.以A为原点，AB所在直线为x轴，AD所在直线为z轴，过点A且与平面ABCD垂直的直线为y轴,建立如图所示的空间直角坐标系，则A(0,0,0),B(\sqrt{5},0,0),D(0,0,\sqrt{5}),E\left(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5},\sqrt{5}\right),所以\overrightarrow{AE}=\left(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5},\sqrt{5}\right)，\overrightarrow{AB}=(\sqrt{5},0,0),\overrightarrow{AD}=(0,0,\sqrt{5}),设平面EAB的一个法向量为m=(a,b,c),则\left\{\begin{matrix}\overrightarrow{AE} \cdot \boldsymbol{m}=0,\\ \overrightarrow{AB} \cdot \boldsymbol{m}=0,\end{matrix}\right.即\left\{\begin{matrix}\frac{\sqrt{5}}{5}a+\frac{2\sqrt{5}}{5}b+\sqrt{5}c=0,\\ \sqrt{5}a=0,\end{matrix}\right.令c=1,则a=0,b=-\frac{5}{2}，所以m=\left(0,-\frac{5}{2},1\right).(10分)设直线AD与平面EAB所成的角为\theta ，\theta \in \left[0,\frac{\pi }{2}\right],则\\sin   \theta =|\\cos  \langle \overrightarrow{AD},m\rangle |=\frac{\sqrt{5}}{\sqrt{5} \times \frac{\sqrt{29}}{2}}=\frac{2\sqrt{29}}{29},(11分)所以\\cos   \theta =\sqrt{1-\\\sin   ^2\theta }=\frac{5\sqrt{29}}{29},所以AD与平面EAB所成角的余弦值为\frac{5\sqrt{29}}{29}.(12分)$<img_2823>



Final answer: \frac{5\sqrt{29}}{29}
Subfield: Solid Geometry


Question: $将一块2 m\times 6 m 的矩形钢板按如图所示的方式划线，要求①至⑦全为矩形，沿线裁去阴影部分，把剩余部分焊接成一个以⑦为底，⑤⑥为盖的长方体水箱(不计损耗)，设水箱的高为x m，容积为y m^{3}.$<img_2682>
$(1)写出y关于x的函数解析式；$

Solution:
$由水箱的高为x m，得水箱底面的宽为(2-2x)m，长为\frac{6-2x}{2}＝(3-x)m，易知0故y＝(2-2x)(3-x)x=2x^{3}-8x^{2}＋6x(0<x<1).$

Final answer: $y = 2x^3 - 8x^2 + 6x$
Subfield: Solid Geometry


Question: $将一块2 m\times 6 m 的矩形钢板按如图所示的方式划线，要求①至⑦全为矩形，沿线裁去阴影部分，把剩余部分焊接成一个以⑦为底，⑤⑥为盖的长方体水箱(不计损耗)，设水箱的高为x m，容积为y m^{3}.$![Image](https://cdntk.53zaixian.com/tq/1663347141765332994/img/compression/M_6d9c4a6f7c9d4bb1c1d017b7f68461e7.jpg)
$(1)写出y关于x的函数解析式；$

Solution:
$由水箱的高为x m，得水箱底面的宽为(2-2x)m，长为\frac{6-2x}{2}＝(3-x)m，易知0故y＝(2-2x)(3-x)x=2x^{3}-8x^{2}＋6x(0<x<1).$

Final answer: $y = 2x^3 - 8x^2 + 6x$
Subfield: Solid Geometry


Question: $将一块2 m\times 6 m 的矩形钢板按如图所示的方式划线，要求①至⑦全为矩形，沿线裁去阴影部分，把剩余部分焊接成一个以⑦为底，⑤⑥为盖的长方体水箱(不计损耗)，设水箱的高为x m，容积为y m^{3}.$<img_2682>
$(2)当x取何值时，水箱的容积最大？$

