Câu hỏi
Cho \(A\left( {1;4;2} \right),\,\,B\left( { - 1;2;4} \right),\,\,\left( \Delta \right):\,\,\dfrac{{x - 1}}{{ - 1}} = \dfrac{{y + 2}}{1} = \dfrac{z}{2}\). Tìm \(M \in \left( \Delta \right)\) để \({\left( {M{A^2} + M{B^2}} \right)_{\min }}\).
- A \(M\left( { - 1;0;4} \right)\)
- B \(M\left( { - 1;0; - 4} \right)\)
- C \(M\left( {1;0; - 4} \right)\)
- D \(M\left( {1;0;4} \right)\)
Lời giải chi tiết:
* \(M \in \left( \Delta \right) \Rightarrow M\left( { - t + 1;t - 2;2t} \right)\).
* \(T = M{A^2} + M{B^2} = \left[ {{t^2} + {{\left( {6 - t} \right)}^2} + {{\left( {2t - 2} \right)}^2} + {{\left( { - t + 2} \right)}^2} + {{\left( {t - 4} \right)}^2} + {{\left( {2t - 4} \right)}^2}} \right]\)
\( = 12{t^2} - 48t + 76\)
* \({T_{\min }} \Leftrightarrow t = - \dfrac{b}{{2a}} = \dfrac{{4.8}}{{2.12}} = 2\) \( \Rightarrow M\left( { - 1;0;4} \right)\).
Chọn A.