Trong mặt phẳng \(Oxy\), cho \(\overrightarrow a = \left( {{a_1};{a_2}} \right),\overrightarrow b = \left( {{b_1};{b_2}} \right)\). Mệnh đề nào sau đây sai?
A.\(\overrightarrow a \bot \overrightarrow b \Leftrightarrow {a_1}{b_1} + {a_2}{b_2} = 0\).
B.\(\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{{a_1}{b_1} + {a_2}{b_2}}}{{\sqrt {a_1^2 + a_2^2} + \sqrt {b_1^2 + b_2^2} }}\).
C.\(\left| {\overrightarrow a } \right| = \sqrt {a_1^2 + a_2^2} ;\left| {\overrightarrow b } \right| = \sqrt {b_1^2 + b_2^2} \).
D.\(\overrightarrow a = {a_2}\overrightarrow j + {a_1}\overrightarrow i ;\,\,\,\overrightarrow b = {b_1}\overrightarrow i + {b_2}\overrightarrow j \).
A.\(\overrightarrow a \bot \overrightarrow b \Leftrightarrow {a_1}{b_1} + {a_2}{b_2} = 0\).
B.\(\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{{a_1}{b_1} + {a_2}{b_2}}}{{\sqrt {a_1^2 + a_2^2} + \sqrt {b_1^2 + b_2^2} }}\).
C.\(\left| {\overrightarrow a } \right| = \sqrt {a_1^2 + a_2^2} ;\left| {\overrightarrow b } \right| = \sqrt {b_1^2 + b_2^2} \).
D.\(\overrightarrow a = {a_2}\overrightarrow j + {a_1}\overrightarrow i ;\,\,\,\overrightarrow b = {b_1}\overrightarrow i + {b_2}\overrightarrow j \).
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