Solution:
$由(1)得y'＝6x^{2}-16x＋6，0令y'＝0，解得x＝\frac{4+\sqrt{7}}{3}(舍去)或x＝\frac{4-\sqrt{7}}{3}，当x\in \left(0,\frac{4-\sqrt{7}}{3}\right)时，y'>0;当x\in \left(\frac{4-\sqrt{7}}{3},1\right)时，y'<0,所以y＝2x^{3}-8x^{2}＋6x在\left(0,\frac{4-\sqrt{7}}{3}\right)内单调递增，在\left(\frac{4-\sqrt{7}}{3},1\right)内单调递减，所以当x=\frac{4-\sqrt{7}}{3}时，水箱的容积最大.$

Final answer: $\frac{4-\sqrt{7}}{3}$
Subfield: Solid Geometry


Question: $将一块2 m\times 6 m 的矩形钢板按如图所示的方式划线，要求①至⑦全为矩形，沿线裁去阴影部分，把剩余部分焊接成一个以⑦为底，⑤⑥为盖的长方体水箱(不计损耗)，设水箱的高为x m，容积为y m^{3}.$![Image](https://cdntk.53zaixian.com/tq/1663347141765332994/img/compression/M_6d9c4a6f7c9d4bb1c1d017b7f68461e7.jpg)
$(2)当x取何值时，水箱的容积最大？$

Solution:
$由(1)得y'＝6x^{2}-16x＋6，0令y'＝0，解得x＝\frac{4+\sqrt{7}}{3}(舍去)或x＝\frac{4-\sqrt{7}}{3}，当x\in \left(0,\frac{4-\sqrt{7}}{3}\right)时，y'>0;当x\in \left(\frac{4-\sqrt{7}}{3},1\right)时，y'<0,所以y＝2x^{3}-8x^{2}＋6x在\left(0,\frac{4-\sqrt{7}}{3}\right)内单调递增，在\left(\frac{4-\sqrt{7}}{3},1\right)内单调递减，所以当x=\frac{4-\sqrt{7}}{3}时，水箱的容积最大.$

Final answer: $\frac{4-\sqrt{7}}{3}$
Subfield: Solid Geometry


Question: $如图所示，在以长、宽、高分别为AB=3，AD=2，AA_{1}=1的长方体ABCD-A_{1}B_{1}C_{1}D_{1}的八个顶点的两点为起点和终点的向量中：$<img_2027>
$(1)单位向量共有多少个？$

Solution:
$由于长方体的高为1，所以长方体4条高所对应的\overrightarrow{AA_1}，\overrightarrow{A_1A}，\overrightarrow{BB_1}，\overrightarrow{B_1B}，\overrightarrow{CC_1}，\overrightarrow{C_1C}，\overrightarrow{DD_1}，\overrightarrow{D_1D}这8个向量都是单位向量，而其他向量的模均不为1，故单位向量共8个.$

Final answer: $8$
Subfield: Solid Geometry


Question: $如图所示，在以长、宽、高分别为AB=3，AD=2，AA_{1}=1的长方体ABCD-A_{1}B_{1}C_{1}D_{1}的八个顶点的两点为起点和终点的向量中：$![Image](https://cdntk.53zaixian.com/tq/1658661520050679809/img/compression/M_43986c1ae709d9f6da821c1fb6111d4f.jpg)
$(1)单位向量共有多少个？$

Solution:
$由于长方体的高为1，所以长方体4条高所对应的\overrightarrow{AA_1}，\overrightarrow{A_1A}，\overrightarrow{BB_1}，\overrightarrow{B_1B}，\overrightarrow{CC_1}，\overrightarrow{C_1C}，\overrightarrow{DD_1}，\overrightarrow{D_1D}这8个向量都是单位向量，而其他向量的模均不为1，故单位向量共8个.$

Final answer: $8$
Subfield: Solid Geometry


Question: $如图所示，在以长、宽、高分别为AB=3，AD=2，AA_{1}=1的长方体ABCD-A_{1}B_{1}C_{1}D_{1}的八个顶点的两点为起点和终点的向量中：$![Image](https://cdntk.53zaixian.com/tq/1658661520050679809/img/compression/M_43986c1ae709d9f6da821c1fb6111d4f.jpg)
$(2)试写出模为\sqrt{5}的所有向量；$

Solution:
$由于长方体的左、右两个面的对角线长均为\sqrt{5}，故模为\sqrt{5}的向量有\overrightarrow{AD_1}，\overrightarrow{D_1A}，\overrightarrow{A_1D}，\overrightarrow{DA_1}，\overrightarrow{BC_1}，\overrightarrow{C_1B}，\overrightarrow{B_1C}，\overrightarrow{CB_1}.$

