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Câu hỏi: Cho hình hộp chữ nhật $ABCD.{A}'{B}'{C}'{D}'$ có $AB=a, AD=2a,A{A}'=2a$. Khoảng cách từ điểm $A$ đến mặt phẳng $\left( {A}'BD \right)$ bằng
A. $\dfrac{a\sqrt{3}}{2}$.
B. $a$.
C. $\dfrac{a\sqrt{6}}{3}$.
D. $\dfrac{a\sqrt{2}}{2}$.
Trong $\left( ABCD \right)$, kẻ $AK\bot BD$.
Trong $\left( AO{A}' \right)$, kẻ $AH\bot {A}'K$.
Ta có: $\left\{ \begin{aligned}
& AH\bot {A}'K \\
& AH\bot BD\left( BD\bot \left( {A}'AC \right) \right) \\
\end{aligned} \right.\Rightarrow AH\bot \left( {A}'BD \right)$
$\dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{K}^{2}}}+\dfrac{1}{A{{{{A}'}}^{2}}}=\dfrac{1}{A{{{{A}'}}^{2}}}+\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{D}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{\left( 2a \right)}^{2}}}+\dfrac{1}{{{\left( 2a \right)}^{2}}}\Rightarrow AH=\dfrac{a\sqrt{6}}{3}.$
A. $\dfrac{a\sqrt{3}}{2}$.
B. $a$.
C. $\dfrac{a\sqrt{6}}{3}$.
D. $\dfrac{a\sqrt{2}}{2}$.
Trong $\left( AO{A}' \right)$, kẻ $AH\bot {A}'K$.
Ta có: $\left\{ \begin{aligned}
& AH\bot {A}'K \\
& AH\bot BD\left( BD\bot \left( {A}'AC \right) \right) \\
\end{aligned} \right.\Rightarrow AH\bot \left( {A}'BD \right)$
$\dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{K}^{2}}}+\dfrac{1}{A{{{{A}'}}^{2}}}=\dfrac{1}{A{{{{A}'}}^{2}}}+\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{D}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{\left( 2a \right)}^{2}}}+\dfrac{1}{{{\left( 2a \right)}^{2}}}\Rightarrow AH=\dfrac{a\sqrt{6}}{3}.$
Đáp án C.