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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 \(AM\perp NP\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)
\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)
\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)
\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)
\(\Leftrightarrow17k=10\)
\(\Leftrightarrow k=\dfrac{10}{17}\)
Gọi O là trung điểm của AM
BM=BC/2=a/2
\(\Leftrightarrow AM=\dfrac{a\sqrt{3}}{2}\)
\(\Leftrightarrow MO=\dfrac{a\sqrt{3}}{4}\)
Xét ΔOMB vuông tại M có
\(BO^2=OM^2+BM^2\)
\(=a^2\cdot\dfrac{3}{16}+a^2\cdot\dfrac{1}{4}=a^2\cdot\dfrac{7}{16}\)
\(\Leftrightarrow BO=\dfrac{a\sqrt{7}}{4}\)
Xét ΔBMA có BO là đường trung tuyến
nên \(\overrightarrow{BM}+\overrightarrow{BA}=2\cdot\overrightarrow{BO}\)
\(\Leftrightarrow\left|\overrightarrow{BM}+\overrightarrow{BA}\right|=\dfrac{a\sqrt{7}}{2}\)
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
vị trí
Xét ΔBAD có BM là đường trung tuyến
nên \(\overrightarrow{BM}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)\)
\(=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)
\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}\)
\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)
=>\(\overrightarrow{BM}=\dfrac{5}{6}\cdot\overrightarrow{BN}\)
=>B,M,N thẳng hàng
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