Final answer: $\overrightarrow{AD_1}, \overrightarrow{D_1A}, \overrightarrow{A_1D}, \overrightarrow{DA_1}, \overrightarrow{BC_1}, \overrightarrow{C_1B}, \overrightarrow{B_1C}, \overrightarrow{CB_1}$
Subfield: Solid Geometry


Question: $如图所示，在以长、宽、高分别为AB=3，AD=2，AA_{1}=1的长方体ABCD-A_{1}B_{1}C_{1}D_{1}的八个顶点的两点为起点和终点的向量中：$![Image](https://cdntk.53zaixian.com/tq/1658661520050679809/img/compression/M_43986c1ae709d9f6da821c1fb6111d4f.jpg)
$(3)试写出与向量\overrightarrow{AB}相等的所有向量；$

Solution:
$与向量\overrightarrow{AB}相等的向量有\overrightarrow{A_1B_1}，\overrightarrow{DC}，\overrightarrow{D_1C_1}.$

Final answer: $\overrightarrow{A_1B_1}, \overrightarrow{DC}, \overrightarrow{D_1C_1}$
Subfield: Solid Geometry


Question: $如图所示，在以长、宽、高分别为AB=3，AD=2，AA_{1}=1的长方体ABCD-A_{1}B_{1}C_{1}D_{1}的八个顶点的两点为起点和终点的向量中：$![Image](https://cdntk.53zaixian.com/tq/1658661520050679809/img/compression/M_43986c1ae709d9f6da821c1fb6111d4f.jpg)
$(4)试写出向量\overrightarrow{AA_1}的相反向量.$

Solution:
$向量\overrightarrow{AA_1}的相反向量有\overrightarrow{A_1A}，\overrightarrow{B_1B}，\overrightarrow{C_1C}，\overrightarrow{D_1D}.$

Final answer: $\overrightarrow{A_1A}, \overrightarrow{B_1B}, \overrightarrow{C_1C}, \overrightarrow{D_1D}$
Subfield: Solid Geometry


Question: $已知正方形ABCD和正方形ABEF的边长都是1，且平面ABCD\perp 平面ABEF，点M在AC上移动，点N在BF上移动，若CM=BN=a(0<a<\sqrt{2}).$<img_2078>
$(1)求MN的长；$

Solution:
$因为平面ABCD\perp 平面ABEF，平面ABCD\cap 平面ABEF=AB，AB\perp BE，所以BE\perp 平面ABCD，所以AB,BC,BE两两垂直.过点M作MG\perp AB，MH\perp BC，垂足分别为G,H，连接NG，易证NG\perp AB.因为CM=BN=a，所以CH=MH=BG=GN=\frac{\sqrt{2}}{2}a.以B为坐标原点，BA,BE,BC所在直线分别为x轴、y轴、z轴建立如图所示的空间直角坐标系，$<img_1629>$则M\left(\frac{\sqrt{2}}{2}a,0,1-\frac{\sqrt{2}}{2}a\right)，N\left(\frac{\sqrt{2}}{2}a,\frac{\sqrt{2}}{2}a,0\right).|MN|=\sqrt{\left(\frac{\sqrt{2}}{2}a-\frac{\sqrt{2}}{2}a\right)^2+\left(0-\frac{\sqrt{2}}{2}a\right)^2+\left(1-\frac{\sqrt{2}}{2}a-0\right)^2}=\sqrt{a^2-\sqrt{2}a+1}=\sqrt{\left(a-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}.$



Final answer: $\sqrt{\left(a-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}$
Subfield: Solid Geometry


Question: $已知正方形ABCD和正方形ABEF的边长都是1，且平面ABCD\perp 平面ABEF，点M在AC上移动，点N在BF上移动，若CM=BN=a(0<a<\sqrt{2}).$![Image](https://cdntk.53zaixian.com/tq/1658661517251133441/img/compression/M_f566e67f6da61afc99a1890df28db09c.jpg)
$(1)求MN的长；$

