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\(\left[{}\begin{matrix}2x_U-3x_A+x_B=0\\2y_U-3y_A+y_B=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x_U=6\\2y_U=2\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(3;1\right)\)
\(b.\left[{}\begin{matrix}3x_U+2x_A+3x_B=3x_C\\3y_U+2y_A+3y_B=3y_C\end{matrix}\right.\left[{}\begin{matrix}3x_U=1\\3y_U=-31\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(\dfrac{1}{3};-\dfrac{31}{3}\right)\)
Lời giải:
Giả sử 3 vecto trên đôi một ngược hướng nhau
\(\overrightarrow{a}, \overrightarrow{b}\) ngược hướng
$\overrightarrow{c},\overrightarrow{b}$ ngược hướng
$\Rightarrow \overrightarrow{a}, \overrightarrow{c}$ cùng ngược hướng với $\overrightarrow{b}$
$\Rightarrow \overrightarrow{a}, \overrightarrow{c}$ cùng hướng (trái giả sử)
Vậy ít nhất 2 trong số 3 vecto cùng hướng.
Bẹn tự vẽ hình nhé
Vì A' đối xứng với B qua A => AA' =AB
=. \(\overrightarrow{A'A}=\overrightarrow{AB}\)
Vì B' đối xứng với C qua B => \(\overrightarrow{B'B}=\overrightarrow{BC}\)
Vì C' đối xứng với A qua C => \(\overrightarrow{C'C}=\overrightarrow{CA}\)
Ta có: \(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\left(\overrightarrow{OA'}+\overrightarrow{A'A}\right)+\left(\overrightarrow{OB'}+\overrightarrow{B'B}\right)+\left(\overrightarrow{OC'}+\overrightarrow{C'C}\right)\)
\(=\left(\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}\right)+\left(\overrightarrow{A'A}+\overrightarrow{B'B}+\overrightarrow{C'C}\right)\)
Lại có: \(\overrightarrow{A'A}+\overrightarrow{B'B}+\overrightarrow{C'C}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\)\(=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CA}=\overrightarrow{AC}+\overrightarrow{CA}=\overrightarrow{AC}-\overrightarrow{AC}=0\)
\(\Rightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}+0=\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}\)
Đặt \(\overrightarrow{b}=x\cdot\overrightarrow{a}+y\cdot\overrightarrow{c}\)
mà \(\overrightarrow{b}=\left(-1;-1\right);\overrightarrow{a}=\left(4;-2\right);\overrightarrow{c}=\left(2;5\right)\)
nên \(\left\{{}\begin{matrix}4x+2y=-1\\-2x+5y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=-1\\-4x+10y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12y=-3\\4x+2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{4}\\4x=-1-2y=-1-2\cdot\dfrac{-1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{8}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(\overrightarrow{b}=\dfrac{-1}{8}\cdot\overrightarrow{a}+\dfrac{-1}{4}\cdot\overrightarrow{c}\)
Chuyển vế: \(\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}-\overrightarrow{AF}-\overrightarrow{BC}-\overrightarrow{ED}\)\(=\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}+\overrightarrow{FA}+\overrightarrow{CB}+\overrightarrow{DE}\)\(=\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{BD}+\overrightarrow{DE}\right)+\left(\overrightarrow{EF}+\overrightarrow{FA}\right)\)\(=\overrightarrow{AB}+\overrightarrow{BE}+\overrightarrow{EA}\)\(=\overrightarrow{AE}+\overrightarrow{EA}\)
\(=0\)
Suy ra: \(\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}=\overrightarrow{AF}+\overrightarrow{BC}+\overrightarrow{ED}\)
Do G là trọng tâm tam giác
\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
Do I là trung điểm AG
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
Ta có: \(\dfrac{-3}{1}\ne\dfrac{0}{2}\Rightarrow\overrightarrow{a}\) và \(\overrightarrow{b}\) ko cùng phương
b. Đặt \(\overrightarrow{c}=x.\overrightarrow{a}+y.\overrightarrow{b}\)
\(\Rightarrow\left(-1;3\right)=x.\left(1;2\right)+y.\left(-3;0\right)=\left(x-3y;2x\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=-1\\2x=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{5}{6}\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{c}=\dfrac{3}{2}\overrightarrow{a}+\dfrac{5}{6}\overrightarrow{b}\)