Cho hình lập phương A(0;0;0),B(1;0;0),D(0;1;0),A′(0;0;1). Gọi M,N lần lượt là trung điểm của AB,CD. Khoảng cách giữa MN và A′C là:

Gọi C(x;y;z) ta có

\(\overrightarrow {AB} = \overrightarrow {DC} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{1 - 0 = x - 0}\\{0 - 0 = y - 1}\\{0 - 0 = z - 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x = 1}\\{y = 1}\\{z = 0}\end{array}} \right. \Rightarrow C(1;1;0)\)

Lại có

\[\begin{array}{*{20}{l}}{M\left( {\frac{1}{2};0;0} \right),N\left( {\frac{1}{2};1;0} \right) \Rightarrow \overrightarrow {MN} = \left( {0;1;0} \right),\overrightarrow {A'C} = \left( {1;1; - 1} \right),\overrightarrow {MA'} = \left( { - \frac{1}{2};0;1} \right)}\\{ \Rightarrow \left[ {\overrightarrow {MN} ,\overrightarrow {A'C} } \right] = \left( {\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}1\\1\end{array}}&{\begin{array}{*{20}{l}}0\\{ - 1}\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}0\\{ - 1}\end{array}}&{\begin{array}{*{20}{l}}0\\1\end{array}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}0\\1\end{array}}&{\begin{array}{*{20}{l}}1\\1\end{array}}\end{array}} \right|} \right) = \left( { - 1;0; - 1} \right)}\end{array}\]

Vậy

\[d\left( {MN,A'C} \right) = \frac{{\left| {\left[ {\overrightarrow {MN} ,\overrightarrow {A'C} } \right].\overrightarrow {MA'} } \right|}}{{\left| {\left[ {\overrightarrow {MN} ,\overrightarrow {A'C} } \right]} \right|}} = \frac{{\left| {\left( { - 1} \right).\left( { - \frac{1}{2}} \right) + 0.0 + \left( { - 1} \right).1} \right|}}{{\sqrt {{{\left( { - 1} \right)}^2} + {0^2} + {{\left( { - 1} \right)}^2}} }} = \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\]