Solution:
$因为平面ABCD\perp 平面ABEF，平面ABCD\cap 平面ABEF=AB，AB\perp BE，所以BE\perp 平面ABCD，所以AB,BC,BE两两垂直.过点M作MG\perp AB，MH\perp BC，垂足分别为G,H，连接NG，易证NG\perp AB.因为CM=BN=a，所以CH=MH=BG=GN=\frac{\sqrt{2}}{2}a.以B为坐标原点，BA,BE,BC所在直线分别为x轴、y轴、z轴建立如图所示的空间直角坐标系，$![Image](https://cdntk.53zaixian.com/tq/1658661516995280898/img/compression/M_bab3bf645fa4662813099791684775c9.jpg)$则M\left(\frac{\sqrt{2}}{2}a,0,1-\frac{\sqrt{2}}{2}a\right)，N\left(\frac{\sqrt{2}}{2}a,\frac{\sqrt{2}}{2}a,0\right).|MN|=\sqrt{\left(\frac{\sqrt{2}}{2}a-\frac{\sqrt{2}}{2}a\right)^2+\left(0-\frac{\sqrt{2}}{2}a\right)^2+\left(1-\frac{\sqrt{2}}{2}a-0\right)^2}=\sqrt{a^2-\sqrt{2}a+1}=\sqrt{\left(a-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}.$

Final answer: $\sqrt{\left(a-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}$
Subfield: Solid Geometry


Question: $已知正方形ABCD和正方形ABEF的边长都是1，且平面ABCD\perp 平面ABEF，点M在AC上移动，点N在BF上移动，若CM=BN=a(0<a<\sqrt{2}).$![Image](https://cdntk.53zaixian.com/tq/1658661517251133441/img/compression/M_f566e67f6da61afc99a1890df28db09c.jpg)
$(2)当a为何值时，MN的长最小？并求出最小值.$

Solution:
$由(1)得，当a=\frac{\sqrt{2}}{2}时，MN的长最小，且最小值为\frac{\sqrt{2}}{2}.$

Final answer: $当a=\frac{\sqrt{2}}{2}时，MN的长最小，且最小值为\frac{\sqrt{2}}{2}.$
Subfield: Solid Geometry


Question: $已知正方形ABCD和正方形ABEF的边长都是1，且平面ABCD\perp 平面ABEF，点M在AC上移动，点N在BF上移动，若CM=BN=a(0<a<\sqrt{2}).$<img_2078>
$(2)当a为何值时，MN的长最小？并求出最小值.$

Solution:
$由(1)得，当a=\frac{\sqrt{2}}{2}时，MN的长最小，且最小值为\frac{\sqrt{2}}{2}.$



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


Question: $如图，已知在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=3，AD=1，且\angle DAB=\angle BAA_{1}=\angle DAA_{1}=\frac{\pi }{3}.$![Image](https://cdntk.53zaixian.com/tq/1658661527336185858/img/compression/M_94931caa58848ecd9e8390ffa1c89e70.jpg)
$(1)求B_{1}D的长；$

Solution:
$由题可知，\overrightarrow{B_1D}=\overrightarrow{B_1B}+\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1}，那么\overrightarrow{B_1D}^{2}=(\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1})^{2}=\overrightarrow{AD}^{2}+\overrightarrow{AB}^{2}+\overrightarrow{AA_1}^{2}-2\overrightarrow{AD}\cdot \overrightarrow{AB}-2\overrightarrow{AD}\cdot \overrightarrow{AA_1}+2\overrightarrow{AB}\cdot \overrightarrow{AA_1}=1^{2}+2^{2}+3^{2}-2\times (1\times 2+1\times 3-2\times 3)\times \frac{1}{2}=15，因此B_{1}D的长为\sqrt{15}.$

Final answer: $\sqrt{15}$
Subfield: Solid Geometry


Question: $如图，已知在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=3，AD=1，且\angle DAB=\angle BAA_{1}=\angle DAA_{1}=\frac{\pi }{3}.$<img_2798>
$(1)求B_{1}D的长；$

