Biết \(\overrightarrow {{n_{{\Delta _1}}}} = \left( {{a_1};{b_1}} \right);\overrightarrow {{n_{{\Delta _2}}}} = \left( {{a_2};{b_2}} \right)\), công thức tính góc giữa hai đường thẳng \({\Delta _1};{\Delta _2}\) là
A.\(\cos \left( {\overrightarrow {{n_{{\Delta _1}}}} ,\overrightarrow {{n_{{\Delta _2}}}} } \right) = \frac{{{a_1}{b_1} + {a_2}{b_2}}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
B.\(\cos \left( {{\Delta _1},{\Delta _2}} \right) = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
C.\(\left| {\cos \left( {{\Delta _1},{\Delta _2}} \right)} \right| = \frac{{\left| {{a_1}{b_2} - {b_1}{a_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
D.\(\left| {\cos \left( {\overrightarrow {{n_{{\Delta _1}}}} ,\overrightarrow {{n_{{\Delta _2}}}} } \right)} \right| = \frac{{\left| {{a_1}{b_1} + {a_2}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
A.\(\cos \left( {\overrightarrow {{n_{{\Delta _1}}}} ,\overrightarrow {{n_{{\Delta _2}}}} } \right) = \frac{{{a_1}{b_1} + {a_2}{b_2}}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
B.\(\cos \left( {{\Delta _1},{\Delta _2}} \right) = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
C.\(\left| {\cos \left( {{\Delta _1},{\Delta _2}} \right)} \right| = \frac{{\left| {{a_1}{b_2} - {b_1}{a_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
D.\(\left| {\cos \left( {\overrightarrow {{n_{{\Delta _1}}}} ,\overrightarrow {{n_{{\Delta _2}}}} } \right)} \right| = \frac{{\left| {{a_1}{b_1} + {a_2}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\sqrt {a_2^2 + {b_2}^2} }}\)
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Hóa Học lớp 12
