（5分）已知四棱柱$ABCD-A_{1}B_{1}C_{1}D_{1}$底面$ABCD$为平行四边形，$AA_{1}=3$，$BD=4$且$\overrightarrow{A{B}_{1}}\cdot \overrightarrow{BC}-\overrightarrow{A{D}_{1}}\cdot \overrightarrow{DC}=5$，求异面直线$AA_{1}$与$BD$的夹角 ____．

答案：$\arccos  \dfrac{5}{12}$．
分析：由题将$\overrightarrow{A{B}_{1}}\cdot \overrightarrow{BC}-\overrightarrow{A{D}_{1}}\cdot \overrightarrow{DC}=5$转化为$\overrightarrow{AA_{1}}\cdot \overrightarrow{BD}=5$即可求解．
解：如图，

因为${\overrightarrow{AB}}_{1}=\overrightarrow{AB}+\overrightarrow{A{A}_{1}},\overrightarrow{A{D}_{1}}=\overrightarrow{AD}+\overrightarrow{A{A}_{1}}$，又$\overrightarrow{A{B}_{1}}\cdot \overrightarrow{BC}-\overrightarrow{A{D}_{1}}\cdot \overrightarrow{DC}=5$，
$\therefore$$(\overrightarrow{AB}+\overrightarrow{A{A}_{1}})\cdot \overrightarrow{AD}-(\overrightarrow{AD}+\overrightarrow{A{A}_{1}})\cdot \overrightarrow{DC}=5$，
化简得$\overrightarrow{AA_{1}}\cdot \overrightarrow{BD}=5$，
$\therefore$$\overrightarrow{A{A}_{1}}\cdot \overrightarrow{BD}=3\times 4\times \cos  \theta =5$，
$\therefore$$\cos  \theta =\dfrac{5}{12}$．
异面直线$AA_{1}$与$BD$的夹角为$\arccos  \dfrac{5}{12}$．
点评：本题考查向量法求立体几何中的线线角，属于中档题．