Solution:
$由题可知，\overrightarrow{B_1D}=\overrightarrow{B_1B}+\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1}，那么\overrightarrow{B_1D}^{2}=(\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1})^{2}=\overrightarrow{AD}^{2}+\overrightarrow{AB}^{2}+\overrightarrow{AA_1}^{2}-2\overrightarrow{AD}\cdot \overrightarrow{AB}-2\overrightarrow{AD}\cdot \overrightarrow{AA_1}+2\overrightarrow{AB}\cdot \overrightarrow{AA_1}=1^{2}+2^{2}+3^{2}-2\times (1\times 2+1\times 3-2\times 3)\times \frac{1}{2}=15，因此B_{1}D的长为\sqrt{15}.$

Final answer: $\sqrt{15}$
Subfield: Solid Geometry


Question: $如图，已知在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=3，AD=1，且\angle DAB=\angle BAA_{1}=\angle DAA_{1}=\frac{\pi }{3}.$![Image](https://cdntk.53zaixian.com/tq/1658661527336185858/img/compression/M_94931caa58848ecd9e8390ffa1c89e70.jpg)
$(2)求\overrightarrow{CD_1}与\overrightarrow{B_1D}夹角的余弦值.$

Solution:
$连接A_{1}B，由题可知，\overrightarrow{CD_1}=\overrightarrow{BA_1}=\overrightarrow{AA_1}-\overrightarrow{AB}，则|\overrightarrow{CD_1}|=\sqrt{{(\overrightarrow{AA_1}-\overrightarrow{AB})}^2}=\sqrt{{\overrightarrow{AA_1}}^2+{\overrightarrow{AB}}^2-2\overrightarrow{AA_1} \cdot \overrightarrow{AB}}=\sqrt{3^2+2^2-2 \times 3 \times 2 \times \frac{1}{2}}=\sqrt{7}，所以\overrightarrow{CD_1}\cdot \overrightarrow{B_1D}=(\overrightarrow{AA_1}-\overrightarrow{AB})\cdot (\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1})=\overrightarrow{AD}\cdot \overrightarrow{AA_1}-\overrightarrow{AD}\cdot \overrightarrow{AB}+\overrightarrow{AB}^{2}-\overrightarrow{AA_1}^{2}=1\times 3\times \frac{1}{2}-1\times 2\times \frac{1}{2}+2^{2}-3^{2}=-\frac{9}{2}，所以\\cos   \langle \overrightarrow{CD_1}，\overrightarrow{B_1D}\rangle =\frac{\overrightarrow{CD_1} \cdot \overrightarrow{B_1D}}{|\overrightarrow{CD_1}||\overrightarrow{B_1D}|}=\frac{-\frac{9}{2}}{\sqrt{7} \times \sqrt{15}}=-\frac{3\sqrt{105}}{70}.$

Final answer: $-\frac{3\sqrt{105}}{70}$
Subfield: Solid Geometry


Question: $如图，已知在平行六面体ABCD-A_{1}B_{1}C_{1}D_{1}中，AB=2，AA_{1}=3，AD=1，且\angle DAB=\angle BAA_{1}=\angle DAA_{1}=\frac{\pi }{3}.$<img_2798>
$(2)求\overrightarrow{CD_1}与\overrightarrow{B_1D}夹角的余弦值.$

Solution:
$连接A_{1}B，由题可知，\overrightarrow{CD_1}=\overrightarrow{BA_1}=\overrightarrow{AA_1}-\overrightarrow{AB}，则|\overrightarrow{CD_1}|=\sqrt{{(\overrightarrow{AA_1}-\overrightarrow{AB})}^2}=\sqrt{{\overrightarrow{AA_1}}^2+{\overrightarrow{AB}}^2-2\overrightarrow{AA_1} \cdot \overrightarrow{AB}}=\sqrt{3^2+2^2-2 \times 3 \times 2 \times \frac{1}{2}}=\sqrt{7}，所以\overrightarrow{CD_1}\cdot \overrightarrow{B_1D}=(\overrightarrow{AA_1}-\overrightarrow{AB})\cdot (\overrightarrow{AD}-\overrightarrow{AB}-\overrightarrow{AA_1})=\overrightarrow{AD}\cdot \overrightarrow{AA_1}-\overrightarrow{AD}\cdot \overrightarrow{AB}+\overrightarrow{AB}^{2}-\overrightarrow{AA_1}^{2}=1\times 3\times \frac{1}{2}-1\times 2\times \frac{1}{2}+2^{2}-3^{2}=-\frac{9}{2}，所以\\cos   \langle \overrightarrow{CD_1}，\overrightarrow{B_1D}\rangle =\frac{\overrightarrow{CD_1} \cdot \overrightarrow{B_1D}}{|\overrightarrow{CD_1}||\overrightarrow{B_1D}|}=\frac{-\frac{9}{2}}{\sqrt{7} \times \sqrt{15}}=-\frac{3\sqrt{105}}{70}.$

