Đáp án + Giải thích các bước giải:

Ta có:

$\begin {cases} \overrightarrow{AM} = 2\overrightarrow{MB} \\ 3\overrightarrow{NA} + \overrightarrow{NC} = \overrightarrow{0} \end {cases}$

$\Leftrightarrow \begin {cases} \overrightarrow{AB} - \overrightarrow{MB} = 2\overrightarrow{MB} \\ 3\overrightarrow{NA} + \overrightarrow{NA} + \overrightarrow{AC} = \overrightarrow{0} \end {cases}$

$\Leftrightarrow \begin {cases} \overrightarrow{AB} = 3\overrightarrow{MB} \\ 4 \overrightarrow{NA} + \overrightarrow{AC} = \overrightarrow{0} \end {cases}$

$\Leftrightarrow \begin {cases} \overrightarrow{AB} = 3\overrightarrow{MB} \\ \overrightarrow{AC} = 4\overrightarrow{AN} \end {cases}$

$\Rightarrow M$ nằm giữa $A$ và $B$ sao cho $AB= 3MB$ và $N$ nằm giữa $A$ và $C$ sao $AC = 4AN$

$\Rightarrow \overrightarrow{AM} = \dfrac{2}{3}\overrightarrow{AB}$

Ta có: $\begin {cases} \overrightarrow{AM} = \dfrac{2}{3}\overrightarrow{AB} \\ \overrightarrow{AN} = \dfrac{1}{4}\overrightarrow{AC}\end {cases}$

$\Leftrightarrow \overrightarrow{MN} = \overrightarrow{AN} - \overrightarrow{AM} = \dfrac{1}{4}\overrightarrow{AC} - \dfrac{2}{3}\overrightarrow{AB}$

Mà $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$

$\Rightarrow \overrightarrow{MN} = \overrightarrow{AN} - \overrightarrow{AM} = \dfrac{1}{4}\overrightarrow{AB} + \dfrac{1}{4}\overrightarrow{BC}  - \dfrac{2}{3}\overrightarrow{AB}$

$\Leftrightarrow \overrightarrow{MN} = \dfrac{1}{4}\overrightarrow{BC} -\dfrac{5}{12}\overrightarrow{AB}$

$\Leftrightarrow \left(\overrightarrow{MN}\right)^2 = \left(\dfrac{1}{4}\overrightarrow{BC} - \dfrac{5}{12}\overrightarrow{AB}\right)^2$

$\Leftrightarrow MN^2 = \dfrac{1}{16}\left(\overrightarrow{BC}\right)^2 - \dfrac{5}{24}\overrightarrow{AB} \cdot \overrightarrow{BC} + \dfrac{25}{144}\left(\overrightarrow{AB}\right)^2$

$\Leftrightarrow MN^2 = \dfrac{1}{16}BC^2 - \dfrac{5}{24}\left|\overrightarrow{AB}\right|\left|\overrightarrow{BC}\right| \cos 90^\circ + \dfrac{25}{144}AB^2$

$\Leftrightarrow MN^2 = \dfrac{1}{16}BC^2 + \dfrac{25}{144}AB^2$

$\Leftrightarrow MN^2 = \dfrac{1}{16}\left(2a\right)^2 + \dfrac{25}{144}a^2 = \dfrac{61}{144}a^2$

$\Leftrightarrow MN = \dfrac{a\sqrt{61}}{12}$