Cho hình lăng trụ tam giác đều ABC.A'B'C' có cạnh đáy bằng 2a, góc giữa

Đặt AA' = x > 0 ta có:

$\overrightarrow{AB\text{'}}.\overrightarrow{BC\text{'}}=\left(\overrightarrow{BB\text{'}}-\overrightarrow{BA}\right)\left(\overrightarrow{BC}+\overrightarrow{BB\text{'}}\right)$

     $=\overrightarrow{BB\text{'}}.\overrightarrow{BC}+{\overrightarrow{BB\text{'}}}^2-\overrightarrow{BA}.\overrightarrow{BC}-\overrightarrow{BA}.\overrightarrow{BB\text{'}}$

     $=\overrightarrow{BB\text{'}}.\left(\overrightarrow{BC}-\overrightarrow{BA}\right)+{\overrightarrow{BB\text{'}}}^2-\overrightarrow{BA}.\overrightarrow{BC}$

     $=\overrightarrow{BB\text{'}}.\overrightarrow{AC}+{\overrightarrow{BB\text{'}}}^2-\overrightarrow{BA}.\overrightarrow{BC}$

     $={\overrightarrow{BB\text{'}}}^2-\overrightarrow{BA}.\overrightarrow{BC}$ (do $BB\text{'}\perp AC$)

     $=x^2-BA.BC.\cos {60}^0+BB{\text{'}}^2$

     $=x^2-2a^2$

Ta có: $AB\text{'}=BC\text{'}=\sqrt{x^2+4a^2}$ (định lí Pytago)

$\Rightarrow \overrightarrow{AB\text{'}}.\overrightarrow{BC\text{'}}=AB\text{'}.BC\text{'}.\cos \left(\overrightarrow{AB\text{'}};\overrightarrow{BC\text{'}}\right)=\frac{1}{2}\left(x^2+4a^2\right)$

$\Rightarrow \frac{1}{2}\left(x^2+4a^2\right)=x^2-2a^2$

$\iff x^2+4a^2=2x^2-4a^2$

$\iff x^2=8a^2\iff x=2\sqrt{2}a$

Vậy thể tích khối lăng trụ là $V_{ABC.A\text{'}B\text{'}C\text{'}}=AA\text{'}.S_{\Delta ABC}=2\sqrt{2}a.\frac{{\left(2a\right)}^2\sqrt{3}}{4}=2\sqrt{6}{a}^3.$

Chọn D.