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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: \(\overrightarrow{b}=\dfrac{-1}{8}\cdot\overrightarrow{a}+\dfrac{-1}{4}\cdot\overrightarrow{c}\)
b: \(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\)
\(\left[{}\begin{matrix}2x_U-3x_A+x_B=0\\2y_U-3y_A+y_B=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x_U=6\\2y_U=2\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(3;1\right)\)
\(b.\left[{}\begin{matrix}3x_U+2x_A+3x_B=3x_C\\3y_U+2y_A+3y_B=3y_C\end{matrix}\right.\left[{}\begin{matrix}3x_U=1\\3y_U=-31\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(\dfrac{1}{3};-\dfrac{31}{3}\right)\)
a.
\(\overrightarrow{a}.\overrightarrow{b}=2.\left(-3\right)+\left(-1\right).4=-10\)
b.
\(\overrightarrow{a}.\overrightarrow{b}=2.\left(-3\right)+5.1=-1\)