解:(1)，${\displaystyle \overrightarrow{BC}=(-2，-1，2)}$
    设${\displaystyle \overrightarrow{c}=(-2\lambda ，-\lambda ，2\lambda )}$
    ${\displaystyle \left|\overrightarrow{c}\right|=\sqrt{{(-2\lambda )}^2+{(-\lambda )}^2+{\left(2\lambda \right)}^2+}=3\left|\lambda \right|=3}$
    ${\displaystyle \therefore \lambda =\pm 1}$
    ${\displaystyle \therefore \overrightarrow{c}=(-2，-1，2)}$或${\displaystyle \overrightarrow{c}=(2，1，-2)}$
    

    (2)${\displaystyle \overrightarrow{a}=(-1+2，1-0，2-2)=(1，1，0)}$
    ${\displaystyle \overrightarrow{b}=(-3+2，0-0，4-2)=(-1，0，2)}$
    
    ${\displaystyle =\frac{(1，1，0)·(-1，0，2)}{(\sqrt{2}\times \sqrt{5})}}$
    ${\displaystyle =-\frac{\sqrt{10}}{10}}$
    ${\displaystyle \therefore \overrightarrow{a}}$和${\displaystyle \overrightarrow{b}}$的夹角为${\displaystyle \pi -arccos\left(\frac{\sqrt{10}}{10}\right)}$
　

    (3)${\displaystyle k\overrightarrow{a}+\overrightarrow{b}=(k-1，k，2)}$

    ${\displaystyle k\overrightarrow{a}-2\overrightarrow{b}=(k+2，k，-4)}$

    ${\displaystyle \therefore (k-1，k，2)·(k+2，k，-4)=(k-1)(k+2)+k^2-8=0}$

    解得${\displaystyle k=-\frac{5}{2}}$或${\displaystyle k=2}$

    评注:对于空间向量的平行、垂直与夹角，可联系平面向量进行研究．

    解:方法一：(1)由知${\displaystyle \angle C_{1}G{D}_1}$为异面直线${\displaystyle AD}$与${\displaystyle C_{1}G}$所成的角    

连结${\displaystyle C_{1}F}$，因为${\displaystyle AE}$和${\displaystyle C_{1}F}$分别是平行平面${\displaystyle AB{B}_1{A}_1}$和${\displaystyle C{C}_1{D}_{1}D}$与平面${\displaystyle AE{C}_{1}G}$的交线     ，由此得${\displaystyle D_{1}F=BF=\sqrt{3}}$，再由${\displaystyle △F{D}_{1}G~△FDA}$     ${\displaystyle \therefore D_{1}G=\sqrt{3}}$     在${\displaystyle Rt△C_1{D}_{1}G}$中，由${\displaystyle C_1{D}_1=1}$得${\displaystyle \angle C_{1}G{D}_1=\frac{\pi }{6}}$ 　     (2)作${\displaystyle D_{1}H\perp C_{1}G}$于${\displaystyle H}$，由三垂线定理知${\displaystyle FH\perp C_{1}G}$     ${\displaystyle \therefore \angle D_{1}HF}$为二面角${\displaystyle F-C_{1}G-D_1}$即二面角${\displaystyle A-C_{1}G-A_1}$的平面角     在${\displaystyle Rt△H{D}_{1}G}$中，由${\displaystyle D_{1}G=\sqrt{3}}$，${\displaystyle \angle HG{D}_1=\frac{\pi }{6}}$得${\displaystyle D_{1}H=\frac{\sqrt{3}}{2}}$     ${\displaystyle tan\angle H{D}_{1}G=\frac{D_{1}F}{D_{1}H}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2}$

    方法二：如图，以${\displaystyle A_1}$为坐标原点 ，${\displaystyle A_1{B}_1}$、${\displaystyle A_1{D}_1}$、${\displaystyle A_{1}A}$所在直线分别为${\displaystyle x、y、z}$轴，建立空间直角坐标系，于是${\displaystyle A(0，0，\sqrt{3}+1)}$，${\displaystyle C_1(1，1，0)}$，${\displaystyle D(0，1，\sqrt{3}+1)}$，${\displaystyle E(1，0，1)}$   

