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Câu hỏi: Trên hình chóp S.ABCD có đáy ABCD là hình vuông tâm O cạnh 2a. Hình chiếu của S trên mặt đáy là trung điểm H của OA; góc giữa hai mặt phẳng $\left( SCD \right)$ và $\left( ABCD \right)$ bằng $45{}^\circ $. Tính khoảng cách giữa hai đường thẳng AB và SC.
A. $a\sqrt{2}.$
B. $\dfrac{3a\sqrt{2}}{2}.$
C. $\dfrac{3a\sqrt{2}}{4}.$
D. $a\sqrt{6}.$
Trong $\left( ABCD \right)$ kẻ $HM\bot CD\left( M\in CD \right)$.
Ta có: $\left\{ \begin{aligned}
& CD\bot SH\left( SH\bot \left( ABCD \right) \right) \\
& CD\bot HM \\
\end{aligned} \right.\Rightarrow CD\bot \left( SHM \right)\Rightarrow CD\bot SM$
$\begin{aligned}
& \left\{ \begin{aligned}
& \left( SCD \right)\cap \left( ABCD \right)=CD \\
& \left( SCD \right)\supset SM\bot CD \\
& \left( ABCD \right)\supset HM\bot CD \\
\end{aligned} \right. \\
& \Rightarrow \widehat{\left( SCD \right),\left( ABCD \right)}=\widehat{\left( SM,HM \right)}=\widehat{SMH}=45{}^\circ \\
\end{aligned}$
Trong $\left( SHM \right)$ kẻ $HK\bot SM\left( K\in SM \right)$ ta có:
$\left\{ \begin{aligned}
& HK\bot SM \\
& HK\bot CD \\
\end{aligned} \right.\Rightarrow HK\bot \left( SCD \right)$.
Ta có: $AB//CD\Rightarrow d\left( AB;SC \right)=d\left( AB;\left( SCD \right) \right)=d\left( A;\left( SCD \right) \right)$.
$AH\cap \left( SCD \right)=C\Rightarrow \dfrac{d\left( A;\left( SCD \right) \right)}{d\left( H;\left( SCD \right) \right)}=\dfrac{AC}{HC}=\dfrac{4}{3}\Rightarrow d\left( A;\left( SCD \right) \right)=\dfrac{4}{3}d\left( H;\left( SCD \right) \right)=\dfrac{4}{3}HK$.
Áp dụng định lí Ta-lét ta có: $\dfrac{HM}{AD}=\dfrac{HC}{AC}=\dfrac{3}{4}\Rightarrow HM=\dfrac{3}{4}AD=\dfrac{3a}{2}$.
Xét tam giác vuông HMK: $HK=HM.\sin 45{}^\circ =\dfrac{3a}{2}.\dfrac{\sqrt{2}}{2}=\dfrac{3a\sqrt{2}}{4}$.
Vậy $d\left( AB;SC \right)=\dfrac{4}{3}.\dfrac{3a\sqrt{2}}{4}=a\sqrt{2}.$
A. $a\sqrt{2}.$
B. $\dfrac{3a\sqrt{2}}{2}.$
C. $\dfrac{3a\sqrt{2}}{4}.$
D. $a\sqrt{6}.$
Trong $\left( ABCD \right)$ kẻ $HM\bot CD\left( M\in CD \right)$.
Ta có: $\left\{ \begin{aligned}
& CD\bot SH\left( SH\bot \left( ABCD \right) \right) \\
& CD\bot HM \\
\end{aligned} \right.\Rightarrow CD\bot \left( SHM \right)\Rightarrow CD\bot SM$
$\begin{aligned}
& \left\{ \begin{aligned}
& \left( SCD \right)\cap \left( ABCD \right)=CD \\
& \left( SCD \right)\supset SM\bot CD \\
& \left( ABCD \right)\supset HM\bot CD \\
\end{aligned} \right. \\
& \Rightarrow \widehat{\left( SCD \right),\left( ABCD \right)}=\widehat{\left( SM,HM \right)}=\widehat{SMH}=45{}^\circ \\
\end{aligned}$
Trong $\left( SHM \right)$ kẻ $HK\bot SM\left( K\in SM \right)$ ta có:
$\left\{ \begin{aligned}
& HK\bot SM \\
& HK\bot CD \\
\end{aligned} \right.\Rightarrow HK\bot \left( SCD \right)$.
Ta có: $AB//CD\Rightarrow d\left( AB;SC \right)=d\left( AB;\left( SCD \right) \right)=d\left( A;\left( SCD \right) \right)$.
$AH\cap \left( SCD \right)=C\Rightarrow \dfrac{d\left( A;\left( SCD \right) \right)}{d\left( H;\left( SCD \right) \right)}=\dfrac{AC}{HC}=\dfrac{4}{3}\Rightarrow d\left( A;\left( SCD \right) \right)=\dfrac{4}{3}d\left( H;\left( SCD \right) \right)=\dfrac{4}{3}HK$.
Áp dụng định lí Ta-lét ta có: $\dfrac{HM}{AD}=\dfrac{HC}{AC}=\dfrac{3}{4}\Rightarrow HM=\dfrac{3}{4}AD=\dfrac{3a}{2}$.
Xét tam giác vuông HMK: $HK=HM.\sin 45{}^\circ =\dfrac{3a}{2}.\dfrac{\sqrt{2}}{2}=\dfrac{3a\sqrt{2}}{4}$.
Vậy $d\left( AB;SC \right)=\dfrac{4}{3}.\dfrac{3a\sqrt{2}}{4}=a\sqrt{2}.$
Đáp án A.