已知 $E$ 为棱长为 $a$ 的正方体 $ABCD - {A_1}{B_1}{C_1}{D_1}$ 的棱 $AB$ 的中点，求点 $B$ 到平面 ${A_1}EC$ 的距离．
【难度】
【出处】
2004年复旦大学保送生招生测试
【标注】
【答案】
$\dfrac{{\sqrt 6 }}{6}a$
【解析】
用等体积法．
因为$${A_1}C = \sqrt 3 a,{A_1}E = EC = \dfrac{{\sqrt 5 }}{2}a,$$所以$${S_{\triangle {A_1}EC}} = \sqrt {{{\left( {\dfrac{{\sqrt 5 }}{2}a} \right)}^2} - {{\left( {\dfrac{{\sqrt 3 }}{2}a} \right)}^2}} \cdot \dfrac{{\sqrt 3 }}{2}a = \dfrac{{\sqrt 6 }}{4}{a^2}.$$故\[\begin{split}d\left( {B,{A_1}EC} \right) &= \dfrac{{{S_{\triangle EBC}}}}{{{S_{\triangle {A_1}EC}}}} \cdot d\left( {{A_1} , EBC} \right) \\&= \dfrac{{\dfrac{1}{4}{a^2}}}{{\dfrac{{\sqrt 6 }}{4}{a^2}}} \cdot a = \dfrac{{\sqrt 6 }}{6}a.\end{split}\]
因为$${A_1}C = \sqrt 3 a,{A_1}E = EC = \dfrac{{\sqrt 5 }}{2}a,$$所以$${S_{\triangle {A_1}EC}} = \sqrt {{{\left( {\dfrac{{\sqrt 5 }}{2}a} \right)}^2} - {{\left( {\dfrac{{\sqrt 3 }}{2}a} \right)}^2}} \cdot \dfrac{{\sqrt 3 }}{2}a = \dfrac{{\sqrt 6 }}{4}{a^2}.$$故\[\begin{split}d\left( {B,{A_1}EC} \right) &= \dfrac{{{S_{\triangle EBC}}}}{{{S_{\triangle {A_1}EC}}}} \cdot d\left( {{A_1} , EBC} \right) \\&= \dfrac{{\dfrac{1}{4}{a^2}}}{{\dfrac{{\sqrt 6 }}{4}{a^2}}} \cdot a = \dfrac{{\sqrt 6 }}{6}a.\end{split}\]
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