(1)${\displaystyle \therefore \overrightarrow{AD}=(0，1，0)}$，${\displaystyle \overrightarrow{E{C}_1}=(0，1，-1)}$     ${\displaystyle \because E{C}_1}$和${\displaystyle AF}$分别是平行平面${\displaystyle B{B}_1{C}_{1}C}$和${\displaystyle A{A}_1{D}_{1}D}$与平面${\displaystyle AE{C}_{1}G}$的交线         设${\displaystyle G(0，y，0)}$，则${\displaystyle \overrightarrow{AG}=(0，y，-\sqrt{3}-1)}$            由得${\displaystyle -\frac{1}{y}=\frac{-1}{-\sqrt{3}-1}}$，于是${\displaystyle y=\sqrt{3}+1}$     ${\displaystyle \therefore G(0，\sqrt{3}+1，0)}$，则${\displaystyle \overrightarrow{C_{1}G}=(-1，\sqrt{3}，0)}$     设异面直线${\displaystyle AD}$与${\displaystyle C_{1}G}$所成的角的大小为${\displaystyle \theta }$，则         ${\displaystyle \therefore \theta =\frac{\pi }{6}}$ 　     (2)作${\displaystyle A_{1}H\perp C_{1}G}$交${\displaystyle G{C}_1}$的延长线于${\displaystyle H}$，由三垂线定理知${\displaystyle AH\perp GH}$     ${\displaystyle \therefore \angle AH{A}_1}$为二面角${\displaystyle A-C_{1}G-A_1}$的平面角     设${\displaystyle H(a，b，0)}$，则${\displaystyle \overrightarrow{A_{1}H}=(a，b，0)}$，${\displaystyle \overrightarrow{C_{1}H}=(a-1，b-1，0)}$     由${\displaystyle A_{1}H\perp C_{1}G}$得${\displaystyle \overrightarrow{A_{1}H}·\overrightarrow{C_{1}G}=0}$     由此得${\displaystyle a-\sqrt{3}b=0}$                            ①     又由${\displaystyle H、C_1、G}$共线得     从而${\displaystyle \frac{a-1}{-1}=\frac{b-1}{\sqrt{3}}}$，于是${\displaystyle \sqrt{3}a+b-(\sqrt{3}+1)=0}$     ②     联立①和②得${\displaystyle a=\frac{\sqrt{3}+3}{4}}$，${\displaystyle b=\frac{\sqrt{3}+1}{4}}$     ${\displaystyle \therefore H(\frac{\sqrt{3}+3}{4}，\frac{\sqrt{3}+1}{4}，0)}$     由${\displaystyle \left|\overrightarrow{A_{1}H}\right|=\sqrt{{\left(\frac{\sqrt{3}+3}{4}\right)}^2+{\left(\frac{\sqrt{3}+1}{4}\right)}^2}=\frac{\sqrt{3}+1}{2}}$，${\displaystyle \left|\overrightarrow{A_{1}A}\right|=\sqrt{3}+1}$     得${\displaystyle tan\angle AH{A}_1=\frac{\left|\overrightarrow{A_{1}A}\right|}{\left|\overrightarrow{A_{1}H}\right|}=\frac{\sqrt{3}+1}{\frac{\sqrt{3}+1}{2}}=2}$

    评注:利用向量求空间角，关键在于利用向量表示出空间角，借助向量进行计算．

    (1)求异面直线所成的角：设表示两条直线的向量为${\displaystyle \overrightarrow{n_1}}$与${\displaystyle \overrightarrow{n_2}}$，且${\displaystyle \overrightarrow{n_1}}$与${\displaystyle \overrightarrow{n_2}}$之间的夹角为${\displaystyle \theta }$．若${\displaystyle \theta }$为锐角，则两条异面直线所成的角为${\displaystyle \theta }$；若${\displaystyle \theta }$为钝角，则两条异面直线所成的角${\displaystyle \pi -\theta }$；
　

    (2)求直线与平面所成的角：求斜线与平面所成角可采用如下方法：设表示斜线的一个向量为${\displaystyle \overrightarrow{n_1}}$，表示平面的一个法向量为${\displaystyle \overrightarrow{n_2}}$，且${\displaystyle \overrightarrow{n_1}}$与${\displaystyle \overrightarrow{n_2}}$之间的夹角为${\displaystyle \theta }$．若${\displaystyle \theta }$为锐角，则斜线与平面所成的角为${\displaystyle \frac{\pi }{2}-\theta }$，若${\displaystyle \theta }$为钝角，则两条异面直线所成的角为${\displaystyle \theta -\frac{\pi }{2}}$；
　

    (3)求二面角：设${\displaystyle \overrightarrow{n_1}}$是平面${\displaystyle \alpha }$的法向量，${\displaystyle \overrightarrow{n_2}}$是平面${\displaystyle \beta }$的法向量．

