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b) \(S=\frac{1}{2}\sqrt{AB^2.AC^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)
\(=\frac{1}{2}\sqrt{AB^2.AC^2-AB^2.AC^2.cos^2A}\)
\(=\frac{1}{2}\sqrt{AB^2AC^2.sin^2A}\)
\(=\frac{1}{2}.AB.AC.\sin A\) (đpcm)
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b) Ta có :
\(IB=2IC\Leftrightarrow IB=2\left(IB+BC\right)\Leftrightarrow-IB=2BC\Leftrightarrow BI=2BC\)
\(JC=-\frac{1}{2}JA\Leftrightarrow JB+BC=-\frac{1}{2}\left(JB+BA\right)\)
\(\Leftrightarrow\frac{3}{2}JB=-\frac{1}{2}BA-BC\Leftrightarrow JB=-\frac{1}{3}BA-\frac{2}{3}BC\)
\(\Rightarrow BJ=\frac{1}{3}BA+\frac{2}{3}BC\)
\(\Rightarrow IJ=BJ-BI=\frac{1}{3}BA+\frac{2}{3}BC-2BC=\frac{1}{3}BA-\frac{4}{3}BC\)
\(KA=-KB\Leftrightarrow KB+BA=-KB\Leftrightarrow2KB=-BA\)
\(\Rightarrow2BK=BA\Leftrightarrow BK=\frac{1}{2}BA\)
\(\Rightarrow JK=BK-BJ=\frac{1}{2}BA-\frac{2}{3}BC=\frac{1}{6}BA-\frac{2}{3}BC\)
\(=\frac{1}{2}\left(\frac{1}{3}BA-\frac{4}{3}BC\right)=\frac{1}{2}IJ\)
Vậy \(I,J,K\)thẳng hàng
Đok đề cứ thấy sai sai... Sao cho J lại thoả mãn \(\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AC}-\frac{2}{3}\overrightarrow{AB}\) :))
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Rightarrow\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)^2=0\)
\(\Rightarrow-2\left(\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}\right)=GA^2+GB^2+GC^2\)
\(\Rightarrow\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}=-\frac{1}{2}\left(\frac{2}{3}m_a^2+\frac{2}{3}m_b^2+\frac{2}{3}m_c^2\right)\)
\(=-\frac{1}{6}\left(AB^2+BC^2+CA^2\right)\)
Hình như đề bài sai dấu?
