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\(\overrightarrow{BM}=\overrightarrow{BC}-2\overrightarrow{AB}\Leftrightarrow\overrightarrow{BI}+\overrightarrow{IM}=\overrightarrow{BC}-2\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)
\(\Leftrightarrow\frac{1}{2}\overrightarrow{BC}+\overrightarrow{IM}=\overrightarrow{BC}-2\overrightarrow{AC}+2\overrightarrow{BC}\Rightarrow\overrightarrow{IM}=\frac{5}{2}\overrightarrow{BC}-2\overrightarrow{AC}\)
\(\overrightarrow{CI}+\overrightarrow{IN}=x\overrightarrow{AC}-\overrightarrow{BC}\Rightarrow-\frac{1}{2}\overrightarrow{BC}+\overrightarrow{IN}=x\overrightarrow{AC}-\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{IN}=-\frac{1}{2}\overrightarrow{BC}+x\overrightarrow{AC}=-\frac{1}{5}\left(\frac{5}{2}\overrightarrow{BC}-5x.\overrightarrow{AC}\right)\)
Để MN qua I hay I;M;N thẳng hàng \(\Leftrightarrow5x=2\Rightarrow x=\frac{2}{5}\)
Lời giải:
$\overrightarrow{CM}.\overrightarrow{BN}=(\overrightarrow{CA}+\overrightarrow{AM})(\overrightarrow{BA}+\overrightarrow{AN})$
$=\overrightarrow{CA}.\overrightarrow{BA}+\overrightarrow{CA}.\overrightarrow{AN}+\overrightarrow{AM}.\overrightarrow{BA}+\overrightarrow{AM}.\overrightarrow{AN}$
$=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{CA}.\frac{1}{4}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{AB}.\overrightarrow{BA}+\frac{1}{5}\overrightarrow{AB}.\frac{1}{4}\overrightarrow{AC}$
$=\frac{21}{20}\overrightarrow{AB}.\overrightarrow{AC}-\frac{1}{4}AC^2-\frac{1}{5}AB^2$
$=\frac{21}{20}\cos A.|\overrightarrow{AB}|.|\overrightarrow{AC}|-\frac{1}{4}AC^2-\frac{1}{5}AB^2$
$=\frac{21}{20}.\frac{1}{2}.5.8-\frac{1}{4}.8^2-\frac{1}{5}.5^2=0$
$\Rightarrow CM\perp BN$
a)\(\overrightarrow{AB}+\overrightarrow{AN}=\overrightarrow{AM}\)
b)\(\overrightarrow{BA}+\overrightarrow{CN}=2\overrightarrow{BA}\)
c)Có \(\overrightarrow{AB}=\overrightarrow{NC}\)=>bt trở thành \(\overrightarrow{NC}+MC+\overrightarrow{MN}=\overrightarrow{MC}-\overrightarrow{MC}=vt0\)
d)có vt BA+vt BC=vtBN
bt trở thành vtMN-vtMN=vt0
hok tốt!
\(\overrightarrow{AM}-\overrightarrow{AN}=\overrightarrow{NM}\)
\(\overrightarrow{MN}-\overrightarrow{NC}=\overrightarrow{CM}\)
a) Ta có:
+) \(\overrightarrow {MB} = \dfrac{1}{2}\overrightarrow {BC} \Rightarrow \overrightarrow {MB} \) và \(\overrightarrow {BC} \) cùng hướng; tỉ số độ dài \(\dfrac{{BC}}{{MB}} = 2\)
\( \Rightarrow M\) nằm ngoài đoạn thẳng BC sao cho \(MB = \dfrac{1}{2}BC\)
+) \({\overrightarrow {AN} = 3\overrightarrow {NB} \Rightarrow \overrightarrow {AB} + \overrightarrow {BN} = 3\overrightarrow {NB} \Rightarrow 4\overrightarrow {NB} = \overrightarrow {AB} \Leftrightarrow \overrightarrow {NB} = \dfrac{1}{4}\overrightarrow {AB} }\)
\( \Rightarrow N\) thuộc đoạn thẳng AB và \(NB=\dfrac{{1}}{{4}} AB\)
+) \(\overrightarrow {CP} = \overrightarrow {PA} \Leftrightarrow \overrightarrow {PC} + \overrightarrow {PA} = \overrightarrow 0 \)
\( \Rightarrow P\) là trung điểm của CA
b) \(\overrightarrow {MN} = \overrightarrow {MB} + \overrightarrow {BN} = \frac{1}{2}\overrightarrow {BC} + \frac{1}{4}\overrightarrow {BA} \)
\(\begin{array}{l}\overrightarrow {MP} = \overrightarrow {MC} + \overrightarrow {CP} = \overrightarrow {MC} + \frac{1}{2}\overrightarrow {CA} \\= \frac{3}{2}\overrightarrow {BC} + \frac{1}{2}\left( {\overrightarrow {BA} - \overrightarrow {BC} } \right)\\ = \overrightarrow {BC} + \frac{1}{2}\overrightarrow {BA} \end{array}\)
c) Ta có:
\(\overrightarrow {MN} = \frac{1}{2}\overrightarrow {BC} + \frac{1}{4}\overrightarrow {BA} ;\) \(\overrightarrow {MP} = \overrightarrow {BC} + \frac{1}{2}\overrightarrow {BA} \)
\( \Rightarrow \overrightarrow {MP} = 2\overrightarrow {MN} \)
Vậy \(M,N,P\) thẳng hàng
a, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{BC}^2\)
\(\Leftrightarrow AC^2+AB^2-2\overrightarrow{AB}.\overrightarrow{AC}=BC^2\)
\(\Leftrightarrow2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{AB^2+AC^2-BC^2}{2}=\dfrac{5^2+8^2-7^2}{2}=20\)
b, \(2\overrightarrow{CA}.\overrightarrow{CB}=CA^2+CB^2-BC^2=CA^2\)
\(\Rightarrow\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2}{2}=\dfrac{8^2}{2}=32\)
Lời giải:
a)
\(\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{BC}\)
\(\Rightarrow (\overrightarrow{AC}-\overrightarrow{AB})^2=\overrightarrow{BC}^2\Leftrightarrow AB^2+AC^2-2\overrightarrow{AC}.\overrightarrow{AB}=BC^2\)
\(\Leftrightarrow 2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\) (đpcm)
Ta có:
\(\overrightarrow{AB}.\overrightarrow{AC}=\frac{AB^2+AC^2-BC^2}{2}=\frac{5^2+8^2-7^2}{2}=20\)
\(\cos \angle A=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|}=\frac{20}{5.8}=\frac{1}{2}\)
\(\Rightarrow \angle A=60^0\)
b)
Tương tự phần a, \(\overrightarrow{CA}.\overrightarrow{CB}=\frac{CA^2+CB^2-AB^2}{2}=\frac{8^2+7^2-5^2}{2}=44\)