    ①若两个平面的二面角如图(1)所示的示意图，则${\displaystyle \overrightarrow{n_1}}$与${\displaystyle \overrightarrow{n_2}}$之间的夹角就是欲求的二面角；

    ②若两个平面的二面角如图(2)所示的示意图，设${\displaystyle \overrightarrow{n_1}}$与${\displaystyle \overrightarrow{n_2}}$之间的夹角为${\displaystyle \theta }$，则两个平面的二面角为${\displaystyle \pi -\theta }$．

    解:方法一：(1)不妨设正三角形${\displaystyle ABC}$的边长为${\displaystyle 3}$

证明：在图(1)中，取${\displaystyle BE}$的中点${\displaystyle D}$，连结${\displaystyle DF}$     ${\displaystyle \because AE:EB=CF:FA=1:2}$     ${\displaystyle \therefore AF=AD=2}$，而${\displaystyle \angle A=60°}$     ${\displaystyle \therefore △ADF}$是正三角形     又${\displaystyle AE=DE=1}$     ${\displaystyle \therefore EF\perp AD}$     在图(2)中，${\displaystyle A_{1}E\perp EF}$，${\displaystyle BE\perp EF}$     ${\displaystyle \therefore \angle A_{1}EB}$为二面角${\displaystyle A_1-EF-B}$的平面角     由题设条件知此二面角为直二面角     ${\displaystyle \therefore A_{1}E\perp BE}$     又${\displaystyle BE\cap EF=E}$     ${\displaystyle \therefore A_{1}E\perp }$平面${\displaystyle BEF}$，即${\displaystyle A_{1}E\perp }$平面${\displaystyle BEP}$ 　     (2)在图(2)中，${\displaystyle \because A_{1}E}$不垂直于${\displaystyle A_{1}B}$     ${\displaystyle \therefore A_{1}E}$是平面${\displaystyle A_{1}BP}$的斜线     又${\displaystyle A_{1}E\perp }$平面${\displaystyle BEP}$     ${\displaystyle \therefore A_{1}E\perp BP}$     从而${\displaystyle BP}$垂直于${\displaystyle A_{1}E}$在平面${\displaystyle A_{1}BP}$的射影(三垂线逆定理)     设${\displaystyle A_{1}E}$在平面${\displaystyle A_{1}BP}$的射影为${\displaystyle A_{1}Q}$，且${\displaystyle A_{1}Q}$交${\displaystyle BP}$于点${\displaystyle Q}$，则${\displaystyle \angle E{A}_{1}Q}$就是${\displaystyle A_{1}E}$与平面${\displaystyle A_{1}BP}$所成的角，且${\displaystyle BP\perp A_{1}Q}$     在${\displaystyle △EBP}$中，${\displaystyle \because BE=BP=2，\angle EBP=60°}$     ${\displaystyle \therefore △EBP}$是等边三角形     ${\displaystyle \therefore BE=EP}$     又${\displaystyle A_{1}E\perp }$平面${\displaystyle BEP}$     ${\displaystyle \therefore A_{1}B=A_{1}P}$     ${\displaystyle \therefore Q}$为${\displaystyle BP}$的中点，且${\displaystyle EQ=\sqrt{3}}$     又${\displaystyle A_{1}E=1}$，在${\displaystyle Rt△A_{1}EQ}$中，${\displaystyle tan\angle E{A}_{1}Q=\frac{EQ}{A_{1}E}=\sqrt{3}}$     ${\displaystyle \therefore \angle E{A}_{1}Q=60°}$     ${\displaystyle \therefore }$直线${\displaystyle A_{1}E}$与平面${\displaystyle A_{1}BP}$所成角的大小为${\displaystyle 60°}$

