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a) \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\dfrac{-1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
b) CG.CAN??
1: MC=2MB
MC+MB=BC
Do đó: BC=2MB+BM=3BM
=>\(BM=\frac13BC\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\frac13\cdot\overrightarrow{BC}=\overrightarrow{AB}+\frac13\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}\)
\(=\frac13\left(2\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)
2: Xét ΔMAC có MD là đường trung tuyến
nên \(\overrightarrow{MD}=\frac12\left(\overrightarrow{MA}+\overrightarrow{MC}\right)=\frac12\left(\frac23\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}+\frac23\cdot\overrightarrow{BC}\right)\)
\(=\frac12\left(\frac23\overrightarrow{AC}+\frac13\cdot\overrightarrow{AC}\right)=\frac12\cdot\overrightarrow{AC}=\frac12\left(-\overrightarrow{BA}+\overrightarrow{BC}\right)\)
3: \(\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac12\cdot\overrightarrow{BD}\)
\(=\overrightarrow{AB}+\frac14\cdot\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{BC}\)
\(=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{BA}+\frac14\cdot\overrightarrow{AC}=\frac12\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)
\(=\frac14\left(2\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)
=>\(\frac{\overrightarrow{AE}}{\overrightarrow{AM}}=\frac14:\frac13=\frac34\)
=>A,E,M thẳng hàng
a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
Do G là trọng tâm tam giác
\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
Do I là trung điểm AG
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
a: Ta có: M là trung điểm của BC
=>\(\overrightarrow{MB}=\overrightarrow{CM}\)
\(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AM}\)
\(\overrightarrow{AC}+\overrightarrow{MB}=\overrightarrow{AC}+\overrightarrow{CM}=\overrightarrow{AM}\)
Do đó: \(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{MB}\)
b:
ΔABC cân tại A
=>\(\hat{ABC}=\hat{ACB}=\frac{180^0-\hat{BAC}}{2}=\frac{180^0-120^0}{2}=30^0\)
ΔABC cân tại A
mà AM là đường trung tuyến
nên AM⊥BC
Xét ΔAMB vuông tại M có \(\sin B=\frac{AM}{AB}\)
=>\(\frac{AM}{6}=\sin30=\frac12\)
=>AM=3(cm)
\(\left|\overrightarrow{AB}-\overrightarrow{CM}\right|=\left|\overrightarrow{AB}+\overrightarrow{MC}\right|\)
\(=\left|\overrightarrow{AB}+\overrightarrow{BM}\right|=\left|\overrightarrow{AM}\right|=AM\) =3(cm)
Sửa đề: Chứng minh \(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{MB}\)
\(\overrightarrow{AB}-\overrightarrow{MB}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AM}\)
\(\overrightarrow{AC}-\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{CM}=\overrightarrow{AC}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{MB}=\overrightarrow{AC}-\overrightarrow{MC}\)
=>\(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{MB}\)
Gọi M là trung điểm BC, theo tính chất trọng tâm:
\(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)
Mà I là trung điểm AG \(\Rightarrow\overrightarrow{IG}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AM}\Rightarrow\overrightarrow{GI}=-\dfrac{1}{3}\overrightarrow{AM}\)
Lại có: M là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}\)
Nên ta có:
\(\overrightarrow{AB}+\overrightarrow{AC}+6\overrightarrow{GI}=\overrightarrow{AM}+\overrightarrow{MB}+\overrightarrow{AM}+\overrightarrow{MC}+6.\left(-\dfrac{1}{3}\right)\overrightarrow{AM}\)
\(=2\overrightarrow{AM}-2\overrightarrow{AM}=\overrightarrow{0}\) (đpcm)
ΔBAC vuông tại B
=>\(BA^2+BC^2=AC^2\)
=>\(AC^2=\left(3a\right)^2+\left(4a\right)^2=9a^2+16a^2=25a^2=\left(5a\right)^2\)
=>AC=5a
\(\overrightarrow{MA}+\overrightarrow{MB}-\overrightarrow{MC}=\overrightarrow{0}\)
=>\(\overrightarrow{MB}+\overrightarrow{MA}+\overrightarrow{CM}=\overrightarrow{0}\)
=>\(\overrightarrow{MB}+\overrightarrow{CA}=\overrightarrow{0}\)
=>\(\overrightarrow{MB}=-\overrightarrow{CA}=\overrightarrow{AC}\)
=>AMBC là hình bình hành
Xét ΔABC vuông tại B có sin BAC=\(\frac{BC}{CA}=\frac{4a}{5a}=\frac45\) ; cos BAC=\(\frac{BA}{AC}=\frac{3a}{5a}=\frac35\)
\(\sin CAM=\sin\left(CAB+BAM\right)\)
\(=\sin90^0\cdot cosBAC+cos90^0\cdot\sin BAC=cosBAC\) =3/5
=>\(cosCAM=-\sqrt{1-\sin^2CAM}=-\frac45\)
N là trung điểm của AC
=>\(AN=\frac{AC}{2}=2,5a\)
AMBC là hình bình hành
=>AM=BC=4a
Xét ΔAMN có \(cosNAM=\frac{AN^2+AM^2-MN^2}{2\cdot AN\cdot AM}\)
=>\(\frac{\left(2.5a\right)^2+\left(4a\right)^2-MN^2}{2\cdot2,5a\cdot4a}=\frac{-4}{5}\)
=>\(6,25a^2+16a^2-MN^2=-\frac45\cdot5a\cdot4a=-4a\cdot4a=-16a^2\)
=>\(MN^2=6.25a^2+16a^2+16a^2=38,25a^2\)
=>\(MN=\sqrt{38,25a^2}=a\cdot\sqrt{\frac{153}{4}}=a\cdot\frac{3\sqrt{17}}{2}\)