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Câu hỏi: Cho hình lăng trụ đứng $A B C \cdot A^{\prime} B^{\prime} C^{\prime}$ có đáy $A B C$ là tam giác cân với $A B=A C=1$, góc $\widehat{B A C}=$ $120^{\circ}$, cạnh $A A^{\prime}=2$. Tính khoảng cách $d$ giữa hai đường thẳng $A B^{\prime}$ và $B C$.
A. $d=\dfrac{1}{\sqrt{17}}$.
B. $d=\dfrac{4}{\sqrt{17}}$.
C. $d=\dfrac{2}{\sqrt{17}}$.
D. $d=\dfrac{6}{\sqrt{17}}$.
Ta có: $\left\{\begin{array}{l}B C / / B^{\prime} C^{\prime} \\ B^{\prime} C^{\prime} \subset\left(A B^{\prime} C^{\prime}\right)\end{array} \Rightarrow B C / /\left(A B^{\prime} C^{\prime}\right) \Rightarrow \mathrm{d}\left(B C ; A B^{\prime}\right)=\mathrm{d}\left(B C ;\left(A B^{\prime} C^{\prime}\right)\right)=\mathrm{d}\left(B ;\left(A B^{\prime} C^{\prime}\right)\right)=\right.$
h. Mà $V_{B \prime^{\prime} A B C^{\prime}}=V_{L T}-V_{A \cdot A^{\prime} B^{\prime} C^{\prime}}-V_{C^{\prime} \cdot A B C} \Rightarrow V_{B \prime^{\prime} \cdot A B C^{\prime}}=V_{L T}-\dfrac{1}{3} V_{L T}-\dfrac{1}{3} V_{L T}=\dfrac{2}{3} V_{L T}$.
Ta có $V_{L T}=\dfrac{1}{2} A B \cdot A C \cdot \sin 120 \cdot A A^{\prime}=\dfrac{\sqrt{3}}{2} \Rightarrow V_{B \cdot A B C^{\prime}}=\dfrac{1}{3} \cdot \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}$.
$B C=\sqrt{A B^2+A C^2-2 A B A C \cdot \cos 120^{\circ}}=\sqrt{3}, A B^{\prime}=A C^{\prime}=\sqrt{A B^2+B B^{\prime 2}}=\sqrt{5}$.
$B^{\prime} C^{\prime}=\sqrt{3} \Rightarrow S_{A B \prime C^{\prime}}=\dfrac{\sqrt{51}}{4}$.
Ta có $V_{B . A B \prime C^{\prime}}=\dfrac{1}{3} \mathrm{~d}\left(B ;\left(A B^{\prime} C^{\prime}\right)\right) \cdot S_{A B \prime C^{\prime}} \Rightarrow \mathrm{d}\left(B ;\left(A B^{\prime} C^{\prime}\right)\right)=\dfrac{3 V_{B . A B \prime C^{\prime}}}{S_{A B C^{\prime}}}=\dfrac{2}{\sqrt{17}}$.
A. $d=\dfrac{1}{\sqrt{17}}$.
B. $d=\dfrac{4}{\sqrt{17}}$.
C. $d=\dfrac{2}{\sqrt{17}}$.
D. $d=\dfrac{6}{\sqrt{17}}$.
h. Mà $V_{B \prime^{\prime} A B C^{\prime}}=V_{L T}-V_{A \cdot A^{\prime} B^{\prime} C^{\prime}}-V_{C^{\prime} \cdot A B C} \Rightarrow V_{B \prime^{\prime} \cdot A B C^{\prime}}=V_{L T}-\dfrac{1}{3} V_{L T}-\dfrac{1}{3} V_{L T}=\dfrac{2}{3} V_{L T}$.
Ta có $V_{L T}=\dfrac{1}{2} A B \cdot A C \cdot \sin 120 \cdot A A^{\prime}=\dfrac{\sqrt{3}}{2} \Rightarrow V_{B \cdot A B C^{\prime}}=\dfrac{1}{3} \cdot \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}$.
$B C=\sqrt{A B^2+A C^2-2 A B A C \cdot \cos 120^{\circ}}=\sqrt{3}, A B^{\prime}=A C^{\prime}=\sqrt{A B^2+B B^{\prime 2}}=\sqrt{5}$.
$B^{\prime} C^{\prime}=\sqrt{3} \Rightarrow S_{A B \prime C^{\prime}}=\dfrac{\sqrt{51}}{4}$.
Ta có $V_{B . A B \prime C^{\prime}}=\dfrac{1}{3} \mathrm{~d}\left(B ;\left(A B^{\prime} C^{\prime}\right)\right) \cdot S_{A B \prime C^{\prime}} \Rightarrow \mathrm{d}\left(B ;\left(A B^{\prime} C^{\prime}\right)\right)=\dfrac{3 V_{B . A B \prime C^{\prime}}}{S_{A B C^{\prime}}}=\dfrac{2}{\sqrt{17}}$.
Đáp án C.