(3)在图(3)中，过${\displaystyle F}$作${\displaystyle FM\perp A_{1}P}$于${\displaystyle M}$，连结${\displaystyle QM}$、${\displaystyle QF}$     ${\displaystyle \because CF=CP=1}$，${\displaystyle \angle C=60°}$     ${\displaystyle \therefore △FCP}$是正三角形，且${\displaystyle PF=1}$     又${\displaystyle PQ=\frac{1}{2}BP=1}$     ${\displaystyle \therefore PF=PQ}$              ①     ${\displaystyle \because A_{1}E\perp }$平面${\displaystyle BEP}$，${\displaystyle EQ=EF=\sqrt{3}}$     ${\displaystyle \therefore A_{1}F\perp A_{1}Q}$     ${\displaystyle \therefore △A_{1}FP\cong △A_{1}QP}$       ${\displaystyle \therefore \angle A_{1}PF=\angle A_{1}PQ}$       ②     由①②及${\displaystyle MP}$为公共边知${\displaystyle △FMP\cong △QMP}$     ${\displaystyle \therefore \angle QMP=\angle FMP=90°}$，且${\displaystyle MF=MQ}$     ${\displaystyle \therefore \angle FMQ}$为二面角${\displaystyle B-A_{1}P-F}$的平面角     在${\displaystyle Rt△A_{1}QP}$中，${\displaystyle A_{1}Q=A_{1}F=2}$，${\displaystyle FQ=1}$     ${\displaystyle \therefore A_{1}P=\sqrt{5}}$     ${\displaystyle \because MQ\perp A_{1}P}$      ${\displaystyle \therefore MQ=\frac{A_{1}Q·PQ}{A_{1}P}=\frac{2\sqrt{5}}{5}}$     ${\displaystyle \therefore MF=\frac{2\sqrt{5}}{5}}$     在${\displaystyle △FCQ}$中，${\displaystyle FC=1}$，${\displaystyle QC=2}$，${\displaystyle \angle C=60°}$，由余弦定理得${\displaystyle QF=\sqrt{3}}$     在${\displaystyle △FMQ}$中，${\displaystyle cos\angle FMQ=\frac{M{F}^2+M{Q}^2-Q{F}^2}{2MF·MQ}=-\frac{7}{8}}$     ${\displaystyle \therefore }$二面角${\displaystyle B-A_{1}P-F}$的大小为${\displaystyle \pi -arccos\left(\frac{7}{8}\right)}$

    方法二：不妨设正三角形${\displaystyle ABC}$的边长为${\displaystyle 3}$，

    (1)同方法一

    (2)如图(1)，由方法一知${\displaystyle A_{1}E\perp }$平面${\displaystyle BEF}$，${\displaystyle BE\perp EF}$，建立如图(4)所示的空间直角坐标系${\displaystyle O-xyz}$，则${\displaystyle E(0，0，0)}$，${\displaystyle A_1(0，0，1)}$，${\displaystyle B(2，0，0)}$，${\displaystyle F(0，\sqrt{3}，0)}$

连结${\displaystyle DP}$，${\displaystyle \because AF=BP=2}$，${\displaystyle AE=BD=1}$，${\displaystyle \angle A=\angle B}$     ${\displaystyle \therefore △FEA\cong △PDB}$，${\displaystyle PD=EF=\sqrt{3}}$     由图(1)知且${\displaystyle PF=DE=1}$     ${\displaystyle \therefore P(1，\sqrt{3}，0)}$     ${\displaystyle \therefore \overrightarrow{A_{1}B}=(2，0，-1)，\overrightarrow{BP}=(-1，\sqrt{3}，0)}$     ${\displaystyle \therefore }$对于平面${\displaystyle A_{1}BP}$内任一非零向量${\displaystyle \overrightarrow{a}}$，存在不全为零的实数${\displaystyle \lambda }$、${\displaystyle \mu }$      使得${\displaystyle \overrightarrow{a}=\lambda \overrightarrow{A_{1}B}+\mu \overrightarrow{BP}=(2\lambda -\mu ，\sqrt{3}\mu ，-\lambda )}$     又${\displaystyle \overrightarrow{A_{1}E}=(0，0，-1)}$         ${\displaystyle =\frac{\overrightarrow{A_{1}E}·\overrightarrow{a}}{\left|\overrightarrow{A_{1}E}\right|·\left|\overrightarrow{a}\right|}}$     ${\displaystyle =\frac{\lambda }{\sqrt{5{\lambda }^2-4\lambda \mu +4{\mu }^2}}}$     ${\displaystyle \because }$直线${\displaystyle A_{1}E}$与平面${\displaystyle A_{1}BP}$所成的角是向量${\displaystyle \overrightarrow{A_{1}E}}$与平面${\displaystyle A_{1}BP}$内非零向量夹角中最小的角      ${\displaystyle \therefore }$可设${\displaystyle \lambda >0}$         又${\displaystyle 5-4\frac{\mu }{\lambda }+4{\left(\frac{\mu }{\lambda }\right)}^2=4{(\frac{\mu }{\lambda }-\frac{1}{2})}^2+4}$的最小值为${\displaystyle 4}$     的最大值为${\displaystyle \frac{1}{2}}$     ${\displaystyle 即\overrightarrow{A_{1}E}}$与${\displaystyle \overrightarrow{a}}$夹角中最小的角为${\displaystyle 60°}$     ${\displaystyle \therefore }$直线${\displaystyle A_{1}E}$与平面${\displaystyle A_{1}BP}$所成角的大小为${\displaystyle 60°}$

