Đáp án:
a) \(\angle A = 90^\circ \)
Giải thích các bước giải:
\(\begin{array}{l}
a)vtpt:{\overrightarrow n _d} = \left( {2;1} \right) \to vtcp:{\overrightarrow u _d} = \left( {1; - 2} \right)\\
vtpt:{\overrightarrow n _\Delta } = \left( {1; - 2} \right) \to vtcp:{\overrightarrow u _\Delta } = \left( {2;1} \right)\\
\to \cos A = \dfrac{{\left| {{{\overrightarrow u }_\Delta }.{{\overrightarrow u }_d}} \right|}}{{\left| {{{\overrightarrow u }_\Delta }} \right|.\left| {{{\overrightarrow u }_d}} \right|}} = \dfrac{{\left| {2.1 - 2.1} \right|}}{{\sqrt 5 .\sqrt 5 }} = 0\\
\to \angle A = 90^\circ \\
b)vtpt:{\overrightarrow n _d} = \left( {2; - 3} \right) \to vtcp:{\overrightarrow u _d} = \left( {3;2} \right)\\
vtpt:{\overrightarrow n _\Delta } = \left( {1;1} \right) \to vtcp:{\overrightarrow u _\Delta } = \left( {1; - 1} \right)\\
\to \cos A = \dfrac{{\left| {{{\overrightarrow u }_\Delta }.{{\overrightarrow u }_d}} \right|}}{{\left| {{{\overrightarrow u }_\Delta }} \right|.\left| {{{\overrightarrow u }_d}} \right|}} = \dfrac{{\left| {3.1 - 2.1} \right|}}{{\sqrt {10} .\sqrt 2 }} = \dfrac{1}{{2\sqrt 5 }}\\
\to \angle A = {\cos ^{ - 1}}\left( {\dfrac{1}{{2\sqrt 5 }}} \right)\\
c)vtpt:{\overrightarrow n _d} = \left( {4;1} \right) \to vtcp:{\overrightarrow u _d} = \left( {1; - 4} \right)\\
vtpt:{\overrightarrow n _\Delta } = \left( {3;1} \right) \to vtcp:{\overrightarrow u _\Delta } = \left( {1; - 3} \right)\\
\to \cos A = \dfrac{{\left| {{{\overrightarrow u }_\Delta }.{{\overrightarrow u }_d}} \right|}}{{\left| {{{\overrightarrow u }_\Delta }} \right|.\left| {{{\overrightarrow u }_d}} \right|}} = \dfrac{{\left| {1.1 - 3.\left( { - 4} \right)} \right|}}{{\sqrt {17} .\sqrt {10} }} = \dfrac{{13}}{{\sqrt {170} }}\\
\to \angle A = {\cos ^{ - 1}}\left( {\dfrac{{13}}{{\sqrt {170} }}} \right)\\
d)vtpt:{\overrightarrow n _d} = \left( {2; - 5} \right) \to vtcp:{\overrightarrow u _d} = \left( {5;2} \right)\\
vtpt:{\overrightarrow n _\Delta } = \left( {1;4} \right) \to vtcp:{\overrightarrow u _\Delta } = \left( {4; - 1} \right)\\
\to \cos A = \dfrac{{\left| {{{\overrightarrow u }_\Delta }.{{\overrightarrow u }_d}} \right|}}{{\left| {{{\overrightarrow u }_\Delta }} \right|.\left| {{{\overrightarrow u }_d}} \right|}} = \dfrac{{\left| {4.5 - 2.1} \right|}}{{\sqrt {29} .\sqrt {17} }} = \dfrac{{18}}{{\sqrt {493} }}\\
\to \angle A = {\cos ^{ - 1}}\left( {\dfrac{{18}}{{\sqrt {493} }}} \right)
\end{array}\)
