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\(AC=\sqrt{AB^2+BC^2}=a\sqrt{5}\)
\(BD=\sqrt{AD^2+AB^2}=a\sqrt{2}\)
\(\overrightarrow{AC}.\overrightarrow{BD}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\left(\overrightarrow{BA}+\overrightarrow{AD}\right)\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AB}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BA}+\overrightarrow{BC}.\overrightarrow{AD}\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AD}.2\overrightarrow{AD}=-\overrightarrow{AB}^2+2\overrightarrow{AD}^2\)
\(=-a^2+2a^2=a^2\)
\(cos\left(\overrightarrow{AC};\overrightarrow{BD}\right)=\dfrac{\overrightarrow{AC}.\overrightarrow{BD}}{AC.BD}=\dfrac{a^2}{a\sqrt{2}.a\sqrt{5}}=\dfrac{1}{\sqrt{10}}\)
Tất cả biểu thức đều là vecto, cái nào là độ dài thì nằm trong trị tuyệt đối:
\(\left|BD\right|=\sqrt{AB^2+AD^2}=a\sqrt{5}\)
\(\left|AC\right|=\sqrt{AB^2+BC^2}=a\sqrt{13}\)
a/ \(AB.BD=-BA.BD=-\left|AB\right|.\left|BD\right|.cos\widehat{ABD}\)
\(=-2a.a\sqrt{5}.\frac{2a}{a\sqrt{5}}=-4a^2\)
\(BC.BD=\left|BC\right|.\left|BD\right|.cos\widehat{DBC}=3a.a\sqrt{5}.\frac{a}{a\sqrt{5}}=3a^2\)
\(AC.BD=AC\left(BA+AD\right)=AC.BA+AC.AD\)
\(=AC.AD-AC.AB=\left|AC\right|.\left|AD\right|.cos\widehat{DAC}-\left|AB\right|.\left|AC\right|.cos\widehat{BAC}\)
\(=a.a\sqrt{13}.\frac{3a}{a\sqrt{13}}-2a.a\sqrt{13}.\frac{2a}{a\sqrt{13}}=-a^2\)
\(AC.IJ=\frac{1}{2}AC\left(AD+BC\right)=\frac{1}{2}AC.AD+\frac{1}{2}AC.BC\)
Ta có \(AC.AD=3a^2\) (ngay bên trên)
\(AC.BC=CA.CB=\left|CA\right|.\left|CB\right|.cos\widehat{BCA}=a\sqrt{13}.3a.\frac{3a}{a\sqrt{13}}=9a^2\)
\(\Rightarrow AC.IJ=6a^2\)
a) ta có : \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NB}+\overrightarrow{DM}+\overrightarrow{MN}+\overrightarrow{NC}\)
\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{DM}\right)+\left(\overrightarrow{NB}+\overrightarrow{NC}\right)=2\overrightarrow{MN}\left(đpcm\right)\)
b) ta có : \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AI}+\overrightarrow{IJ}+\overrightarrow{JB}+\overrightarrow{CI}+\overrightarrow{IJ}+\overrightarrow{JD}\)
\(=2\overrightarrow{IJ}+\left(\overrightarrow{AI}+\overrightarrow{CI}\right)+\left(\overrightarrow{JB}+\overrightarrow{JD}\right)=2\overrightarrow{IJ}\left(đpcm\right)\)
bn dùng định lí ta lét chứng minh được \(\overrightarrow{MJ}=\overrightarrow{IN}=\dfrac{1}{2}\overrightarrow{AB}\)
C) ta có : \(\overrightarrow{MN}+\overrightarrow{IJ}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{IA}+\overrightarrow{AB}+\overrightarrow{BJ}\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{MA}+\overrightarrow{BJ}\right)+\left(\overrightarrow{BN}+\overrightarrow{IA}\right)\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{DM}+\overrightarrow{JD}\right)+\left(\overrightarrow{NC}+\overrightarrow{CI}\right)=2\overrightarrow{AB}+\overrightarrow{JM}+\overrightarrow{NI}\) \(=2\overrightarrow{AB}+\overrightarrow{BA}=\overrightarrow{AB}\left(đpcm\right)\)d) ta có : \(\overrightarrow{IM}+\overrightarrow{IN}=\overrightarrow{IJ}+\overrightarrow{JM}+\overrightarrow{IN}=\overrightarrow{IJ}\left(đpcm\right)\)
a) Chữa đề: \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(Ta\text{ }có:\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{DA}+\overrightarrow{AB}\\ =\overrightarrow{CB}+\overrightarrow{DA}+\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\)\(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{DC}\\ =2\overrightarrow{CM}+2\overrightarrow{NC}=2\left(\overrightarrow{NC}+\overrightarrow{CM}\right)=2\overrightarrow{NM}\)
Vậy \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(\text{b) }\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{BC}=-\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{CA}+\overrightarrow{CB}\right)\\ =-\left[\left(\overrightarrow{DA}+\overrightarrow{DB}\right)+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\right]\\ =-\left(2\overrightarrow{DM}+2\overrightarrow{CM}\right)=2\left(\overrightarrow{MD}+\overrightarrow{MC}\right)=4\left(\overrightarrow{MN}\right)\)
\(\text{c) }2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{DA}\right)+\left(\overrightarrow{AI}+\overrightarrow{NA}\right)\right]\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{DB}\right)+\overrightarrow{NI}\right]=2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)\)
Mà IN là dường trung bình \(\Delta BCD\)
\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)
\(\left|\overrightarrow{MA}+\overrightarrow{MC}-\overrightarrow{MN}\right|=\left|\overrightarrow{MA}+\overrightarrow{MD}+\overrightarrow{DC}-\overrightarrow{MN}\right|\)\(=\left|\overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}-\frac{1}{2}\overrightarrow{AB}\right|=\left|\overrightarrow{DC}-\frac{3}{4}\overrightarrow{DC}\right|=\frac{1}{A}DC=\frac{a}{2}\)