Lời giải

a) Ta có \(\overrightarrow {AB} .\overrightarrow {AC} = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\,\overrightarrow {AC} } \right)\)

\( = AB.AC.\cos \widehat {BAC} = 2.3.\cos 60^\circ = 3\).

Vậy \(\overrightarrow {AB} .\overrightarrow {AC} = 3\).

b) Vì M là trung điểm BC nên ta có \(2\overrightarrow {AM} = \overrightarrow {AB} + \overrightarrow {AC} \).

\( \Leftrightarrow \overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \).

Ta có \(\overrightarrow {BD} = \overrightarrow {AD} - \overrightarrow {AB} = \frac{7}\overrightarrow {AC} - \overrightarrow {AB} \).

Vậy \(\overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \) và \(\overrightarrow {BD} = \frac{7}\overrightarrow {AC} - \overrightarrow {AB} \).

c) Ta có \(\overrightarrow {AM} .\overrightarrow {BD} = \left( {\frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} } \right).\left( {\frac{7}\overrightarrow {AC} - \overrightarrow {AB} } \right)\)

\( = \frac{7}\overrightarrow {AB} .\overrightarrow {AC} - \frac{1}{2}.{\overrightarrow {AB} ^2} + \frac{7}.{\overrightarrow {AC} ^2} - \frac{1}{2}\overrightarrow {AB} .\overrightarrow {AC} \)

\( = - \frac{5}\overrightarrow {AB} .\overrightarrow {AC} - \frac{1}{2}.A{B^2} + \frac{7}.A{C^2}\)

\( = - \frac{5}.3 - \frac{1}{2}{.2^2} + \frac{7}{.3^2} = 0\).

Do đó \(\overrightarrow {AM} \bot \overrightarrow {BD} \).