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

* Hướng dẫn giải

Đáp án D

Ta có $B^{\text{'}}G\perp \left(ABC\right)\Rightarrow \left(\widehat{B{B}^{\text{'}},\left(ABC\right)}\right)=\left(\widehat{B{B}^{\text{'}},BG}\right)=\widehat{B^{\text{'}}BG}$.

Theo giả thiết ta có $\widehat{B^{\text{'}}BG}=60°$.

Gọi M là trung điểm BC. Kẻ $AH\perp B^{\text{'}}M$, $H\in B^{\text{'}}M$.

$\Rightarrow AM\perp BC$. Mà

$\left\{\begin{array}{l}BC\perp AM\\ BC\perp B^{\text{'}}G\end{array}\right.\Rightarrow BC\perp \left(A{B}^{\text{'}}M\right)\Rightarrow BC\perp AH$

+) $\left\{\begin{array}{l}AH\perp B^{\text{'}}M\\ AH\perp BC\end{array}\right.$

$\Rightarrow AH\perp \left(BC{C}^{\text{'}}{B}^{\text{'}}\right)\Rightarrow d\left(A,\left(BC{C}^{\text{'}}{B}^{\text{'}}\right)\right)=AH$

+) $\Delta ABC$ đều cạnh a nên ta có $AM=\frac{a\sqrt{3}}{2}$, $BG=\frac{a\sqrt{3}}{3}$, $GM=\frac{a\sqrt{3}}{6}$ 

+)  $B^{\text{'}}G=GB.\tan 60°=\frac{a\sqrt{3}}{3}.\sqrt{3}=a$

+)  $B^{\text{'}}M=\sqrt{B^{\text{'}}{G}^2+G{M}^2}=\sqrt{a^2+{\left(\frac{a\sqrt{3}}{6}\right)}^2}=\frac{a\sqrt{39}}{6}$

+)  $B^{\text{'}}G.AM=AH.B^{\text{'}}M\Rightarrow AH=\frac{B^{\text{'}}G.AM}{B^{\text{'}}M}=\frac{a.\frac{a\sqrt{3}}{2}}{\frac{a\sqrt{39}}{6}}=\frac{3a}{\sqrt{13}}$

Vậy $d\left(A,\left(BC{C}^{\text{'}}{B}^{\text{'}}\right)\right)=AH=\frac{3a}{\sqrt{13}}$.