Ta có: sin C = \(\frac{{AB}}{{BC}} = \frac{a}{{2a}} = \frac{1}{2}\)⇒ \(\widehat C = 30^\circ \)⇒ \(\widehat B = 60^\circ \)

Áp dụng định lý Pitago ta có:

AC = \(\sqrt {4{a^2} - {a^2}} = a\sqrt 3 \)

\[\overrightarrow {BC} .\overrightarrow {CA} + \overrightarrow {BA} .\overrightarrow {AC} = BC.CA.\cos \left( {\overrightarrow {BC} ;\overrightarrow {CA} } \right) + BA.AC.\cos \left( {\overrightarrow {BA} ;\overrightarrow {AC} } \right)\]

= \(2a.a\sqrt 3 .\cos 30^\circ + a.a\sqrt 3 .\cos 90^\circ \)

= \(2{a^2}\sqrt 3 .\frac{{\sqrt 3 }}{2} = 3{a^2}\).