Final answer: $-\frac{3\sqrt{105}}{70}$
Subfield: Solid Geometry


Question: $如图,在正三棱柱ABC-A_{1}B_{1}C_{1}中,底面边长为\sqrt{2}.$<img_1997>
$(1)若侧棱长为1,求证:AB_{1}\perp BC_{1};$

Solution:
$证明:\overrightarrow{AB_1}=\overrightarrow{AB}+\overrightarrow{BB_1},\overrightarrow{BC_1}=\overrightarrow{BB_1}+\overrightarrow{BC}.在正三棱柱ABC-A_{1}B_{1}C_{1}中,BB_{1}\perp 平面ABC,\triangle ABC为正三角形,所以\overrightarrow{BB_1}\cdot \overrightarrow{AB}=0,\overrightarrow{BB_1}\cdot \overrightarrow{BC}=0,\langle \overrightarrow{AB},\overrightarrow{BC}\rangle =\pi -\langle \overrightarrow{BA},\overrightarrow{BC}\rangle =\pi -\frac{\pi }{3}=\frac{2\pi }{3}.所以\overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=(\overrightarrow{AB}+\overrightarrow{BB_1})\cdot (\overrightarrow{BB_1}+\overrightarrow{BC})=\overrightarrow{AB}\cdot \overrightarrow{BB_1}+\overrightarrow{AB}\cdot \overrightarrow{BC}+\overrightarrow{BB_1}^{2}+\overrightarrow{BB_1}\cdot \overrightarrow{BC}=|\overrightarrow{AB}||\overrightarrow{BC}|\\cos  \langle \overrightarrow{AB},\overrightarrow{BC}\rangle +\overrightarrow{BB_1}^{2}=-1+1=0,所以\overrightarrow{AB_1}\perp \overrightarrow{BC_1},即AB_{1}\perp BC_{1}.$
Subfield: Solid Geometry


Question: $如图,在正三棱柱ABC-A_{1}B_{1}C_{1}中,底面边长为\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1658661529831796739/img/compression/M_722b3642efa4bda30aaebea6f5baa629.jpg)
$(1)若侧棱长为1,求证:AB_{1}\perp BC_{1};$

Solution:
$证明:\overrightarrow{AB_1}=\overrightarrow{AB}+\overrightarrow{BB_1},\overrightarrow{BC_1}=\overrightarrow{BB_1}+\overrightarrow{BC}.在正三棱柱ABC-A_{1}B_{1}C_{1}中,BB_{1}\perp 平面ABC,\triangle ABC为正三角形,所以\overrightarrow{BB_1}\cdot \overrightarrow{AB}=0,\overrightarrow{BB_1}\cdot \overrightarrow{BC}=0,\langle \overrightarrow{AB},\overrightarrow{BC}\rangle =\pi -\langle \overrightarrow{BA},\overrightarrow{BC}\rangle =\pi -\frac{\pi }{3}=\frac{2\pi }{3}.所以\overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=(\overrightarrow{AB}+\overrightarrow{BB_1})\cdot (\overrightarrow{BB_1}+\overrightarrow{BC})=\overrightarrow{AB}\cdot \overrightarrow{BB_1}+\overrightarrow{AB}\cdot \overrightarrow{BC}+\overrightarrow{BB_1}^{2}+\overrightarrow{BB_1}\cdot \overrightarrow{BC}=|\overrightarrow{AB}||\overrightarrow{BC}|\\cos  \langle \overrightarrow{AB},\overrightarrow{BC}\rangle +\overrightarrow{BB_1}^{2}=-1+1=0,所以\overrightarrow{AB_1}\perp \overrightarrow{BC_1},即AB_{1}\perp BC_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图,在正三棱柱ABC-A_{1}B_{1}C_{1}中,底面边长为\sqrt{2}.$![Image](https://cdntk.53zaixian.com/tq/1658661529831796739/img/compression/M_722b3642efa4bda30aaebea6f5baa629.jpg)
$(2)若AB_{1}与BC_{1}的夹角为\frac{\pi }{3},求侧棱长.$