    (3)如图(4)，过${\displaystyle F}$作${\displaystyle FM\perp A_{1}P}$于${\displaystyle M}$，过${\displaystyle M}$作${\displaystyle MN\perp A_{1}P}$交${\displaystyle BP}$于${\displaystyle N}$，则${\displaystyle \angle FMN}$为二面角${\displaystyle B-A_{1}P-F}$的平面角 

设${\displaystyle M(x，y，z)}$，则${\displaystyle \overrightarrow{MF}=(-x，\sqrt{3}-y，-z)}$     ${\displaystyle \because \overrightarrow{MF}\perp \overrightarrow{A_{1}P}}$     ${\displaystyle \therefore \overrightarrow{MF}·\overrightarrow{A_{1}P}=0}$，又${\displaystyle \overrightarrow{A_{1}P}=(1，\sqrt{3}，-1)}$     ${\displaystyle \therefore x+\sqrt{3}(y-\sqrt{3})-z=0}$        ①     ${\displaystyle \because A_1、M、P}$三点共线     ${\displaystyle \therefore }$存在${\displaystyle \lambda \in R}$，使得${\displaystyle \overrightarrow{A_{1}M}=\lambda \overrightarrow{A_{1}P}}$     ${\displaystyle \because \overrightarrow{A_{1}M}=(x，y，z-1)}$     ${\displaystyle \therefore (x，y，z-1)=\lambda (1，\sqrt{3}，-1)}$     从而${\displaystyle \{\begin{array}{l}x=\lambda \\ y=\sqrt{3}\lambda \\ z-1=-\lambda \end{array}}$，代入①得${\displaystyle \lambda =\frac{4}{5}}$     ${\displaystyle \therefore M(\frac{4}{5}，\frac{4\sqrt{3}}{5}，\frac{1}{5})}$，同理可得${\displaystyle N(\frac{3}{2}，\frac{\sqrt{3}}{2}，0)}$     ${\displaystyle \therefore \overrightarrow{MF}=(-\frac{4}{5}，\frac{\sqrt{3}}{5}，-\frac{1}{5})}$，${\displaystyle \overrightarrow{MN}=(\frac{7}{10}，-\frac{3\sqrt{3}}{10}，-\frac{1}{5})}$          ${\displaystyle =\frac{\overrightarrow{MF}·\overrightarrow{MN}}{\left|\overrightarrow{MF}\right|·\left|\overrightarrow{MN}\right|}}$     ${\displaystyle =\frac{-\frac{7}{10}}{\frac{2\sqrt{5}}{5}·\frac{2\sqrt{5}}{5}}}$     ${\displaystyle =-\frac{7}{8}}$     ${\displaystyle \therefore }$二面角${\displaystyle B-A_{1}P-F}$的大小为${\displaystyle \pi -arccos\left(\frac{7}{8}\right)}$

    评注:1、利用空间向量的坐标运算解决立体几何中的平行、垂直、夹角、距离等问题，关键是建立恰当的空间直角坐标系，难点是正确表达已知点的坐标．
    2、在立体几何中，要证明两线垂直，可选取适当的基底，然后确定各向量的坐标，通过向量数量积的坐标运算结果为${\displaystyle O}$来证明，是一个很好的办法．
    3、求异面直线所成角的关键是求异面直线上两向量的数量积，而要求两向量的数量积，必须会把所求向量用空间的一组基向量来表示．
    4、在空间图形中，如果线段较多，关系也较为复杂(如平行、垂直、角和距离等均有涉及)．此时如何入手，是用逻辑推理方法解决，还是用向量方法解决，一般较难作出判断，而且在较为综合的问题中，只用某一种方法，有时也难以奏效，常常需要多种方法合理结合，灵活使用，才能达到理想效果．
    图形中如果存在三条两两垂直的线段，则可考虑以它们所在直线为坐标轴建立空间直角坐标系，设法确定点、向量的坐标，然后通过向量的坐标，通过向量坐标运算解决有关问题 ．本题中，第(1)问的解决，实际上用传统方法解决也很方便，${\displaystyle E}$点坐标的确定．也需要通过几何推理和计算，解题中应注意逻辑推导和向量运算的有机结合．

