Cho lăng trụ ABC.A'B'C' có đáy là tam giác

* Hướng dẫn giải

Chọn B

Ta có $AA\text{'}//BB\text{'}\Rightarrow AA\text{'}//\left(BCC\text{'}B\text{'}\right)$ mà $BC\subset \left(BCC\text{'}B\text{'}\right)$

$\Rightarrow d\left(AA\text{'},BC\right)=d\left(AA\text{'},\left(BCC\text{'}B\text{'}\right)\right)=d\left(A,\left(BCC\text{'}B\text{'}\right)\right)=3d\left(H,\left(BCC\text{'}B\text{'}\right)\right)$

Ta có: $A\text{'}H\perp \left(ABC\right)\Rightarrow A\text{'}H\perp BC;BC\perp AB\Rightarrow BC\perp \left(ABB\text{'}A\text{'}\right)\Rightarrow \left(ABB\text{'}A\text{'}\right)\perp \left(BCC\text{'}B\text{'}\right)$

Kẻ $HK\perp BB\text{'}\Rightarrow HK\perp \left(BCC\text{'}B\text{'}\right)\Rightarrow d\left(H;\left(BCC\text{'}B\text{'}\right)\right)=HK$

Gọi $I=A\text{'}H\cap BB\text{'}.$

Ta có $\frac{IH}{IA\text{'}}=\frac{HB}{A\text{'}B\text{'}}=\frac{\frac{a}{3}}{a}=\frac{1}{3}\Rightarrow HI=\frac{1}{2}HA\text{'}=\frac{a\sqrt{2}}{6}$

$\Rightarrow HK=\frac{HB.HI}{\sqrt{H{B}^2+H{I}^2}}=\frac{\frac{a}{3}.\frac{a\sqrt{2}}{6}}{\sqrt{{\left(\frac{a}{3}\right)}^2+{\left(\frac{a\sqrt{2}}{6}\right)}^2}}=\frac{\sqrt{3}}{9}a$

$\Rightarrow d\left(H;\left(BCC\text{'}B\text{'}\right)\right)=\frac{\sqrt{3}}{9}a\Rightarrow d\left(AA\text{'};BC\right)=\frac{\sqrt{3}a}{3}$