Đáp án:
4
Giải thích các bước giải:
Đặt \(\overrightarrow {AP} = k\overrightarrow {AC} \) ta có:
\(\begin{array}{l}\overrightarrow {PG} = \overrightarrow {PA} + \overrightarrow {AG} = - k\overrightarrow {AC} + \dfrac{1}{3}\left( {\overrightarrow {AA} + \overrightarrow {AB} + \overrightarrow {AC} } \right)\\ = - k\overrightarrow {AC} + \dfrac{1}{3}\overrightarrow {AB} + \dfrac{1}{3}\overrightarrow {AC} = \left( {\dfrac{1}{3} - k} \right)\overrightarrow {AC} + \dfrac{1}{3}\overrightarrow {AB} \\\overrightarrow {PN} = \overrightarrow {PA} + \overrightarrow {AB} + \overrightarrow {BN} = - k\overrightarrow {AC} + \overrightarrow {AB} + \dfrac{3}{2}\overrightarrow {BC} \\ = - k\overrightarrow {AC} + \overrightarrow {AB} + \dfrac{3}{2}\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) = \left( {\dfrac{3}{2} - k} \right)\overrightarrow {AC} - \dfrac{1}{2}\overrightarrow {AB} \end{array}\)
Do \(N,P,G\) thẳng hàng nên \(\dfrac{{\dfrac{1}{3} - k}}{{\dfrac{3}{2} - k}} = \dfrac{{\dfrac{1}{3}}}{{ - \dfrac{1}{2}}} \Leftrightarrow - \dfrac{1}{2}\left( {\dfrac{1}{3} - k} \right) = \dfrac{1}{3}\left( {\dfrac{3}{2} - k} \right)\)
\( \Leftrightarrow - \dfrac{1}{6} - \dfrac{1}{2} = - \dfrac{1}{3}k - \dfrac{1}{2}k \Leftrightarrow - \dfrac{2}{3} = - \dfrac{5}{6}k \Leftrightarrow k = \dfrac{4}{5}\)
Vậy \(\overrightarrow {AP} = \dfrac{4}{5}\overrightarrow {AC} \Rightarrow \overrightarrow {AP} = 4\overrightarrow {PC} \Rightarrow \dfrac{{PA}}{{PC}} = 4\)