    解:在长方体${\displaystyle ABCD-A_1{B}_1{C}_1{D}_1}$中， 以${\displaystyle AB}$、${\displaystyle AD}$、${\displaystyle A{A}_1}$所在直线分别为${\displaystyle x、y、z}$轴，建立空间直角坐标系，于是${\displaystyle A(0，0，0)}$，${\displaystyle B(2，0，0)}$，${\displaystyle F(1，0，1)}$

    又${\displaystyle \because AD\perp }$平面${\displaystyle A{A}_1{B}_{1}B}$ 

    ${\displaystyle \therefore BD}$与平面${\displaystyle A{A}_1{B}_{1}B}$所成的角就是${\displaystyle \angle DBA=30°}$

    由${\displaystyle AB=2}$，${\displaystyle AE\perp BD}$，${\displaystyle AE=1}$，${\displaystyle AD=\frac{2\sqrt{3}}{3}}$，得${\displaystyle E(\frac{1}{2}，\frac{\sqrt{3}}{2}，0)}$，${\displaystyle D(0，\frac{2\sqrt{3}}{3}，0)}$

(1)${\displaystyle \because \overrightarrow{AE}=(\frac{1}{2}，\frac{\sqrt{3}}{2}，0)}$，${\displaystyle \overrightarrow{BF}=(-1，0，1)}$               ${\displaystyle =\frac{\overrightarrow{AE}·\overrightarrow{BF}}{\left|\overrightarrow{AE}\right|·\left|\overrightarrow{BF}\right|}}$     ${\displaystyle =\frac{-\frac{1}{2}}{\sqrt{2}}}$     ${\displaystyle =-\frac{\sqrt{2}}{4}}$     即异面直线${\displaystyle AE}$与${\displaystyle BF}$所成的角为${\displaystyle arccos\left(\frac{\sqrt{2}}{4}\right)}$ 　     (2)易知平面${\displaystyle A{A}_{1}B}$的一个法向量${\displaystyle \overrightarrow{m}=(0，1，0)}$，设${\displaystyle \overrightarrow{n}=(x，y，z)}$是平面${\displaystyle BDF}$的一个法向量     ${\displaystyle \because \overrightarrow{BF}=(-1，0，1)}$，${\displaystyle \overrightarrow{BD}=(-2，\frac{2\sqrt{3}}{3}，0)}$      由${\displaystyle \{\begin{array}{l}\overrightarrow{n}\perp \overrightarrow{BF}\\ \overrightarrow{n}\perp \overrightarrow{BD}\end{array}}$     ${\displaystyle \Rightarrow \{\begin{array}{l}\overrightarrow{n}·\overrightarrow{BF}=0\\ \overrightarrow{n}·\overrightarrow{BD}=0\end{array}}$     ${\displaystyle \Rightarrow \{\begin{array}{l}-x+z=0\\ 2x-\frac{2\sqrt{3}}{3}y=0\end{array}}$     ${\displaystyle \Rightarrow \{\begin{array}{l}x=z\\ \sqrt{3}x=y\end{array}}$，取${\displaystyle \overrightarrow{n}=(1，\sqrt{3}，1)}$         ${\displaystyle =\frac{\overrightarrow{m}·\overrightarrow{n}}{\left|\overrightarrow{m}\right|·\left|\overrightarrow{n}\right|}}$     ${\displaystyle =\frac{\sqrt{3}}{1\times \sqrt{5}}}$     ${\displaystyle =\frac{\sqrt{15}}{5}}$     ${\displaystyle \therefore }$平面${\displaystyle BDF}$与平面${\displaystyle A{A}_{1}B}$所成二面角(锐角)的大小是${\displaystyle arccos\left(\frac{\sqrt{15}}{5}\right)}$

    (3)设点${\displaystyle A}$到平面${\displaystyle BDF}$的距离为${\displaystyle d}$，即${\displaystyle AB}$Z在平面${\displaystyle BDF}$的法向量${\displaystyle \overrightarrow{n}}$上的投影长

   

    ${\displaystyle \therefore }$点${\displaystyle A}$到平面${\displaystyle BDF}$的距离 为${\displaystyle \frac{2\sqrt{5}}{5}}$

