Question

Tính góc giữa hai vectơ \(\overrightarrow a \) và \(\overrightarrow b \)trong các trường hợp sau

a) \(\overrightarrow a  = (2; - 3),\overrightarrow b  = (6;4)\)

b) \(\overrightarrow a  = (3;2),\overrightarrow b  = (5; - 1)\)

c) \(\overrightarrow a  = ( - 2; - 2\sqrt 3 ),\overrightarrow b  = (3;\sqrt 3 )\)

Answer 1

a) \(\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{{2.6 + ( - 3).4}}{{\sqrt {{2^2} + {{\left( { - 3} \right)}^2}} .\sqrt {{6^2} + {4^2}} }} = 0 \Rightarrow \overrightarrow a  \bot \overrightarrow b \)

b) \(\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{{3.5 + 2.( - 1)}}{{\sqrt {{3^2} + {2^2}} .\sqrt {{5^2} + {{\left( { - 1} \right)}^2}} }} = \frac{{\sqrt 2 }}{2} \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = 45^\circ \)

c) \(\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{{\left( { - 2} \right).3 + ( - 2\sqrt 3 ).\sqrt 3 }}{{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2\sqrt 3 } \right)}^2}} .\sqrt {{3^2} + {{\sqrt 3 }^2}} }} =  - \frac{{\sqrt 3 }}{2} \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = 150^\circ \)