如图,在直三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(AC=AA_{1}=2\),\(D\)为棱\(CC_{1}\)的中点,\(G\)为棱\(AA_{1}\)上一点,\(AB_{1}∩A_{1}B=O\).
\((1)\)确定\(G\)的位置,使得平面\(C_{1}OG/\!/\)平面\(ABD\),并说明理由;
\((2)\)设二面角\(D-AB-C\)的正切值为\( \dfrac { \sqrt {2}}{2}\),\(AC⊥BC\),\(E\)为线段\(A_{1}B\)上一点,且\(CE\)与平面\(ABD\)所成角的正弦值为\( \dfrac {2 \sqrt {2}}{3}\),求线段\(BE\)的长.