    评注:本题主要考察线线角、面面角、点到平面的距离等基础知识，考察空间想象能力和推理以及利用向量解决问题的能力．

    利用向量法求空间距离的具体途径是：
    (1)欲求两点${\displaystyle P}$、${\displaystyle Q}$之间的距离，可转化为求向量${\displaystyle \overrightarrow{PQ}}$的模；
    (2)欲求点${\displaystyle P}$到直线${\displaystyle AB}$的距离，可在${\displaystyle AB}$上取一点${\displaystyle Q}$，令${\displaystyle \overrightarrow{AQ}=\lambda \overrightarrow{QB}}$，由${\displaystyle \overrightarrow{PQ}\perp \overrightarrow{AB}}$或求${\displaystyle \left|\overrightarrow{PQ}\right|}$的最小值，求得参数${\displaystyle \lambda }$值，以确定${\displaystyle Q}$的位置，则${\displaystyle \overrightarrow{PQ}}$的模${\displaystyle \left|\overrightarrow{PQ}\right|}$，即为点${\displaystyle P}$到直线${\displaystyle AB}$的距离；
    (3)欲求点${\displaystyle P}$到平面${\displaystyle \alpha }$的距离，可设${\displaystyle \overrightarrow{n}}$为平面${\displaystyle \alpha }$的法向量，${\displaystyle Q\in \alpha }$，则${\displaystyle P}$到平面${\displaystyle \alpha }$的距离${\displaystyle d=\frac{|\overrightarrow{PQ}·\overrightarrow{n}|}{\left|\overrightarrow{n}\right|}}$
    (4)欲求两条异面直线${\displaystyle l_1}$、${\displaystyle l_2}$之间的距离，可设与公垂线段${\displaystyle AB}$平行的向量为${\displaystyle \overrightarrow{n}}$，${\displaystyle C}$、${\displaystyle D}$分别是${\displaystyle l_1}$、${\displaystyle l_2}$上的任意两点，则${\displaystyle l_1}$、${\displaystyle l_2}$之间的距离${\displaystyle \left|AB\right|=\frac{|\overrightarrow{CD}·\overrightarrow{n}|}{\left|\overrightarrow{n}\right|}}$．

    (提示内容)

    解:由．

    评注:解决此题

    一、选择题
    1、正方体${\displaystyle ABCD-A_1{B}_1{C}_1{D}_1}$的棱长为${\displaystyle 1}$，以${\displaystyle D}$为原点，${\displaystyle \overrightarrow{DA}}$、${\displaystyle \overrightarrow{DC}}$、${\displaystyle \overrightarrow{D{D}_1}}$所在直线为${\displaystyle x}$、${\displaystyle y}$、${\displaystyle z}$轴建立直角坐标系${\displaystyle D}$${\displaystyle -xyz}$，且${\displaystyle MN}$是${\displaystyle A{B}_1}$与${\displaystyle B{C}_1}$的公垂线，${\displaystyle M}$在${\displaystyle A{B}_1}$上，${\displaystyle N}$在${\displaystyle B{C}_1}$上，则${\displaystyle \overrightarrow{MN}}$等于(   )
    A.${\displaystyle (1，\frac{2}{3}，\frac{2}{3})}$    B.${\displaystyle (\frac{2}{3}，1，\frac{2}{3})}$    C.${\displaystyle (-\frac{1}{3}，\frac{1}{3}，-\frac{1}{3})}$    D.${\displaystyle (\frac{1}{3}，-\frac{1}{3}，\frac{1}{3})}$
    2、在空间直角坐标系中，已知${\displaystyle A(2，3，1)}$，${\displaystyle B(4，1，2)}$，${\displaystyle C(6，3，7)}$，${\displaystyle D(-5，-4，8)}$，${\displaystyle DH\perp }$平面${\displaystyle ABC}$，垂足为${\displaystyle H}$，直线${\displaystyle DH}$交平面${\displaystyle xOy}$于点${\displaystyle M}$，则点${\displaystyle M}$的坐标是(   )
    A.${\displaystyle (4，7，0)}$     B.${\displaystyle (7，4，0)}$     C.${\displaystyle (-4，7，0)}$       D.${\displaystyle (-7，-4，0)}$
    3、直三棱柱${\displaystyle ABC-A_1{B}_1{C}_1}$中，${\displaystyle \angle ACB=90°}$，${\displaystyle \angle BAC=30°}$，${\displaystyle BC=1}$，${\displaystyle A{A}_1=\sqrt{6}}$，${\displaystyle M}$是${\displaystyle C{C}_1}$的中点，则异面直线${\displaystyle A{B}_1}$与${\displaystyle A_{1}M}$所成的角为(   )
    A.${\displaystyle 60°}$      B.${\displaystyle 45°}$      C.${\displaystyle 30°}$      D.${\displaystyle 90°}$
    4、如图，已知空间四边形每条边和对角线长都等于${\displaystyle a}$，点${\displaystyle E、F、G}$分别是${\displaystyle AB、AD、DC}$的中点，则${\displaystyle a^2}$是下列哪个向量的数量积(   )
    A.${\displaystyle 2\overrightarrow{BA}•\overrightarrow{AC}}$      B.${\displaystyle 2\overrightarrow{AD}•\overrightarrow{BD}}$      C.${\displaystyle 2\overrightarrow{FG}•\overrightarrow{CA}}$      D.${\displaystyle 2\overrightarrow{EF}•\overrightarrow{CB}}$

