数学试题讲解

已知平行六面体

A B C D - A_{1} B_{1} C_{1} D_{1}

中,

A A_{1} = 2, B D = 3, \vec{A D_{1}} \cdot \vec{D C} - \vec{A B_{1}} \cdot \vec{B C} = 4

, 则

\cos ⟨ \vec{A A_{1}}, \vec{B D} ⟩ = ()

\frac{2}{3}

- \frac{2}{3}

\frac{3}{4}

- \frac{3}{4}