5. 在四棱柱ABCD−A1B1C1D1ABCD-A_{1}B_{1}C_{1}D_{1}ABCD−A1B1C1D1中,D1E→=kD1A→\overrightarrow{{D_1}E}=k\overrightarrow{{D_1}A}D1E=kD1A,D1F→=kD1B→\overrightarrow{{D_1}F}=k\overrightarrow{{D_1}B}D1F=kD1B,D1G→=kD1C→\overrightarrow{{D_1}G}=k\overrightarrow{{D_1}C}D1G=kD1C,D1H→=kD1D→\overrightarrow{{D_1}H}=k\overrightarrow{{D_1}D}D1H=kD1D.(1)当k=34k=\frac{3}{4}k=43时,试用AB→,AD→,AA1→\overrightarrow{AB},\overrightarrow{AD},\overrightarrow{A{A_1}}AB,AD,AA1表示AF→\overrightarrow{AF}AF;(2)证明:EEE,FFF,GGG,HHH四点共面;(3)判断直线D1C1D_{1}C_{1}D1C1能否是平面D1ABD_{1}ABD1AB和平面D1DCD_{1}DCD1DC的交线,并说明理由.{width="1.25in"