Cho hình lăng trụ tam giác đều ABC.A'B'C' có tất cả các cạnh bằng a (căn3 a^3)/24

* Hướng dẫn giải

Đáp án B

Tọa độ hóa với $O\equiv N,Ox\equiv NB\text{'},Oy\equiv NA\text{'},Oz\equiv NK$ và chuẩn hóa vớí $a=2$.

Ta có

$\begin{array}{l}A\text{'}\left(0;\sqrt{3};0\right),\left\{\begin{array}{l}A\left(0;\sqrt{3};2\right)\\ B\left(1;0;2\right)\end{array}\right.\\ \Rightarrow M\left(\frac{1}{2};\frac{\sqrt{3}}{2};2\right)\end{array}$

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}\overrightarrow{NA\text{'}}=\left(0;\sqrt{3};0\right)\\ \overrightarrow{NM}=\left(\frac{1}{2};\frac{\sqrt{3}}{2};2\right)\end{array}\right.\Rightarrow \overrightarrow{n_{\left(A\text{'}MN\right)}}\\ =\left[\overrightarrow{NA\text{'}}.\overrightarrow{NM}\right]=\left(2\sqrt{3};0;-\frac{\sqrt{3}}{2}\right)\end{array}$

$\Rightarrow \left(A\text{'}MN\right):4\text{x}-z=0$ 

Lại có

$$\begin{gathered} B\left(1;0;2\right),K\left(0;0;2\right) \\ \Rightarrow \overrightarrow{KB}=\left(1;0;0\right)\Rightarrow BC:\left\{\begin{array}{l}x=t\\ y=0\\ z=2\end{array}\right. \end{gathered}$$

Mà

$P=BC\cap \left(A;MN\right)\Rightarrow P\left(\frac{1}{2};0;2\right)$ 

$\begin{array}{l}{V}_{MBP.A\text{'}B\text{'}N\text{'}}=V_{M.A\text{'}B\text{'}N}+V_{M.BPNB}=V_{A.A\text{'}B\text{'}N}+\frac{1}{2}{V}_{A.BPNB\text{'}}\\ V_{A.A\text{'}B\text{'}N}=\frac{1}{2}{V}_{A.A\text{'}B\text{'}C\text{'}}=\frac{1}{6}{V}_{ABC.A\text{'}B\text{'}C\text{'}}\\ S_{BPNB\text{'}}=\frac{1}{2}{S}_{BCC\text{'}B\text{'}}-S_{NPK}=\frac{1}{2}{S}_{BCC\text{'}B\text{'}}-\frac{1}{8}{S}_{BCC\text{'}B\text{'}}\\ =\frac{3}{8}{S}_{BCC\text{'}B\text{'}}=\frac{3}{4}{S}_{BCB\text{'}}\Rightarrow V_{A.BPNB\text{'}}=\frac{3}{4}{V}_{A.BCB\text{'}}=\frac{1}{4}{V}_{ABC.A\text{'}B\text{'}C\text{'}}\\ \Rightarrow V_{MBP.A\text{'}B\text{'}N}=\frac{7}{24}{V}_{ABC.A\text{'}B\text{'}C\text{'}}=\frac{7}{24}A\text{'}A.S_{ABC}=\frac{7}{24}a.\frac{a^2\sqrt{3}}{4}=\frac{7{\text{a}}^3\sqrt{3}}{96}\end{array}$