Câu 2:
\(\begin{array}{l}\overrightarrow {BA} = \left( {4; - 6} \right),\overrightarrow {BC} = \left( {5; - 7} \right) \Rightarrow \overrightarrow {BA} .\overrightarrow {BC} = 4.5 + \left( { - 6} \right).\left( { - 7} \right) = 62\\ \Rightarrow BA = \sqrt {{4^2} + {{\left( { - 6} \right)}^2}} = 2\sqrt {13} ,BC = \sqrt {{5^2} + {{\left( { - 7} \right)}^2}} = \sqrt {74} \\ \Rightarrow \cos \widehat B = \dfrac{{\overrightarrow {BA} .\overrightarrow {BC} }}{{\left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|}} = \dfrac{{62}}{{2\sqrt {13} .\sqrt {74} }} = \dfrac{{31}}{{\sqrt {962} }}\end{array}\)
Câu 3:
Ta có: \(\overrightarrow {AB} = \left( { - 1;4} \right),\overrightarrow {AC} = \left( { - 4;1} \right),\overrightarrow {BC} = \left( { - 3; - 3} \right)\)
\(\begin{array}{l} \Rightarrow \overrightarrow {AC} + \overrightarrow {BC} = \left( { - 7; - 2} \right)\\ \Rightarrow \overrightarrow {AB} .\left( {\overrightarrow {AC} + \overrightarrow {BC} } \right) = \left( { - 1} \right).\left( { - 7} \right) + 1.\left( { - 2} \right) = 5\end{array}\)
