Question

15．四棱柱ABCD-A₁B₁C₁D₁各棱長(zhǎng)均為1，∠A₁AB=∠A₁AD=∠BAD=60°，則點(diǎn)B與點(diǎn)D₁兩點(diǎn)間的距離為$\sqrt{2}$．

Answer 1

分析 由已知得$\overrightarrow{B{D}_{1}}$=$\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{D{D}_{1}}$，由此能求出點(diǎn)B與點(diǎn)D₁兩點(diǎn)間的距離．

解答 解：∵四棱柱ABCD-A₁B₁C₁D₁各棱長(zhǎng)均為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，
∴|$\overrightarrow{B{D}_{1}}$|=$\sqrt{2}$．
∴點(diǎn)B與點(diǎn)D₁兩點(diǎn)間的距離為$\sqrt{2}$．
故答案為：$\sqrt{2}$．

點(diǎn)評(píng) 本題考查兩點(diǎn)間距離的求法，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意向量法的合理運(yùn)用．