                                                      
    5、如图，${\displaystyle P}$是二面角${\displaystyle \alpha -l-\beta }$棱上的一点，分别在${\displaystyle \alpha }$、${\displaystyle \beta }$平面上引射线${\displaystyle PM、PN}$，如果${\displaystyle \angle BPM=\angle BPN=45°}$，${\displaystyle \angle MPN=}$${\displaystyle 60°}$，那么二面角${\displaystyle \alpha -AB-\beta }$的大小为(   )

    A.${\displaystyle 60°}$      B.${\displaystyle 70°}$      C.${\displaystyle 80°}$      D.${\displaystyle 90°}$

                                                    
    二、填空题
    6、如图，在棱长为${\displaystyle 1}$的正方体${\displaystyle ABCD-A_1{B}_1{C}_1{D}_1}$中，${\displaystyle O}$为正方形${\displaystyle ABCD}$的中心，${\displaystyle E、F}$分别为${\displaystyle AB、BC}$的中点，则异面直线${\displaystyle C_{1}O}$与${\displaystyle EF}$的距离为______

                                                       
    7、设${\displaystyle \overrightarrow{OA}=(1，1，-2)}$，${\displaystyle \overrightarrow{OB}=(3，2，8)}$，${\displaystyle \overrightarrow{OC}=(0，1，0)}$，则线段${\displaystyle AB}$的中点${\displaystyle P}$到点${\displaystyle C}$的距离为______
    三、解答题
    8、(05·福建20)如图，直二面角${\displaystyle D-AB-E}$中，四边形${\displaystyle ABCD}$是边长为${\displaystyle 2}$的正方形，${\displaystyle AE=EB}$，${\displaystyle F}$为${\displaystyle CE}$上的点，且${\displaystyle BF\perp }$平面${\displaystyle ACE}$

(1)求证：${\displaystyle AE\perp }$平面${\displaystyle BCE}$     (2)求二面角${\displaystyle B-AC-E}$的大小     (3)求点${\displaystyle D}$到平面${\displaystyle ACE}$的距离

    9、(05·重庆理20)如图，在三棱柱${\displaystyle ABC-A_1{B}_1{C}_1}$中，${\displaystyle AB\perp }$侧面${\displaystyle B{B}_1{C}_{1}C}$，${\displaystyle E}$为棱${\displaystyle C{C}_1}$上异于${\displaystyle C、C_1}$的一点，${\displaystyle EA\perp EB}$，已知${\displaystyle AB=\sqrt{2}}$，${\displaystyle B{B}_1=2}$，${\displaystyle BC=1}$，${\displaystyle \angle BC{C}_1=\frac{\pi }{3}}$，求：

(1)异面直线${\displaystyle AB}$与${\displaystyle E{B}_1}$的距离     (2)二面角${\displaystyle A-E{B}_1-A_1}$的平面角的正切值

    10、(探究创新题)如图所示的多面体是由底面为${\displaystyle ABCD}$的长方体被截面${\displaystyle AE{C}_{1}F}$所截而得到的，其中${\displaystyle AB=4}$，${\displaystyle BC=}$${\displaystyle 2}$，${\displaystyle C{C}_1=3}$，${\displaystyle BE=1}$

(1)求${\displaystyle BF}$与平面${\displaystyle BC{C}_{1}E}$所成的角     (2)求截面${\displaystyle AE{C}_{1}F}$与平面${\displaystyle ABCD}$所成二面角的大小     (3)在线段${\displaystyle C{C}_1}$上是否存在一点${\displaystyle C_2}$，使${\displaystyle C_2}$在平面${\displaystyle AE{C}_{1}F}$上的射影恰为${\displaystyle △E{C}_{1}F}$的重心