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Từ giả thiết ta có PN là đường trung bình tam giác ABC
\(\Rightarrow\overrightarrow{PN}=\dfrac{1}{2}\overrightarrow{BC}=\overrightarrow{BM}\)
Do đó:
\(\overrightarrow{BM}+\overrightarrow{NC}=\overrightarrow{PN}+\overrightarrow{NC}=\overrightarrow{PC}\)
b.
Theo tính chất trọng tâm: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{2}{3}\left(\overrightarrow{AG}+\overrightarrow{GM}\right)\)
\(\Rightarrow\dfrac{1}{3}\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{GM}\Rightarrow2\overrightarrow{MG}=-\overrightarrow{AG}=\overrightarrow{GA}\)
\(\Rightarrow\overrightarrow{GB}+\overrightarrow{GC}+2\overrightarrow{MG}=\overrightarrow{GC}+\overrightarrow{GB}+\overrightarrow{GA}=\overrightarrow{0}\)
Ta có:
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{MB}+4\overrightarrow{MC}\)
\(=6\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{IB}+4\overrightarrow{IC}\)
\(=6\overrightarrow{MI}+4\overrightarrow{IG}+4\overrightarrow{IC}\)
\(=6\overrightarrow{MI}\)
\(\Rightarrow M,I,N\) thẳng hàng
Ta có \(\overrightarrow{IB}=\overrightarrow{BA}\Rightarrow\hept{\begin{cases}I\in AB\\\overrightarrow{AI}=2\overrightarrow{AB}\end{cases}}\). Tương tự \(\hept{\begin{cases}J\in\left[AC\right]\\\overrightarrow{AJ}=\frac{AJ}{AC}\overrightarrow{AC}=\frac{2}{5}\overrightarrow{AC}\end{cases}}\)
Do đó \(\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\frac{2}{5}\overrightarrow{AC}-2\overrightarrow{AB}\)(đpcm).
giải giúp t câu này nha : tính vecto IG theo vecto AB và vecto AC (các b vẽ hình ra hộ t nhé)
Ta có: \(\overrightarrow{IA}-2\cdot\overrightarrow{IB}+4\cdot\overrightarrow{IC}=\overrightarrow{0}\)
=>\(\overrightarrow{IA}-2\left(\overrightarrow{IA}+\overrightarrow{AB}\right)+4\left(\overrightarrow{IA}+\overrightarrow{AC}\right)=\overrightarrow{0}\)
=>\(3\cdot\overrightarrow{IA}-2\cdot\overrightarrow{AB}+4\cdot\overrightarrow{AC}=\overrightarrow{0}\)
=>\(3\cdot\overrightarrow{IA}=2\cdot\overrightarrow{AB}-4\cdot\overrightarrow{AC}\)
=>\(\overrightarrow{IA}=\frac23\cdot\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\)
\(P=\overrightarrow{IA}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\left(\frac23\cdot\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\right)\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\frac23\cdot\left(\overrightarrow{AB}\right)^2-\frac23\cdot\overrightarrow{AB}\cdot\overrightarrow{AC}-\frac43\cdot\left(\overrightarrow{AC}\right)^2\)
\(=\frac23\cdot AB^2-\frac23\cdot AB\cdot AC\cdot cosBAC-\frac43\cdot AC^2\)
\(=\frac23\cdot AB^2-\frac23\cdot AB^2\cdot cos60-\frac43\cdot AB^2=-\frac23\cdot AB^2-\frac23\cdot AB^2\cdot\frac12\)
\(=-AB^2=-a^2\)
Do G là trọng tâm tam giác \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
Ta có:
\(\overrightarrow{GM}-\overrightarrow{NG}+\overrightarrow{GP}=\left(\overrightarrow{GA}+\overrightarrow{AM}\right)-\left(\overrightarrow{NB}+\overrightarrow{BG}\right)+\left(\overrightarrow{GC}+\overrightarrow{CP}\right)\)
\(=\overrightarrow{GA}-\overrightarrow{BG}+\overrightarrow{GC}+\overrightarrow{AM}-\overrightarrow{BG}+\overrightarrow{CP}\)
\(=\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{AM}-\overrightarrow{BG}+\overrightarrow{CP}\)
\(=\overrightarrow{AM}-\overrightarrow{BG}+\overrightarrow{CP}\)
\(4\cdot\overrightarrow{CI}+\overrightarrow{AC}=\overrightarrow{0}\)
=>\(4\cdot\overrightarrow{CI}=-\overrightarrow{AC}=\overrightarrow{CA}\)
=>CA=4CI
\(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\frac14\cdot\overrightarrow{CA}\)
\(=-\overrightarrow{AB}+\overrightarrow{AC}-\frac14\cdot\overrightarrow{AC}=-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\)
\(\overrightarrow{BJ}=\frac12\cdot\overrightarrow{AC}-\frac23\cdot\overrightarrow{AB}\)
\(=\frac23\left(-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{BI}\)
=>B,I,J thẳng hàng