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Ta có: IJ−→=IA−→+AB−→−+BJ−→IJ→=IA→+AB→+BJ→
IJ−→=ID−→+DC−→−+CJ−→IJ→=ID→+DC→+CJ→
⇒IJ−→=12(AB−→−+DC−→−)⇒IJ→=12(AB→+DC→)
Xét:
HK−→−.IJ→=12(OK−→−−OH−→−).(AB−→−+DC−→−)=12(OK−→−.AB−→−+OK−→−.DC−→−−OH−→−.AB−→−−OH−→−.DC−→−)=12(OK−→−.AB−→−−OH−→−.DC−→−)=12[(OC−→−+CK−→−).(OB−→−−OA−→−)−(OA−→−+AH−→−).(OC−→−−OD−→−)]=12[(OB−→−−OA−→−−AH−→−).OC−→−−(CK−→−+OC−→−−OD−→−).OA−→−]=12[(HA−→−+AO−→−+OB−→−).OC−→−−(DO−→−+OC−→−+CK−→−).OA−→−]=12(HB−→−.OC−→−−DK−→−.OA−→−)=0⇔HK⊥IJ
\(\overrightarrow{CD}+\overrightarrow{CB}=\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{CA}\)
a) Theo bài ra ta có: \(\overrightarrow{AM}=\dfrac{1}{2}.\overrightarrow{AB}\)
\(\overrightarrow{AN}=3.\overrightarrow{NC}\) => \(\overrightarrow{AN}=3.\left(\overrightarrow{AC}-\overrightarrow{AN}\right)\) => \(4.\overrightarrow{AN}=3.\overrightarrow{AC}\)
=> \(\overrightarrow{AN}=\dfrac{3}{4}.\overrightarrow{AC}\)
=> \(\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\dfrac{3}{4}.\overrightarrow{AC}-\dfrac{1}{2}.\overrightarrow{AB}\)
b) Xét tam giác ABC, theo định lý Talet có: \(\dfrac{CN}{CA}=\dfrac{CP}{CB}=\dfrac{1}{3}\)
=> NP// AB => \(\dfrac{NP}{AB}=\dfrac{CN}{CA}=\dfrac{1}{4}\) => \(\overrightarrow{NP}=\dfrac{1}{4}.\overrightarrow{AB}\)
=> \(\overrightarrow{MP}=\overrightarrow{MN}+\overrightarrow{NP}=\dfrac{3}{4}.\overrightarrow{AC}-\dfrac{1}{2}.\overrightarrow{AB}+\dfrac{1}{4}.\overrightarrow{AB}=\dfrac{-1}{2}.\overrightarrow{AB}+\dfrac{3}{4}.\overrightarrow{AC}\)
a: vecto OA+vecto OB+vecto OC+vecto OD
=2 vecto OI+2 vecto OJ
=vecto 0
c: vecto MA+vecto MB+vecto MC+vecto MD
=2 vecto MI+2 vecto MJ
=4 vecto MO