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Câu hỏi: Cho lăng trụ $ABC.A'B'C'$ có đáy là tam giác vuông tại $B$ với $AB=a.$ Hình chiếu vuông góc của đỉnh $A'$ lên mặt phẳng $\left( ABC \right)$ là điểm $H$ trên cạnh $AB$ sao cho $HA=2HB.$ Biết $A'H=\dfrac{a\sqrt{2}}{3}.$ Tính khoảng cách giữa hai đường thẳng $AA'$ và $BC$ theo $a.$
A. $\dfrac{a\sqrt{3}}{6}.$
B. $\dfrac{a\sqrt{3}}{3}.$
C. $\dfrac{a\sqrt{3}}{2}.$
D. $\dfrac{2a\sqrt{3}}{3}.$
Ta có $AA'//BB'\Rightarrow AA'//\left( BCC'B' \right)$ mà $BC\subset \left( BCC'B' \right)$
$\Rightarrow d\left( AA',BC \right)=d\left( AA',\left( BCC'B' \right) \right)=d\left( A,\left( BCC'B' \right) \right)=3d\left( H,\left( BCC'B' \right) \right)$
Ta có: $A'H\bot \left( ABC \right)\Rightarrow A'H\bot BC;BC\bot AB\Rightarrow BC\bot \left( ABB'A' \right)\Rightarrow \left( ABB'A' \right)\bot \left( BCC'B' \right)$
Kẻ $HK\bot BB'\Rightarrow HK\bot \left( BCC'B' \right)\Rightarrow d\left( H;\left( BCC'B' \right) \right)=HK$
Gọi $I=A'H\cap BB'.$
Ta có $\dfrac{IH}{IA'}=\dfrac{HB}{A'B'}=\dfrac{\dfrac{a}{3}}{a}=\dfrac{1}{3}\Rightarrow HI=\dfrac{1}{2}HA'=\dfrac{a\sqrt{2}}{6}$
$\Rightarrow HK=\dfrac{HB.HI}{\sqrt{H{{B}^{2}}+H{{I}^{2}}}}=\dfrac{\dfrac{a}{3}.\dfrac{a\sqrt{2}}{6}}{\sqrt{{{\left( \dfrac{a}{3} \right)}^{2}}+{{\left( \dfrac{a\sqrt{2}}{6} \right)}^{2}}}}=\dfrac{\sqrt{3}}{9}a$
$\Rightarrow d\left( H;\left( BCC'B' \right) \right)=\dfrac{\sqrt{3}}{9}a\Rightarrow d\left( AA';BC \right)=\dfrac{\sqrt{3}a}{3}$
A. $\dfrac{a\sqrt{3}}{6}.$
B. $\dfrac{a\sqrt{3}}{3}.$
C. $\dfrac{a\sqrt{3}}{2}.$
D. $\dfrac{2a\sqrt{3}}{3}.$
Ta có $AA'//BB'\Rightarrow AA'//\left( BCC'B' \right)$ mà $BC\subset \left( BCC'B' \right)$
$\Rightarrow d\left( AA',BC \right)=d\left( AA',\left( BCC'B' \right) \right)=d\left( A,\left( BCC'B' \right) \right)=3d\left( H,\left( BCC'B' \right) \right)$
Ta có: $A'H\bot \left( ABC \right)\Rightarrow A'H\bot BC;BC\bot AB\Rightarrow BC\bot \left( ABB'A' \right)\Rightarrow \left( ABB'A' \right)\bot \left( BCC'B' \right)$
Kẻ $HK\bot BB'\Rightarrow HK\bot \left( BCC'B' \right)\Rightarrow d\left( H;\left( BCC'B' \right) \right)=HK$
Gọi $I=A'H\cap BB'.$
Ta có $\dfrac{IH}{IA'}=\dfrac{HB}{A'B'}=\dfrac{\dfrac{a}{3}}{a}=\dfrac{1}{3}\Rightarrow HI=\dfrac{1}{2}HA'=\dfrac{a\sqrt{2}}{6}$
$\Rightarrow HK=\dfrac{HB.HI}{\sqrt{H{{B}^{2}}+H{{I}^{2}}}}=\dfrac{\dfrac{a}{3}.\dfrac{a\sqrt{2}}{6}}{\sqrt{{{\left( \dfrac{a}{3} \right)}^{2}}+{{\left( \dfrac{a\sqrt{2}}{6} \right)}^{2}}}}=\dfrac{\sqrt{3}}{9}a$
$\Rightarrow d\left( H;\left( BCC'B' \right) \right)=\dfrac{\sqrt{3}}{9}a\Rightarrow d\left( AA';BC \right)=\dfrac{\sqrt{3}a}{3}$
Đáp án B.