Cho hình hộp chữ nhật \(ABCD.{A_1}{B_1}{C_1}{D_1}\) có ba kích thước AB = a, AD = 2a, AA1 = 3a.

Vẽ \(AH \bot BD\), ta có:

\(\left\{ \begin{array}{l}
{A_1}A \bot \left( {do{\rm{ }}{A_1}A \bot \left( {ABCD} \right)} \right)\\
AH \bot BD
\end{array} \right. \Rightarrow \left( {{A_1}AH} \right) \bot BD\)

Vẽ \(AI \bot {A_1}H \Rightarrow AI \bot BD\)

\( \Rightarrow AI \bot \left( {{A_1}BD} \right) \Rightarrow d\left( {A,{A_1}BD} \right) = AI\)

Ta có \(AH = \frac{{AB.AD}}{{BD}} = \frac{{a.2a}}{{\sqrt {{a^2} + 4{a^2}} }} = \frac{{2a\sqrt 5 }}{5}\)

\(\frac{1}{{A{I^2}}} = \frac{1}{{{A_1}{A^2}}} + \frac{1}{{A{H^2}}} \Rightarrow AI = \frac{{6a}}{7}\)