Phương pháp giải:

\(\overrightarrow a  = \left( {{a_1};{a_2}} \right)\,\,;\,\,\,\overrightarrow b  = \left( {{b_1};{b_2}} \right) \Rightarrow \overrightarrow a .\overrightarrow b  = {a_1}.{b_1} + {a_2}.{b_2}\)

\(\overrightarrow a  = \left( {{a_1};{a_2}} \right) \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {a_1^2 + a_2^2} \)

\(\overrightarrow a .\overrightarrow b  = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right)\)

Lời giải chi tiết:

\(\begin{array}{l}\vec a = m\vec u + \vec v = m\left( {2;\;5} \right) + \left( { - 3;\;1} \right) = \left( {2m - 3;\;5m + 1} \right)\\ \Rightarrow \left| {\vec a} \right| = \sqrt {{{\left( {2m - 3} \right)}^2} + {{\left( {5m + 1} \right)}^2}}  = \sqrt {29{m^2} - 2m + 10} \\\overrightarrow b  = \left( {1;1} \right) \Rightarrow \left| {\overrightarrow b } \right| = \sqrt 2 \\\overrightarrow a .\overrightarrow b  = 2m - 3 + 5m + 1 = 7m - 2\end{array}\)

 Mặt khác: \(\overrightarrow a .\overrightarrow b  = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \left( {\overrightarrow a ,\;\overrightarrow b } \right) = \sqrt {29{m^2} - 2m + 10} .\sqrt 2 .\cos {45^o} = \sqrt {29{m^2} - 2m + 10} .\)

\(\begin{array}{l} \Leftrightarrow 7m - 2 = \sqrt {29{m^2} - 2m + 10}  \Leftrightarrow \left\{ \begin{array}{l}7m - 2 \ge 0\\49{m^2} - 28m + 4 = 29{m^2} - 2m + 10\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m \ge \frac{2}{7}\\20{m^2} - 26m - 6 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ge \frac{2}{7}\\\left[ \begin{array}{l}m = \frac{3}{2}\\m =  - \frac{1}{5}\end{array} \right.\end{array} \right. \Leftrightarrow m = \frac{3}{2}.\end{array}\)

Chọn A.