Đáp án:
Giải thích các bước giải:
Gọi I là trung điểm AB
\(\begin{array}{l}
a.\overrightarrow {GC} = \frac{2}{3}\overrightarrow {IC} = \frac{2}{3}\overrightarrow {IB} + \frac{2}{3}\overrightarrow {BC} = \frac{2}{3}.\frac{1}{2}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {GC} - \frac{2}{3}\overrightarrow {GB} \\
\to \frac{1}{3}\overrightarrow {GC} = \frac{1}{3}\overrightarrow {GB} - \frac{1}{3}\overrightarrow {GA} - \frac{2}{3}\overrightarrow {GB} = - \frac{1}{3}\overrightarrow {GA} - \frac{1}{3}\overrightarrow {GB} \\
\to \overrightarrow {GC} = - \overrightarrow {GA} - \overrightarrow {GB} \\
b.\overrightarrow {MN} = \frac{5}{7}\overrightarrow {MA} = \frac{5}{7}\overrightarrow {MB} + \frac{5}{7}\overrightarrow {BA} = \frac{5}{7}.\frac{3}{4}\overrightarrow {CB} + \frac{5}{7}\overrightarrow {GA} - \frac{5}{7}\overrightarrow {GB} \\
= \frac{{15}}{{28}}\overrightarrow {GB} - \frac{{15}}{{28}}\overrightarrow {GC} + \frac{5}{7}\overrightarrow {GA} - \frac{5}{7}\overrightarrow {GB} \\
= \frac{{ - 5}}{{28}}\overrightarrow {GB} + \frac{5}{7}\overrightarrow {GA} + \frac{{15}}{{28}}\overrightarrow {GA} + \frac{{15}}{{28}}\overrightarrow {GB} \\
= \frac{5}{{14}}\overrightarrow {GB} + \frac{5}{4}\overrightarrow {GA}
\end{array}\)
