a) Ta có:
\(\overrightarrow {CM} = 2\overrightarrow {MB} \,\,\, \Rightarrow \,\,\overrightarrow {AM} - \overrightarrow {AC} = 2(\overrightarrow {AB} - \overrightarrow {AM} )\)
\(\Rightarrow \,\,\overrightarrow {AM} = {2 \over 3}\overrightarrow {AB} + {1 \over 3}\overrightarrow {AC} \)
Mặt khác \(\overrightarrow {BN} = 2\overrightarrow {NA} \,\, \Rightarrow \,\,\overrightarrow {AN} - \overrightarrow {AB} = - 2\overrightarrow {AN} \)
\(\Rightarrow \,\,\overrightarrow {AN} = {1 \over 3}\overrightarrow {AB} \)
\( \Rightarrow \,\,\overrightarrow {CN} = \overrightarrow {AN} - \overrightarrow {AC} = {1 \over 3}\overrightarrow {AB} - \overrightarrow {AC} \)
b) Ta có
\(\eqalign{
& \overrightarrow {AM} \bot \overrightarrow {CN} \Leftrightarrow \,\,\overrightarrow {AM} .\overrightarrow {CN} = 0\cr& \Leftrightarrow \,\,\left( {{2 \over 3}\overrightarrow {AB} + {1 \over 3}\overrightarrow {AC} } \right)\left( {{1 \over 3}\overrightarrow {AB} - \overrightarrow {AC} } \right) \cr&\;\;\;\;\;= 0 \cr
& \Leftrightarrow \,\,{2 \over 9}A{B^2} - {2 \over 3}\overrightarrow {AB} .\overrightarrow {AC} + {1 \over 9}\overrightarrow {AC} .\,\overrightarrow {AB} - {1 \over 3}A{C^2}\cr&\;\;\;\;\; = 0 \cr
& \Leftrightarrow \,\,{2 \over 9}{c^2} - {1 \over 3}{b^2} = 0 \cr
& \ \Leftrightarrow \,\,2{c^2} = 3{b^2} \cr} \)