Câu hỏi:
Cho \(A\left( {5;4;3} \right),\,\,B\left( {6;7;2} \right),\,\,\left( \Delta \right):\,\,\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{1}\). Tìm \(M \in \left( \Delta \right)\) để \({S_{\Delta ABM\,\,\min }}\).
- A \(M\left( {5;3;4} \right)\)
- B \(M\left( {3;5;4} \right)\)
- C \(M\left( {4;3;5} \right)\)
- D \(M\left( { - 3; - 5; - 4} \right)\)
Phương pháp giải:
Lời giải chi tiết:
* Vì \(M \in \left( \Delta \right) \Rightarrow M\left( {2t + 1;3t + 2;t + 3} \right)\)
* \(\left\{ \begin{array}{l}\overrightarrow {AM} = \left( {2t - 4;3t - 2;t} \right)\\\overrightarrow {AB} = \left( {1;3; - 1} \right)\end{array} \right. \Rightarrow \overrightarrow n = \left[ {\overrightarrow {AM} ;\overrightarrow {AB} } \right]\)
\( \Rightarrow \overrightarrow n = \left( { - 6t + 2;3t - 4;3t - 10} \right)\)
\(\begin{array}{l}*\,\,{S_{\Delta MAB}} = \dfrac{1}{2}\left| {\overrightarrow n } \right| = \dfrac{1}{2}\sqrt {{{\left( {6t - 2} \right)}^2} + {{\left( {3t - 4} \right)}^2} + {{\left( {3t - 10} \right)}^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\sqrt {54{t^2} - 108t + 120} \end{array}\)
* \({S_{\Delta MAB\,\,\min }} \Leftrightarrow t = - \dfrac{b}{{2a}} = \dfrac{{108}}{{2.54}} = 1\).
\( \Rightarrow M\left( {3;5;4} \right)\).
Chọn C.
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