Solution:
$由(1)知\overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=|\overrightarrow{AB}||\overrightarrow{BC}|\\cos  \langle \overrightarrow{AB},\overrightarrow{BC}\rangle +\overrightarrow{BB_1}^{2}=\overrightarrow{BB_1}^{2}-1.又|\overrightarrow{AB_1}|=\sqrt{(\overrightarrow{AB}+\overrightarrow{BB_1})^2}=\sqrt{{\overrightarrow{AB}}^2+{\overrightarrow{BB_1}}^2}=\sqrt{2+{\overrightarrow{BB_1}}^2}=|\overrightarrow{BC_1}|,所以\\cos  \langle \overrightarrow{AB_1},\overrightarrow{BC_1}\rangle =\frac{{\overrightarrow{BB_1}}^2-1}{2+{\overrightarrow{BB_1}}^2}=\frac{1}{2},所以|\overrightarrow{BB_1}|=2,即侧棱长为2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图,在正三棱柱ABC-A_{1}B_{1}C_{1}中,底面边长为\sqrt{2}.$<img_1997>
$(2)若AB_{1}与BC_{1}的夹角为\frac{\pi }{3},求侧棱长.$

Solution:
$由(1)知\overrightarrow{AB_1}\cdot \overrightarrow{BC_1}=|\overrightarrow{AB}||\overrightarrow{BC}|\\cos  \langle \overrightarrow{AB},\overrightarrow{BC}\rangle +\overrightarrow{BB_1}^{2}=\overrightarrow{BB_1}^{2}-1.又|\overrightarrow{AB_1}|=\sqrt{(\overrightarrow{AB}+\overrightarrow{BB_1})^2}=\sqrt{{\overrightarrow{AB}}^2+{\overrightarrow{BB_1}}^2}=\sqrt{2+{\overrightarrow{BB_1}}^2}=|\overrightarrow{BC_1}|,所以\\cos  \langle \overrightarrow{AB_1},\overrightarrow{BC_1}\rangle =\frac{{\overrightarrow{BB_1}}^2-1}{2+{\overrightarrow{BB_1}}^2}=\frac{1}{2},所以|\overrightarrow{BB_1}|=2,即侧棱长为2.$

Final answer: $2$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AC\cap BD=O，底面ABCD为菱形，边长为2，PC\perp BD，PA=PC，且\angle ABC=60^\circ ，异面直线PB与CD的夹角为60^\circ .$<img_3123>
$(1)求证：PO\perp 平面ABCD；$

Solution:
$证明：\because 四边形ABCD是菱形，\therefore AC\perp BD，又\because PC\perp BD，PC\cap AC=C，\therefore BD\perp 平面APC.\because PO\subset 平面APC，\therefore BD\perp PO.\because PA=PC，O为AC的中点，\therefore PO\perp AC，又AC\cap BD=O，\therefore PO\perp 平面ABCD.$
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AC\cap BD=O，底面ABCD为菱形，边长为2，PC\perp BD，PA=PC，且\angle ABC=60^\circ ，异面直线PB与CD的夹角为60^\circ .$![Image](https://cdntk.53zaixian.com/tq/1658661579048116226/img/compression/M_85afe5c2e745dadf143f4628cc469d63.jpg)
$(1)求证：PO\perp 平面ABCD；$

Solution:
$证明：\because 四边形ABCD是菱形，\therefore AC\perp BD，又\because PC\perp BD，PC\cap AC=C，\therefore BD\perp 平面APC.\because PO\subset 平面APC，\therefore BD\perp PO.\because PA=PC，O为AC的中点，\therefore PO\perp AC，又AC\cap BD=O，\therefore PO\perp 平面ABCD.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在四棱锥P-ABCD中，AC\cap BD=O，底面ABCD为菱形，边长为2，PC\perp BD，PA=PC，且\angle ABC=60^\circ ，异面直线PB与CD的夹角为60^\circ .$<img_3123>
$(2)若E是线段OC的中点，求点E到直线BP的距离.$

