a) Vẽ vecto \(\overrightarrow {BD} = \overrightarrow {CB} \). Ta có:
\((\overrightarrow {CB} ,\overrightarrow {BA} ) = (\overrightarrow {BD} ,\overrightarrow {BA} ) = \widehat {DBA} = {120^o}\)
Vậy \(\overrightarrow {CB} .\overrightarrow {BA} = \left| {\overrightarrow {CB} } \right|.\left| {\overrightarrow {BA} } \right|\cos (\overrightarrow {CB} ,\overrightarrow {BA} ) = a.a.\cos {120^o} = {a^2}.\left( { - \frac{1}{2}} \right) = - \frac{{{a^2}}}{2}.\)
b) Vì \(AH \bot BC\) nên \[(\overrightarrow {AH} ,\overrightarrow {BC} ) = {90^o}\], suy ra \(\cos (\overrightarrow {AH} ,\overrightarrow {BC} ) = \cos {90^o} = 0.\)
Vậy \(\overrightarrow {AH} .\overrightarrow {BC} = \left| {\overrightarrow {AH} } \right|.\left| {\overrightarrow {BC} } \right|.\cos (\overrightarrow {AH} ,\overrightarrow {BC} ) = 0.\)