Question

1．如圖，圓臺(tái)的高為4，上、下底面半徑分別為3、5，M、N分別在上、下底面圓周上，且＜$\overrightarrow{{O}_{2}M}$，$\overrightarrow{{O}_{1}N}$＞=120°，則|$\overrightarrow{MN}$|等于（　�。�

• A． $\sqrt{65}$
• B． 5$\sqrt{2}$
• C． $\sqrt{35}$
• D． 5

Answer 1

分析 用$\overrightarrow{M{O}_{2}}$，$\overrightarrow{{O}_{2}{O}_{1}}$，$\overrightarrow{{O}_{1}N}$表示出$\overrightarrow{MN}$，計(jì)算${\overrightarrow{MN}}^{2}$再開(kāi)方即可得出答案．

解答 解：∵O₂M⊥O₁O₂，O₁N⊥O₁O₂，
∴${\overrightarrow{M{O}_{2}}}_{\;}$•$\overrightarrow{{O}_{2}{O}_{1}}$=0，$\overrightarrow{{O}_{2}{O}_{1}}•\overrightarrow{{O}_{1}N}$=0，
又${\overrightarrow{M{O}_{2}}}_{\;}$$•\overrightarrow{{O}_{1}N}$=3×5×cos60°=$\frac{15}{2}$．
∵$\overrightarrow{MN}$=$\overrightarrow{M{O}_{2}}+\overrightarrow{{O}_{2}{O}_{1}}$+$\overrightarrow{{O}_{1}N}$，
∴$\overrightarrow{MN}$²=（$\overrightarrow{M{O}_{2}}+\overrightarrow{{O}_{2}{O}_{1}}$+$\overrightarrow{{O}_{1}N}$）²
=$\overrightarrow{M{O}_{2}}$²+$\overrightarrow{{O}_{2}{O}_{1}}$²+$\overrightarrow{{O}_{1}N}$²+2${\overrightarrow{M{O}_{2}}}_{\;}$•$\overrightarrow{{O}_{2}{O}_{1}}$+2$\overrightarrow{{O}_{2}{O}_{1}}•\overrightarrow{{O}_{1}N}$+2${\overrightarrow{M{O}_{2}}}_{\;}$$•\overrightarrow{{O}_{1}N}$=9+16+25+15=65，
∴|$\overrightarrow{MN}$|=$\sqrt{65}$．
故選A．

點(diǎn)評(píng) 本題考查了空間向量的數(shù)量積運(yùn)算，屬于中檔題．