Question

15．四棱柱ABCD-A₁B₁C₁D₁各棱长均为1，∠A₁AB=∠A₁AD=∠BAD=60°，则点B与点D₁两点间的距离为$\sqrt{2}$．

Answer 1

分析 由已知得$\overrightarrow{B{D}_{1}}$=$\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{D{D}_{1}}$，由此能求出点B与点D₁两点间的距离．

解答 解：∵四棱柱ABCD-A₁B₁C₁D₁各棱长均为1，∠A₁AB=∠A₁AD=∠BAD=60°，
∴$\overrightarrow{B{D}_{1}}$=$\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{D{D}_{1}}$，
∴${\overrightarrow{B{D}_{1}}}^{2}$=（$\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{D{D}_{1}}$）²=${\overrightarrow{BA}}^{2}+{\overrightarrow{AD}}^{2}+{\overrightarrow{D{D}_{1}}}^{2}$+2$\overrightarrow{BA}•\overrightarrow{AD}$+2$\overrightarrow{BA}•\overrightarrow{D{D}_{1}}$+2$\overrightarrow{AD}•\overrightarrow{D{D}_{1}}$
=1+1+1+2×1×1×cos120°+2×1×1×cos120°+2×1×1×cos60°
=2，
∴|$\overrightarrow{B{D}_{1}}$|=$\sqrt{2}$．
∴点B与点D₁两点间的距离为$\sqrt{2}$．
故答案为：$\sqrt{2}$．

点评 本题考查两点间距离的求法，是基础题，解题时要认真审题，注意向量法的合理运用．