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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) \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\dfrac{-1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
b) CG.CAN??
Xét ΔAEH có AI là đường trung tuyến
nên \(\overrightarrow{AI}=\frac12\left(\overrightarrow{AE}+\overrightarrow{AH}\right)\)
\(=\frac12\left(\frac34\cdot\overrightarrow{AB}+\frac12\cdot\overrightarrow{AC}\right)=\frac12\cdot\frac14\left(3\cdot\overrightarrow{AB}+2\cdot\overrightarrow{AC}\right)=\frac18\left(3\cdot\overrightarrow{AB}+2\cdot\overrightarrow{AC}\right)\)
\(\overrightarrow{AF}=\overrightarrow{AB}+\overrightarrow{BF}=\overrightarrow{AB}+\frac25\cdot\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\frac25\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac35\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AC}=\frac15\left(3\cdot\overrightarrow{AB}+2\cdot\overrightarrow{AC}\right)\)
=>\(\overrightarrow{AI}=\frac58\cdot\overrightarrow{AF}\)
=>A,I,F thẳng hàng
a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}\)(D là trung điểm của BC) (1)
\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AK}\)(K là trung điểm của MN) (2)
Lấy (1) trừ (2) có: \(\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=2\left(\overrightarrow{AD}-\overrightarrow{AK}\right)\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\overrightarrow{AB}+\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)=\(\overrightarrow{KD}\)
AN=2NC
=>\(AN=\frac23AC\)
Xét ΔAMN có AK là đường trung tuyến
nên \(\overrightarrow{AK}=\frac12\left(\overrightarrow{AM}+\overrightarrow{AN}\right)\)
\(=\frac12\left(\frac12\cdot\overrightarrow{AB}+\frac23\cdot\overrightarrow{AC}\right)=\frac14\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}\)
a: vecto MN=vecto MC+vecto CA+vecto AN
=vecto CA+4 vecto BC+3/4 vecto AB
=vecto CA+4(vecto BA+vecto AC)+3/4vecto AB
=vecto CA-4vecto AB-4vecto CA+3/4vecto AB
=-3 vecto CA-13/4vecto AB
=-13/4vecto AB+3 vecto AC
=>-13/4 vecto AB=vecto MN-3 vecto AC
=>vecto AB=-4/13 vecto MN+12/13 vecto AC
b: vecto AE=vecto AN+vecto NE
=3/4vecto AB+1/2vecto NM
=3/4vecto AB-1/2 vecto MN
=3/4*vecto AB-1/2(-13/4vecto AB+3vecto AC)
=3/4*vecto AB+13/8vecto AB-3/2vecto AC
=19/8vecto AB-3/2vecto AC