Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
\overrightarrow {AN} + \overrightarrow {AB} - \frac{2}{3}\overrightarrow {BD} = \overrightarrow {AD} + \overrightarrow {DN} + \overrightarrow {AB} - \frac{2}{3}\left( {\overrightarrow {BA} + \overrightarrow {AD} } \right)\\
= \overrightarrow {AD} + \frac{1}{3}\overrightarrow {AB} + \overrightarrow {AB} + \frac{2}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {AD} = 2\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AD} \\
\[\begin{array}{l}
{\left| {2\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AD} } \right|^2} = 4{\overrightarrow {AB} ^2} + \frac{4}{3}\overrightarrow {AB} .\overrightarrow {AD} + \frac{1}{9}{\overrightarrow {AD} ^2} = \frac{{37}}{9}.{\left( {6a} \right)^2} = 148{a^2}\\
\Rightarrow \left| {\overrightarrow {AN} + \overrightarrow {AB} - \frac{2}{3}\overrightarrow {BD} } \right| = 2\sqrt {37} a
\end{array}\]
