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a: \(\overrightarrow{IB}=2\cdot\overrightarrow{IC}\)
=>C là trung điểm của IB
=>IC=CB=1/2IB
\(\overrightarrow{IK}=\overrightarrow{IB}+\overrightarrow{BK}\)
\(=2\cdot\overrightarrow{CB}+\frac12\cdot\overrightarrow{BA}=2\left(\overrightarrow{CA}+\overrightarrow{AB}\right)-\frac12\cdot\overrightarrow{AB}\)
\(=-2\cdot\overrightarrow{AC}+2\cdot\overrightarrow{AB}-\frac12\cdot\overrightarrow{AB}=-2\cdot\overrightarrow{AC}+\frac32\cdot\overrightarrow{AB}\)
b: JA=2JC
=>\(JA=\frac23AC;JC=\frac13AC\)
\(\overrightarrow{IJ}=\overrightarrow{IC}+\overrightarrow{CJ}=\overrightarrow{CB}+\frac13\cdot\overrightarrow{CA}\)
\(=\overrightarrow{CA}+\overrightarrow{AB}+\frac13\cdot\overrightarrow{CA}=\frac43\cdot\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\)
\(=\frac23\left(\frac32\cdot\overrightarrow{AB}-2\cdot\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{IK}\)
=>I,J,K thẳng hàng
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
a: \(\overrightarrow{MC}=\frac13\cdot\overrightarrow{MB}\)
=>\(MC=\frac13MB\) và C nằm giữa M và B
MC+CB=MB
=>\(CB=MB-MC=MB-\frac13MB=\frac23MB\)
Ta có: \(\overrightarrow{NA}+3\cdot\overrightarrow{NC}=\overrightarrow{0}\)
=>\(\overrightarrow{NA}=-3\cdot\overrightarrow{NC}\)
=>N nằm giữa A và C và NA=3NC
NA+NC=AC
=>AC=NC+3NC=4NC
=>\(CN=\frac14CA\)
=>\(NA=3\cdot\frac14\cdot AC=\frac34AC\)
\(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{0}\)
=>\(\overrightarrow{PA}=-\overrightarrow{PB}\)
=>P nằm giữa A và B và PA=PB
=>P là trung điểm của AB
=>\(AP=PB=\frac{AB}{2}\)
\(\overrightarrow{MP}=\overrightarrow{MB}+\overrightarrow{BP}\)
\(=-\overrightarrow{BM}+\overrightarrow{BP}=-\frac32\cdot\overrightarrow{BC}+\frac12\cdot\overrightarrow{BA}\)
\(=-\frac32\cdot\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\frac12\cdot\overrightarrow{BA}=-\frac32\cdot\overrightarrow{BA}-\frac32\cdot\overrightarrow{AC}+\frac12\cdot\overrightarrow{BA}\)
\(=-\overrightarrow{BA}-\frac32\cdot\overrightarrow{AC}=\overrightarrow{AB}-\frac32\cdot\overrightarrow{AC}\) (2)
\(\overrightarrow{NP}=\overrightarrow{NA}+\overrightarrow{AP}\)
\(=-\frac34\cdot\overrightarrow{AC}+\frac12\cdot\overrightarrow{AB}=\frac12\cdot\left(\overrightarrow{AB}-\frac32\cdot\overrightarrow{AC}\right)\) (1)
b: Từ (1),(2) suy ra \(\overrightarrow{NP}=\frac12\cdot\overrightarrow{MP}\)
=>N,M,P thẳng hàng
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 \(AM\perp NP\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)
\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)
\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)
\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)
\(\Leftrightarrow17k=10\)
\(\Leftrightarrow k=\dfrac{10}{17}\)
a: \(\overrightarrow{AE}=\dfrac{2}{3}\overrightarrow{EC}\)
=>E nằm giữa A và C và AE=2/3EC
Ta có: AE+EC=AC(E nằm giữa A và C)
=>\(AC=\dfrac{2}{3}EC+EC=\dfrac{5}{3}EC\)
=>\(\dfrac{AE}{AC}=\dfrac{\dfrac{2}{3}EC}{\dfrac{5}{3}EC}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)
=>\(AE=\dfrac{2}{5}AC\)
=>\(\overrightarrow{AE}=\dfrac{2}{5}\cdot\overrightarrow{AC}\)
\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}\)
\(=-\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC}\)
b: \(\left|\overrightarrow{IA}+\overrightarrow{IG}\right|=\left|\overrightarrow{IA}-\overrightarrow{IG}\right|\)
=>\(\left[{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IA}-\overrightarrow{IG}\\\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IG}-\overrightarrow{IA}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2\cdot\overrightarrow{IG}=\overrightarrow{0}\\2\cdot\overrightarrow{IA}=\overrightarrow{0}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}I\equiv G\\I\equiv A\end{matrix}\right.\)
vecto AI=1/2*vecto AM
=1/4(vecto AB+vecto AC)
=1/4(vecto AC+vecto CB+vecto AC)
=1/4(-2vecto CA+vecto CB)
=-1/2*vecto CA+1/4*vecto CB
=>m=-1/2; n=1/4