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Gọi G là trọng tâm tam giác \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\)
\(\overrightarrow{MA}^2+\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MA}.\overrightarrow{MC}=0\)
\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)=0\)
\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right)=0\)
\(\Leftrightarrow3\overrightarrow{MA}.\overrightarrow{MG}=0\)
\(\Rightarrow\) M thuộc đường tròn đường kính AG
Bán kính: \(R=\dfrac{1}{2}AG=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}\)
a) Ta có:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+k\overrightarrow{BC}\)
\(=\overrightarrow{AB}+k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
\(=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)
b) \(\overrightarrow{NP}=\overrightarrow{AP}-\overrightarrow{AN}\)
\(=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{3}{4}\overrightarrow{AB}\)
Để \(AM\perp NP\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)
\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)
\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)
\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)
\(\Leftrightarrow17k=10\)
\(\Leftrightarrow k=\dfrac{10}{17}\)
\(\overrightarrow{AB}.\overrightarrow{CB}+\overrightarrow{AC}.\overrightarrow{BC}=12\)
\(\Leftrightarrow\overrightarrow{BC}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=12\)
\(\Leftrightarrow\overrightarrow{BC}.\overrightarrow{BC}=12\)
\(\Rightarrow BC^2=12\Rightarrow BC=2\sqrt{3}\)
\(\overrightarrow{BM}=\dfrac{1}{3}\overrightarrow{MC}=\dfrac{1}{3}\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\Rightarrow\overrightarrow{BM}=\dfrac{1}{4}\overrightarrow{BC}\)
\(k\overrightarrow{AN}=\overrightarrow{CN}=\overrightarrow{CA}+\overrightarrow{AN}\Rightarrow\left(1-k\right)\overrightarrow{AN}=\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\)
\(\Rightarrow\overrightarrow{AN}=\dfrac{1}{1-k}\overrightarrow{AB}+\dfrac{1}{1-k}\overrightarrow{AD}\)
\(\overrightarrow{AM}.\overrightarrow{DN}=0\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{BM}\right)\left(\overrightarrow{DA}+\overrightarrow{AN}\right)=0\)
\(\Leftrightarrow\left(\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AD}\right)\left(\dfrac{1}{1-k}\overrightarrow{AB}+\dfrac{k}{1-k}\overrightarrow{AD}\right)=0\)
\(\Rightarrow\dfrac{1}{1-k}AB^2+\dfrac{k}{4\left(1-k\right)}AD^2=0\)
\(\Leftrightarrow\dfrac{1}{1-k}+\dfrac{k}{4\left(1-k\right)}=0\Leftrightarrow k=-4\)
Đáp án B
\(\overrightarrow{AM}.\overrightarrow{AB}=AM^2=\overrightarrow{AM}^2\)
\(\Leftrightarrow\overrightarrow{AM}\left(\overrightarrow{AB}-\overrightarrow{AM}\right)=0\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{MB}=0\)
\(\Rightarrow AM\perp BM\)
\(\Rightarrow\) Quỹ tích là đường tròn đường kính AB
ta có: I là trung điểm của AB
=>\(IA=IB=\dfrac{AB}{2}\)
M là trung điểm của IB
=>\(MI=MB=\dfrac{IB}{2}=\dfrac{AB}{4}\)
AM=AI+IM=1/2AB+1/4AB=3/4AB
=>AM=MB
=>\(\overrightarrow{AM}=3\overrightarrow{MB}\)
=>\(\overrightarrow{AM}-3\overrightarrow{MB}=\overrightarrow{0}\)
=>\(\overrightarrow{AM}+3\overrightarrow{BM}=\overrightarrow{0}\)
=>Chọn C
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