Đáp án:
$k = \dfrac{2}{5}$
Giải thích các bước giải:
$\begin{array}{l}
\overrightarrow {AC} .\overrightarrow {AB} = 4{a^2} \Leftrightarrow \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right).\overrightarrow {AB} = 4{a^2}\\
\Leftrightarrow {\overrightarrow {AB} ^2} + \overrightarrow {BC} .\overrightarrow {AB} = 4{a^2} \Leftrightarrow A{B^2} = 4{a^2} \Leftrightarrow AB = 2a\\
\overrightarrow {CA} .\overrightarrow {CB} = 9{a^2} \Leftrightarrow \left( {\overrightarrow {CB} + \overrightarrow {BA} } \right).\overrightarrow {CB} = 9{a^2}\\
\Leftrightarrow {\overrightarrow {CB} ^2} + \overrightarrow {BA} .\overrightarrow {CB} = 9{a^2} \Leftrightarrow C{B^2} = 9{a^2} \Leftrightarrow CB = 3a\\
\overrightarrow {CB} .\overrightarrow {CD} = 6{a^2} \Leftrightarrow \overrightarrow {CB} \left( {\overrightarrow {CB} + \overrightarrow {BA} + \overrightarrow {AD} } \right) = 6{a^2}\\
\Leftrightarrow \overrightarrow {CB} .\overrightarrow {CB} . + \overrightarrow {CB} .\overrightarrow {BA} + \overrightarrow {CB} .\overrightarrow {AD} = 6{a^2}\\
\Leftrightarrow C{B^2} + 0 - CB.AD = 6{a^2}\\
\Leftrightarrow 9{a^2} - 3a.AD = 6{a^2} \Leftrightarrow AD = a\\
\overrightarrow {BM} = \overrightarrow {BA} + \overrightarrow {AM} = \overrightarrow {BA} + k\overrightarrow {AC} \\
= \overrightarrow {BA} + k\left( {\overrightarrow {BC} - \overrightarrow {BA} } \right) = \left( {1 - k} \right)\overrightarrow {BA} + k\overrightarrow {BC} \\
\overrightarrow {CD} = \overrightarrow {AD} - \overrightarrow {AC} = \dfrac{1}{3}\overrightarrow {BC} - \left( {\overrightarrow {BC} - \overrightarrow {BA} } \right) = \overrightarrow {BA} - \dfrac{2}{3}\overrightarrow {BC} \\
\Rightarrow \overrightarrow {BM} .\overrightarrow {CD} = \left( {\left( {1 - k} \right)\overrightarrow {BA} + k\overrightarrow {BC} } \right).\left( {\overrightarrow {BA} - \dfrac{2}{3}\overrightarrow {BC} } \right)\\
= \left( {1 - k} \right)B{A^2} - \dfrac{{2k}}{3}B{C^2} = \left( {1 - k} \right)4{a^2} - \dfrac{{2k}}{3}.9{a^2}\\
BM \bot CD \Leftrightarrow \left( {1 - k} \right)4{a^2} - \dfrac{{2k}}{3}.9{a^2} = 0\\
\Leftrightarrow 4\left( {1 - k} \right) - 6k = 0 \Leftrightarrow 4 - 10k = 0 \Leftrightarrow k = \dfrac{2}{5}
\end{array}$
