Đáp án:
1B
2B
Giải thích các bước giải:
\(\begin{array}{l}
1)\overrightarrow {AG} = \dfrac{1}{3}\left( {\overrightarrow {AA} + \overrightarrow {AB} + \overrightarrow {AC} } \right) = \dfrac{1}{3}\overrightarrow {AB} + \dfrac{1}{3}\overrightarrow {AC} \\
= - \dfrac{1}{3}\overrightarrow {BA} + \dfrac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) = - \dfrac{1}{3}\overrightarrow {BA} - \dfrac{1}{3}\overrightarrow {BA} + \dfrac{1}{3}\overrightarrow {BC} \\
= - \dfrac{2}{3}\overrightarrow {BA} + \dfrac{1}{3}\overrightarrow {BC} \\
2)\left\{ \begin{array}{l}
mx + 3y - 3 = 0\\
3x + my - 3 = 0
\end{array} \right.\\
\left( {{d_1}} \right)\,cat\,\left( {{d_2}} \right) \Leftrightarrow D = {m^2} - 9 \ne 0 \Leftrightarrow m \ne \pm 3\\
\Leftrightarrow \left\{ \begin{array}{l}
3mx + 9y - 9 = 0\\
3mx + {m^2}y - 3m = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
mx + 3y - 3 = 0\\
\left( {{m^2} - 9} \right)y = 3m - 9
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
mx + 3y - 3 = 0\\
\left( {{m^2} - 9} \right)y = 3m - 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{{m + 3}}\\
y = \dfrac{3}{{m + 3}}
\end{array} \right. \Rightarrow A\left( {\dfrac{3}{{m + 3}};\dfrac{3}{{m + 3}}} \right)\\
\Rightarrow OA = \sqrt {\dfrac{9}{{{{\left( {m + 3} \right)}^2}}} + \dfrac{9}{{{{\left( {m + 3} \right)}^2}}}} = \dfrac{{3\sqrt 2 }}{{\left| {m + 3} \right|}}
\end{array}\)
