a) Ta có: \[B{C^2} = {\left| {\overrightarrow {BC} } \right|^2} = {\overrightarrow {BC} ^2} = {\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)^2} = A{C^2} + A{B^2} - 2\overrightarrow {AC} \cdot \overrightarrow {AB} \].

Suy ra: \[\overrightarrow {AB} \cdot \overrightarrow {AC} = \frac{{A{C^2} + A{B^2} - B{C^2}}}{2} = \frac{{{b^2} + {c^2} - {a^2}}}{2}\].

Tương tự ta có: \[\overrightarrow {BC} \cdot \overrightarrow {BA} = \frac{{{a^2} + {c^2} - {b^2}}}{2};\;\;\;\overrightarrow {CA} \cdot \overrightarrow {CB} = \frac{{{a^2} + {b^2} - {c^2}}}{2}\].

Suy ra: \[\;\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} \]\[ = - \overrightarrow {BA} \cdot \overrightarrow {BC} - \overrightarrow {CB} \cdot \overrightarrow {CA} - \overrightarrow {AC} \cdot \overrightarrow {AB} \]

\[ = - \left( {\frac{{{c^2} + {a^2} - {b^2}}}{2} + \frac{{{a^2} + {b^2} - {c^2}}}{2} + \frac{{{b^2} + {c^2} - {a^2}}}{2}} \right) = - \frac{{{a^2} + {b^2} + {c^2}}}{2}\].

b) Ta có: \[\overrightarrow {AB} + \overrightarrow {AC} = 2\overrightarrow {AM} \] với \[M\] là trung điểm của \[BC\].

Vì \[G\] là trọng tâm tam giác \[ABC\] nên \[\overrightarrow {AG} = \frac{2}{3}\overrightarrow {AM} \]. Vậy \[\overrightarrow {AG} = \frac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\].

Suy ra: \[A{G^2} = {\overrightarrow {AG} ^2} = \frac{1}{9}{\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)^2} = \frac{1}{9}\left( {A{B^2} + A{C^2} + 2\overrightarrow {AB} \cdot \overrightarrow {AC} } \right)\]

\[ = \frac{1}{9}\left( {{c^2} + {b^2} + 2 \cdot \frac{{{b^2} + {c^2} - {a^2}}}{2}} \right) = \frac{1}{9}\left( {2{b^2} + 2{c^2} - {a^2}} \right)\].

\[ \Rightarrow AG = \frac{1}{3}\sqrt {2\left( {{b^2} + {c^2}} \right) - {a^2}} \].

Ta có: \[\overrightarrow {AG} \cdot \overrightarrow {BC} = \left| {\overrightarrow {AG} } \right| \cdot \left| {\overrightarrow {BC} } \right| \cdot \cos \left( {\overrightarrow {AG} ,\,\,\overrightarrow {BC} } \right)\]\( \Rightarrow \cos \left( {\overrightarrow {AG} ,\,\,\overrightarrow {BC} } \right) = \frac{{\overrightarrow {AG} \cdot \overrightarrow {BC} }}{{\left| {\overrightarrow {AG} } \right| \cdot \left| {\overrightarrow {BC} } \right|}}\).

Lại có: \[\overrightarrow {AG} \cdot \overrightarrow {BC} = \frac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) = \frac{1}{3}\left( {A{C^2} - A{B^2}} \right) = \frac{1}{3}\left( {{b^2} - {c^2}} \right)\].

Do đó, \[\cos \left( {\overrightarrow {AG} ,\,\overrightarrow {BC} } \right) = \frac{{\frac{1}{3}\left( {{b^2} - {c^2}} \right)}}{{\frac{1}{3}\sqrt {2\left( {{b^2} + {c^2}} \right) - {a^2}} \cdot a}} = \frac{{{b^2} - {c^2}}}{{a\sqrt {2\left( {{b^2} + {c^2}} \right) - {a^2}} }}\].

Cho cấp số cộng ( u n ) có u 1 = 3 ; u 8 = 24 thì u 11 bằng