Nội dung bài giảng
5. Trên mặt phẳng Oxy hãy 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) \(\overrightarrow a = (-2; -2\sqrt3)\), \(\overrightarrow b = (3; \sqrt3)\);
Hướng dẫn:
a) \(\overrightarrow a .\overrightarrow b = 2.6 + \left( { - 3} \right).4 = 0 \Rightarrow \overrightarrow a \bot \overrightarrow b\) hay \(\left( {\overrightarrow a ,\overrightarrow b } \right) = {90^0}\)
b) \(\overrightarrow a .\overrightarrow b = 3.5 + 2\left( { - 1} \right) = 13\)
Mặt khác:
\(\eqalign{
& \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& = \sqrt {{3^2} + {2^2}} .\sqrt {{5^2} + {{\left( { - 1} \right)}^2}} .\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& = \sqrt {26} .\sqrt {13} .\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr} \)
\(\eqalign{
& \Rightarrow 13 = \sqrt {26} .\sqrt {13} \cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& \Rightarrow \cos \left( {\overrightarrow a ,\overrightarrow b } \right) = {1 \over {\sqrt 2 }} \cr
& \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = {45^0} \cr} \)
c) \(\overrightarrow a .\overrightarrow b = - 2.3 + \left( { - 2\sqrt 3 } \right).\sqrt 3 = -12\)
\(\eqalign{
& \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2\sqrt 3 } \right)}^2}} .\sqrt {{3^2} + {{\left( {\sqrt 3 } \right)}^2}} .\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& = 4 .\sqrt {12} .\cos \left( {\overrightarrow a ,\overrightarrow b } \right) \cr
& \Rightarrow \cos \left( {\overrightarrow a ,\overrightarrow b } \right) = {{-12} \over 4.{\sqrt {12} }} = {{-\sqrt3 \over 2}} \cr
& \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = 150^0\cr} \)