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Câu 2 . Cho hình lăng trụ tam giác đều A B C ⋅ A ′ B ′ C ′ A B C \cdot A^{\prime} B^{\prime} C^{\prime} A BC ⋅ A ′ B ′ C ′ a a a M M M A A ′ A A^{\prime} A A ′ B M ⊥ A C ′ B M \perp A C^{\prime} BM ⊥ A C ′ C C C B M C ′ B M C^{\prime} BM C ′ a = 2 2 a=2 \sqrt{2} a = 2 2
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Đáp án: 2.
Giả sử cạnh bên A A ′ = B B ′ = C C ′ = h A A^{\prime}=B B^{\prime}=C C^{\prime}=h A A ′ = B B ′ = C C ′ = h B M → = 1 2 ( B A → + B A ′ → ) = 1 2 ( B A → + B A → + B B ′ → ) = B A → + 1 2 B B ′ → \overrightarrow{B M}=\frac{1}{2}\left(\overrightarrow{B A}+\overrightarrow{B A^{\prime}}\right)=\frac{1}{2}\left(\overrightarrow{B A}+\overrightarrow{B A}+\overrightarrow{B B^{\prime}}\right)=\overrightarrow{B A}+\frac{1}{2} \overrightarrow{B B^{\prime}} BM = 2 1 ( B A + B A ′ ) = 2 1 ( B A + B A + B B ′ ) = B A + 2 1 B B ′ A C ′ → = A A ′ → + A ′ C ′ → \overrightarrow{A C^{\prime}}=\overrightarrow{A A^{\prime}}+\overrightarrow{A^{\prime} C^{\prime}} A C ′ = A A ′ + A ′ C ′ B M → ⋅ A C ′ → = ( B A → + 1 2 B B ′ → ) ( A A ′ → + A ′ C ′ → ) \overrightarrow{B M} \cdot \overrightarrow{A C^{\prime}}=\left(\overrightarrow{B A}+\frac{1}{2} \overrightarrow{B B^{\prime}}\right)\left(\overrightarrow{A A^{\prime}}+\overrightarrow{A^{\prime} C^{\prime}}\right) BM ⋅ A C ′ = ( B A + 2 1 B B ′ ) ( A A ′ + A ′ C ′ ) = B A → ⋅ A A ′ → + B A → ⋅ A ′ C ′ → + 1 2 B B ′ → ⋅ A A ′ → + 1 2 B B ′ → ⋅ A ′ C ′ → =\overrightarrow{B A} \cdot \overrightarrow{A A^{\prime}}+\overrightarrow{B A} \cdot \overrightarrow{A^{\prime} C^{\prime}}+\frac{1}{2} \overrightarrow{B B^{\prime}} \cdot \overrightarrow{A A^{\prime}}+\frac{1}{2} \overrightarrow{B B^{\prime}} \cdot \overrightarrow{A^{\prime} C^{\prime}} = B A ⋅ A A ′ + B A ⋅ A ′ C ′ + 2 1 B B ′ ⋅ A A ′ + 2 1 B B ′ ⋅ A ′ C ′ = 0 + B A ⋅ A C ⋅ cos 120 ∘ + 1 2 B B ′ ⋅ A A ′ ⋅ cos 0 ∘ + 0 =0+B A \cdot A C \cdot \cos 120^{\circ}+\frac{1}{2} B B^{\prime} \cdot A A^{\prime} \cdot \cos 0^{\circ}+0 = 0 + B A ⋅ A C ⋅ cos 12 0 ∘ + 2 1 B B ′ ⋅ A A ′ ⋅ cos 0 ∘ + 0 = a ⋅ a ⋅ ( − 1 2 ) + 1 2 h ⋅ h = − 1 2 a 2 + 1 2 h 2 =a \cdot a \cdot\left(-\frac{1}{2}\right)+\frac{1}{2} h \cdot h=-\frac{1}{2} a^{2}+\frac{1}{2} h^{2} = a ⋅ a ⋅ ( − 2 1 ) + 2 1 h ⋅ h = − 2 1 a 2 + 2 1 h 2 B M ⊥ A C ′ ⇒ B M → ⋅ A C ′ → = 0 ⇔ 1 2 h 2 = 1 2 a 2 ⇔ h = a B M \perp A C^{\prime} \Rightarrow \overrightarrow{B M} \cdot \overrightarrow{A C^{\prime}}=0 \Leftrightarrow \frac{1}{2} h^{2}=\frac{1}{2} a^{2} \Leftrightarrow h=a BM ⊥ A C ′ ⇒ BM ⋅ A C ′ = 0 ⇔ 2 1 h 2 = 2 1 a 2 ⇔ h = a A B C A B C A BC S A B C = a 2 3 4 S_{A B C}=\frac{a^{2} \sqrt{3}}{4} S A BC = 4 a 2 3 A M / / ( B C C ′ ) A M / /\left(B C C^{\prime}\right) A M // ( BC C ′ ) V M . B C C ′ = V A . B C C ′ V_{M . B C C^{\prime}}=V_{A . B C C^{\prime}} V M . BC C ′ = V A . BC C ′ V M . B C C ′ = 3 12 a 3 V_{M . B C C^{\prime}}=\frac{\sqrt{3}}{12} a^{3} V M . BC C ′ = 12 3 a 3 M B = M C ′ = a 5 2 , B C ′ = a 2 M B=M C^{\prime}=\frac{a \sqrt{5}}{2}, B C^{\prime}=a \sqrt{2} MB = M C ′ = 2 a 5 , B C ′ = a 2 S M B C ′ = a 2 6 4 S_{M B C^{\prime}}=\frac{a^{2} \sqrt{6}}{4} S MB C ′ = 4 a 2 6 d ( C , ( B M C ′ ) ) = 3 V C B M C ′ S M B C ′ = 2 2 a d\left(C,\left(B M C^{\prime}\right)\right)=\frac{3 V_{C B M C^{\prime}}}{S_{M B C^{\prime}}}=\frac{\sqrt{2}}{2} a d ( C , ( BM C ′ ) ) = S MB C ′ 3 V CBM C ′ = 2 2 a a = 2 2 a=2 \sqrt{2} a = 2 2 d ( C , ( B M C ′ ) ) = 2 d\left(C,\left(B M C^{\prime}\right)\right)=2 d ( C , ( BM C ′ ) ) = 2
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Theo giả thiết: .
Diện tích tam giác là: .
Vì nên hay .
Ta có . Theo công thức Heron suy ra .
Vậy khoảng cách cần tìm là .
Do nên .