Tính góc giữa hai vectơ \(\overrightarrow a \) và \(\overrightarrow b \) trong các trường hợp sau:
a) \(\overrightarrow a = \left( {1; - 4} \right),\overrightarrow b = \left( {5;3} \right)\)
b) \(\overrightarrow a = \left( {4;3} \right),\overrightarrow b = \left( {6;0} \right)\)
c) \(\overrightarrow a = \left( {2;2\sqrt 3 } \right),\overrightarrow b = \left( { - 3;\sqrt 3 } \right)\)
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Cho hai vectơ \(\overrightarrow a = \left( {{a_1},{a_2}} \right),\overrightarrow b = \left( {{b_1},{b_2}} \right)\). Ta có:
+ \(cos\left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}} = \frac{{{a_1}{a_2} + {a_2}{b_2}}}{{\sqrt {{a_1}^2 + {a_2}^2} .\sqrt {{b_1}^2 + {b_2}^2} }}\)
a) \(cos\left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{1.5 - 4.3}}{{\sqrt {{1^2} + {{\left( { - 4} \right)}^2}} .\sqrt {{5^2} + {3^2}} }} = \frac{{ - 7\sqrt 2 }}{{34}} \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) \approx {106^ \circ }56’\)
b) \(cos\left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{4.6 + 3.0}}{{\sqrt {{4^2} + {6^2}} .\sqrt {{3^2} + {0^2}} }} = \frac{4}{5} \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) \approx {36^ \circ }52’\)
c) \(cos\left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{2.\left( { - 3} \right) + 2\sqrt 3 .\sqrt 3 }}{{\sqrt {{2^2} + {{\left( {2\sqrt 3 } \right)}^2}} .\sqrt {{{\left( { - 3} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} }} = 0 \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) \approx {90^ \circ }\)