b) →IA−2→IB=→0⇒→IA−2→IA−2→AB=→0⇒→IA=−2→AB=−2→a. 3→KA+2→KC=→0⇒3→KA+2→KA+2→AC=→0⇒→AK=25→AC=25→b. Suy ra∙ →IK=→IA+→AK=−2→a+25→b.∙ →BK=→BA+→AK=−→a+25→b.∙ $\overrightarrow{GI}=
\overrightarrow{AI}-
\overrightarrow{AG}=2\overrightarrow{a}+\frac{2}{3}.\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})=\frac{7}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}.$$\bullet$ $\overrightarrow{GK}=
\overrightarrow{AK}-
\overrightarrow{AG}=\frac{2}{5}\overrightarrow{b}+\frac{2}{3}.\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})=\frac{1}{3}\overrightarrow{a}+\frac{11}{15}\overrightarrow{b}.$
b)
→IA−2→IB=→0⇒→IA−2→IA−2→AB=→0⇒→IA=−2→AB=−2→a.
3→KA+2→KC=→0⇒3→KA+2→KA+2→AC=→0⇒→AK=25→AC=25→b. Suy ra
∙ →IK=→IA+→AK=−2→a+25→b.∙ →BK=→BA+→AK=−→a+25→b.∙
$\overrightarrow{GI}=
\overrightarrow{AI}-
\overrightarrow{AG}=2\overrightarrow{a}
-\frac{2}{3}.\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})=\frac{
5}{3}\overrightarrow{a}
-\frac{1}{3}\overrightarrow{b}.$$\bullet$
$\overrightarrow{GK}=
\overrightarrow{AK}-
\overrightarrow{AG}=\frac{2}{5}\overrightarrow{b}
-\frac{2}{3}.\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})=
-\frac{1}{3}\overrightarrow{a}+\frac{1}{15}\overrightarrow{b}.$