Cho tam giác ABC, M là điểm thuộc BC sao cho MC=2MB

* Hướng dẫn giải

Đặt: \(\overrightarrow{AB}=\overrightarrow{a}\,;\,\,\overrightarrow{AC}=\overrightarrow{b}\) và \(\overrightarrow{AK}=t.\overrightarrow{AM}\)

Khi đó: \(\overrightarrow{BK}=\left( \frac{2t}{3}-1 \right)\overrightarrow{a}+\frac{t}{3}.\overrightarrow{b}\)

\(\overrightarrow{BN}=-\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}\)

Do B, N, K thẳng hàng nên \(\exists \,m:\overrightarrow{BK}=m\overrightarrow{BN}\Leftrightarrow \left( \frac{2t}{3}-1 \right)\overrightarrow{a}+\frac{t}{3}.\overrightarrow{b}=m\left( -\overrightarrow{a}+\frac{2}{3}\overrightarrow{b} \right)\)

\( \Leftrightarrow \left\{ \begin{array}{l}
\frac{{2t}}{3} - 1 =  - m\\
\frac{t}{3} = \frac{{2m}}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = \frac{3}{7}\\
t = \frac{6}{7}
\end{array} \right.\)

Suy ra \(\overrightarrow{AK}=\frac{6}{7}.\overrightarrow{AM}\Rightarrow \overrightarrow{AK}=6.\overrightarrow{KM}\Rightarrow AK=6.KM\)

Vậy \(\frac{KM}{KA}=\frac{1}{6}\)