Solution:
$连接BE，以O为原点，OB，OC，OP所在直线分别为x轴、y轴、z轴建立空间直角坐标系，如图.\because AB\parallel CD，\therefore \angle PBA为异面直线PB与CD的夹角，\therefore \angle PBA=60^\circ .在菱形ABCD中，AB=2，\angle ABC=60^\circ ，\therefore OA=1，OB=\sqrt{3}.设PO=a，则PA=\sqrt{a^2+1}，PB=\sqrt{a^2+3}，在\triangle PBA中，由余弦定理得PA^{2}=AB^{2}+BP^{2}-2AB\cdot BP\cdot \\cos  \angle PBA，即a^{2}+1=4+a^{2}+3-2\times 2\times \sqrt{a^2+3}\times \frac{1}{2}，\therefore a=\sqrt{6}(负值舍去)，\therefore B(\sqrt{3},0,0)，P(0,0,\sqrt{6})，E\left(0,\frac{1}{2},0\right)，\therefore \overrightarrow{BE}=\left(-\sqrt{3},\frac{1}{2},0\right)，\overrightarrow{BP}=(-\sqrt{3},0,\sqrt{6})，\therefore |\overrightarrow{BE}|=\frac{\sqrt{13}}{2}，|\overrightarrow{BP}|=3，\therefore 点E到直线BP的距离d=\sqrt{{|\overrightarrow{BE}|}^2-\left(\frac{\overrightarrow{BE} \cdot \overrightarrow{BP}}{|\overrightarrow{BP}|}\right)^2}=\sqrt{\frac{13}{4}-1}=\frac{3}{2}.$<img_1900>

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


Question: $如图，在四棱锥P-ABCD中，AC\cap BD=O，底面ABCD为菱形，边长为2，PC\perp BD，PA=PC，且\angle ABC=60^\circ ，异面直线PB与CD的夹角为60^\circ .$![Image](https://cdntk.53zaixian.com/tq/1658661579048116226/img/compression/M_85afe5c2e745dadf143f4628cc469d63.jpg)
$(2)若E是线段OC的中点，求点E到直线BP的距离.$

Solution:
$连接BE，以O为原点，OB，OC，OP所在直线分别为x轴、y轴、z轴建立空间直角坐标系，如图.\because AB\parallel CD，\therefore \angle PBA为异面直线PB与CD的夹角，\therefore \angle PBA=60^\circ .在菱形ABCD中，AB=2，\angle ABC=60^\circ ，\therefore OA=1，OB=\sqrt{3}.设PO=a，则PA=\sqrt{a^2+1}，PB=\sqrt{a^2+3}，在\triangle PBA中，由余弦定理得PA^{2}=AB^{2}+BP^{2}-2AB\cdot BP\cdot \\cos  \angle PBA，即a^{2}+1=4+a^{2}+3-2\times 2\times \sqrt{a^2+3}\times \frac{1}{2}，\therefore a=\sqrt{6}(负值舍去)，\therefore B(\sqrt{3},0,0)，P(0,0,\sqrt{6})，E\left(0,\frac{1}{2},0\right)，\therefore \overrightarrow{BE}=\left(-\sqrt{3},\frac{1}{2},0\right)，\overrightarrow{BP}=(-\sqrt{3},0,\sqrt{6})，\therefore |\overrightarrow{BE}|=\frac{\sqrt{13}}{2}，|\overrightarrow{BP}|=3，\therefore 点E到直线BP的距离d=\sqrt{{|\overrightarrow{BE}|}^2-\left(\frac{\overrightarrow{BE} \cdot \overrightarrow{BP}}{|\overrightarrow{BP}|}\right)^2}=\sqrt{\frac{13}{4}-1}=\frac{3}{2}.$![Image](https://cdntk.53zaixian.com/tq/1658661579509489666/img/compression/M_48fc0e960d44cdae4b59adf94c12a8e5.jpg)

Final answer: $\frac{3}{2}$