Câu hỏi:
Cho \(A\left( {2;0;1} \right),\,\,B\left( {2; - 1;0} \right),\,\,C\left( {1;0;1} \right),\,\,\left( \Delta \right):\,\,\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}\). Tìm \(M \in \left( \Delta \right)\) để \({\left| {\overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} } \right|_{\min }}\).
- A \(M\left( {3;6;9} \right)\)
- B \(M\left( {\dfrac{{ - 3}}{{14}};\dfrac{{ - 6}}{{14}};\dfrac{{ - 9}}{{14}}} \right)\)
- C \(M\left( {\dfrac{3}{{14}};\dfrac{6}{{14}};\dfrac{9}{{14}}} \right)\)
- D \(M\left( {9;3;6} \right)\)
Phương pháp giải:
Lời giải chi tiết:
* \(M \in \left( \Delta \right) \Rightarrow M\left( {t;2t;3t} \right)\)
* \(P = \left| {\overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} } \right|\)
\(\begin{array}{l} = \left| {\left( {2 - t; - 2t;1 - 3t} \right) + \left( {2 - t; - 1 - 2t; - 3t} \right) + \left( {1 - t; - 2t;1 - 3t} \right)} \right|\\ = \left| {\left( {5 - 3t; - 1 - 6t;2 - 9t} \right)} \right|\\ = \sqrt {{{\left( {5 - 3t} \right)}^2} + {{\left( { - 1 - 6t} \right)}^2} + {{\left( {2 - 9t} \right)}^2}} \\ = \sqrt {126{t^2} - 54t + 30} \end{array}\)
* \({P_{\min }} \Leftrightarrow t = - \dfrac{b}{{2a}} = \dfrac{{5.4}}{{2.126}} = \dfrac{3}{{14}}\).
\( \Rightarrow M\left( {\dfrac{3}{{14}};\dfrac{6}{{14}};\dfrac{9}{{14}}} \right)\).
Chọn B.
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