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            • 1. 如图所示,在空间直角坐标系中\(BC=2\),原点\(O\)是\(BC\)的中点,点\(A\)的坐标是\(( \dfrac { \sqrt {3}}{2}, \dfrac {1}{2},0)\),点\(D\)在平面\(yOz\)上,且\(∠BDC=90^{\circ}\),\(∠DCB=30^{\circ}\),则向量\( \overrightarrow{AD}\)的坐标为\((\)  \()\)
              A.\((- \dfrac { \sqrt {3}}{2},- \dfrac {1}{2}, \dfrac { \sqrt {3}}{2})\)
              B.\((- \dfrac { \sqrt {3}}{2},-1, \dfrac { \sqrt {3}}{2})\)
              C.\((- \dfrac {1}{2},- \dfrac { \sqrt {3}}{2}, \dfrac { \sqrt {3}}{2})\)
              D.\(( \dfrac { \sqrt {3}}{2},1, \dfrac { \sqrt {3}}{2})\)
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