a) Ta có: \(B{C^2} = {\overrightarrow {BC} ^2} = {\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)^2}\)\( = {\overrightarrow {AC} ^2} + {\overrightarrow {AB} ^2} - 2\overrightarrow {AC} .\overrightarrow {AB} \)
Do đó \(\overrightarrow {AB} .\overrightarrow {AC} = \dfrac{{{{\overrightarrow {AC} }^2} + {{\overrightarrow {AB} }^2} - {{\overrightarrow {BC} }^2}}}{2}\)\( = \dfrac{{{8^2} + {5^2} - {7^2}}}{2} = 20\)
Mặt khác \(\overrightarrow {AB} .\overrightarrow {AC} = AB.AC.cosA\)\( \Rightarrow 5.8.cosA = 20\)
Suy ra \(\cos A = \dfrac{{20}}{{40}} = \dfrac{1}{2} \Rightarrow \widehat A = {60^0}\)
b) Ta có: \(B{A^2} = {\overrightarrow {BA} ^2}\)\( = {\left( {\overrightarrow {CA} - \overrightarrow {CB} } \right)^2} = {\overrightarrow {CA} ^2} + {\overrightarrow {CB} ^2} - 2\overrightarrow {CA} .\overrightarrow {CB} \)
Do đó \(\overrightarrow {CA} .\overrightarrow {CB} = \dfrac{1}{2}\left( {{{\overrightarrow {CA} }^2} + {{\overrightarrow {CB} }^2} - {{\overrightarrow {BA} }^2}} \right)\)\( = \dfrac{1}{2}\left( {{8^2} + {7^2} - {5^2}} \right) = 44\)
