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

* Hướng dẫn giải

Đáp án C

Ta dễ dàng chứng minh được $AA\text{'}//\left(BCC\text{'}B\text{'}\right)$

$\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)$

Gọi G là trọng tâm của tam giác ABC. Suy ra $A\text{'}G\perp \left(ABC\right)$.

Ta có  $S_{\Delta ABC}=\frac{a^2\sqrt{3}}{4}$

 $\begin{array}{l}\Rightarrow V_{ABC.A\text{'}B\text{'}C\text{'}}=A\text{'}G.S_{\Delta ABC}\\ \iff A\text{'}G=\frac{V_{ABC.A\text{'}B\text{'}C\text{'}}}{S_{\Delta ABC}}\\ =\frac{a^3\sqrt{3}}{4}:\frac{a^2\sqrt{3}}{4}=a\end{array}$

Lại có

$\begin{array}{l}AM=\frac{a\sqrt{3}}{2}\\ \Rightarrow AG=\frac{2}{3}AM=\frac{a\sqrt{3}}{3}\\ \Rightarrow AA\text{'}=\sqrt{A\text{'}{G}^2+A{G}^2}=\frac{2a\sqrt{3}}{3}\end{array}$

 Ta luôn có $V_{A\text{'}.ABC}=\frac{1}{3}{V}_{ABC.A\text{'}B\text{'}C\text{'}}=\frac{1}{3}.\frac{a^3\sqrt{3}}{4}=\frac{a^3\sqrt{3}}{12}$.

Mà $V_{ABC.A\text{'}B\text{'}C\text{'}}=V_{A\text{'}.ABC}+V_{A\text{'}.BCC\text{'}B\text{'}}$ 

$\begin{array}{l}\Rightarrow V_{A\text{'}.BCC\text{'}B\text{'}}=V_{ABC.A\text{'}B\text{'}C\text{'}}-V_{A\text{'}.ABC}\\ =\frac{a^3\sqrt{3}}{4}-\frac{a^3\sqrt{3}}{12}=\frac{a^3\sqrt{3}}{6}\end{array}$.

Gọi M,M' lần lượt là trung điểm của BC và B'C'. Ta có $BC\perp AM,BC\perp A\text{'}G$ $\Rightarrow BC\perp \left(AMM\text{'}A\text{'}\right)\Rightarrow BC\perp MM\text{'}$. Mà $MM\text{'}//BB\text{'}$ nên $BC\perp BB\text{'}\Rightarrow BCC\text{'}B\text{'}$ là hình chữ nhật

 $$\begin{gathered} \Rightarrow S_{BCC\text{'}B\text{'}}=BB\text{'}.BC \\ =\frac{2a\sqrt{3}}{3}.a=\frac{2a^2\sqrt{3}}{3} \end{gathered}$$ .

 Từ

$\begin{array}{l}{V}_{A\text{'}.BCC\text{'}B\text{'}}=\frac{1}{3}d\left(A\text{'};\left(BCC\text{'}B\text{'}\right)\right).S_{BCC\text{'}B\text{'}}\\ \iff d\left(A\text{'};\left(BCC\text{'}B\text{'}\right)\right)=\frac{3V_{A\text{'}.BCC\text{'}B\text{'}}}{S_{BCC\text{'}B\text{'}}}\end{array}$ 

$\begin{array}{l}\Rightarrow d\left(A\text{'};\left(BCC\text{'}B\text{'}\right)\right)\\ =\frac{a^3\sqrt{3}}{2}:\frac{2a^2\sqrt{3}}{3}=\frac{3a}{4}\end{array}$. Vậy $d\left(AA\text{'};BC\right)=\frac{3a}{4}$.