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Câu 4:
Áp dụng định lý Pytago
\(BC^2=AB^2+AC^2\Rightarrow BC=2\)
Ta có:
\(\overrightarrow{CA}.\overrightarrow{BC}=-\overrightarrow{CA}.\overrightarrow{CB}=-\dfrac{CA^2+CB^2-AB^2}{2}=-\dfrac{2+4-2}{2}=-2\)
Câu 5:
Gọi M là trung điểm BC
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
Mà: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
Câu 6:
\(\left|\overrightarrow{a}-\overrightarrow{b}\right|=3\)
\(a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=9\)
\(\overrightarrow{a}.\overrightarrow{b}=\dfrac{1^2+2^2-9}{2}=-2\)
Câu 7:
\(\left|\overrightarrow{AB}-\overrightarrow{AD}+\overrightarrow{CD}\right|=\left|\overrightarrow{DB}+\overrightarrow{CD}\right|\)
\(=\left|\overrightarrow{DB}-\overrightarrow{DC}\right|=\left|\overrightarrow{CB}\right|=BC=a\)
\(4\cdot\overrightarrow{CI}+\overrightarrow{AC}=\overrightarrow{0}\)
=>\(4\cdot\overrightarrow{CI}=-\overrightarrow{AC}=\overrightarrow{CA}\)
=>CA=4CI
\(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\frac14\cdot\overrightarrow{CA}\)
\(=-\overrightarrow{AB}+\overrightarrow{AC}-\frac14\cdot\overrightarrow{AC}=-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\)
\(\overrightarrow{BJ}=\frac12\cdot\overrightarrow{AC}-\frac23\cdot\overrightarrow{AB}\)
\(=\frac23\left(-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{BI}\)
=>B,I,J thẳng hàng
a) II là điểm trên cạnh BCBC mà: 2CI=3BI⇒BICI=232CI=3BI⇒BICI=23
⇒BICI+BI=23+2⇒BIBC=25⇒BICI+BI=23+2⇒BIBC=25
⇒BI=25BC⇒BI=25BC tương tự IC=35BCIC=35BC
JJ là điểm trên BCBC kéo dài: 5JB=2JC⇒JBJC=255JB=2JC⇒JBJC=25
⇒JBJC−JB=25−2⇒JBBC=23⇒JBJC−JB=25−2⇒JBBC=23
⇒JB=23BC⇒JB=23BC và BC=35JCBC=35JC
→AB=→AI+→IBAB→=AI→+IB→
=→AI−25→BC=AI→−25BC→
=→AI−25.32→JB=AI→−25.32JB→
=→AI−35→JB=AI→−35JB→
=→AI−35(→JA+→AB)=AI→−35(JA→+AB→)
=→AI+35→AJ−35→AB=AI→+35AJ→−35AB→
⇒→AB+35→AB=→AI+35→AJ⇒AB→+35AB→=AI→+35AJ→
⇒→AB=58→AI+38→AJ⇒AB→=58AI→+38AJ→
→AC=→AI+→ICAC→=AI→+IC→
=→AI+35→BC=AI→+35BC→
=→AI+35.35→JC=AI→+35.35JC→
=→AI+925(→JA+→AC)=AI→+925(JA→+AC→)
⇒→AC−925→AC=→AI−925→AJ⇒AC→−925AC→=AI→−925AJ→
⇒→AC=2516→AI−916→AJ⇒AC→=2516AI→−916AJ→
⇒52→AB=2516→AI+1516→AJ⇒52AB→=2516AI→+1516AJ→
và →AC=2516→AI−916→AJAC→=2516AI→−916AJ→
Trừ vế với vế ta có:
52→AB−→AC=32→AJ52AB→−AC→=32AJ→
⇒→AJ=53→AB−23→AC
Ta có: \(\overrightarrow{IA}-2\cdot\overrightarrow{IB}+4\cdot\overrightarrow{IC}=\overrightarrow{0}\)
=>\(\overrightarrow{IA}-2\left(\overrightarrow{IA}+\overrightarrow{AB}\right)+4\left(\overrightarrow{IA}+\overrightarrow{AC}\right)=\overrightarrow{0}\)
=>\(3\cdot\overrightarrow{IA}-2\cdot\overrightarrow{AB}+4\cdot\overrightarrow{AC}=\overrightarrow{0}\)
=>\(3\cdot\overrightarrow{IA}=2\cdot\overrightarrow{AB}-4\cdot\overrightarrow{AC}\)
=>\(\overrightarrow{IA}=\frac23\cdot\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\)
\(P=\overrightarrow{IA}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\left(\frac23\cdot\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\right)\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\frac23\cdot\left(\overrightarrow{AB}\right)^2-\frac23\cdot\overrightarrow{AB}\cdot\overrightarrow{AC}-\frac43\cdot\left(\overrightarrow{AC}\right)^2\)
\(=\frac23\cdot AB^2-\frac23\cdot AB\cdot AC\cdot cosBAC-\frac43\cdot AC^2\)
\(=\frac23\cdot AB^2-\frac23\cdot AB^2\cdot cos60-\frac43\cdot AB^2=-\frac23\cdot AB^2-\frac23\cdot AB^2\cdot\frac12\)
\(=-AB^2=-a^2\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\frac14\cdot\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\frac14\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\frac14\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)
\(=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)
BC=BM+CM
=>4BM=BM+CM
=>CM=3BM
=>\(MC=\frac34BC\)
=>\(\overrightarrow{MC}=\frac34\cdot\overrightarrow{BC}=\frac34\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac34\cdot\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\)