Đáp án:

\[k =  - 4\]

Giải thích các bước giải:

 Đặt \(\left\{ \begin{array}{l}
\overrightarrow {AB}  = \overrightarrow x \\
\overrightarrow {AD}  = \overrightarrow y 
\end{array} \right. \Rightarrow \overrightarrow x .\overrightarrow y  = \overrightarrow {AB} .\overrightarrow {AD}  = \overrightarrow 0 \)

Suy ra 

\(\begin{array}{l}
\left\{ \begin{array}{l}
\overrightarrow {BM}  = \frac{1}{3}\overrightarrow {MC} \\
\overrightarrow {CN}  = k.\overrightarrow {AN} 
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {BM}  = \frac{1}{4}\overrightarrow {BC}  = \frac{1}{4}\overrightarrow y \\
\overrightarrow {CN}  + k\overrightarrow {NC}  = k\left( {\overrightarrow {AN}  + \overrightarrow {NC} } \right)
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {BM}  = \frac{1}{4}\overrightarrow y \\
\overrightarrow {CN}  = \frac{k}{{1 - k}}\overrightarrow {AC}  = \frac{k}{{1 - k}}\left( {\overrightarrow x  + \overrightarrow y } \right)
\end{array} \right.
\end{array}\)

Ta có:

\(\begin{array}{l}
AM \bot DN \Rightarrow \overrightarrow {AM} .\overrightarrow {DN}  = \overrightarrow 0 \\
 \Leftrightarrow \left( {\overrightarrow {AB}  + \overrightarrow {BM} } \right)\left( {\overrightarrow {DC}  + \overrightarrow {CN} } \right) = \overrightarrow 0 \\
 \Leftrightarrow \left( {\overrightarrow x  + \frac{1}{4}\overrightarrow y } \right)\left( {\overrightarrow x  + \frac{k}{{1 - k}}\overrightarrow x  + \frac{k}{{1 - k}}\overrightarrow y } \right) = \overrightarrow 0 \\
 \Leftrightarrow \left( {1 + \frac{k}{{1 - k}}} \right){\overrightarrow x ^2} + \frac{k}{{1 - k}}\overrightarrow x .\overrightarrow y  + \frac{1}{4}\left( {1 + \frac{k}{{1 - k}}} \right)\overrightarrow x .\overrightarrow y  + \frac{1}{4}.\frac{k}{{1 - k}}{\overrightarrow y ^2} = 0\\
\left| {\overrightarrow x } \right| = \left| {\overrightarrow y } \right| = a,\overrightarrow x .\overrightarrow y  = \overrightarrow 0 \\
 \Rightarrow \frac{{1 - k + k}}{{1 - k}}.{a^2} + \frac{k}{{4\left( {1 - k} \right)}}{a^2} = 0\\
 \Leftrightarrow \frac{1}{{1 - k}} + \frac{k}{{4\left( {1 - k} \right)}} = 0\\
 \Leftrightarrow \frac{{k + 4}}{{4\left( {1 - k} \right)}} = 0\\
 \Leftrightarrow k =  - 4
\end{array}\)