\(\overrightarrow{a}=\left(2;0\right);\overrightarrow{b}=\left(-1;\dfrac{1}{2}\right);\overrightarrow{c}=\left(4;-6\right)\)
a: \(\overrightarrow{d}=2\cdot\overrightarrow{a}-3\cdot\overrightarrow{b}+5\cdot\overrightarrow{c}\)
Tọa độ vecto d là:
\(\left\{{}\begin{matrix}x=2\cdot2-3\cdot\left(-1\right)+5\cdot4=4+20+3=27\\y=2\cdot0-3\cdot\dfrac{1}{2}+5\cdot\left(-6\right)=-30-\dfrac{3}{2}=-31,5\end{matrix}\right.\)
Vậy: \(\overrightarrow{d}=\left(27;-31,5\right)\)
b: Đặt \(\overrightarrow{c}=x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}\)
=>\(\left\{{}\begin{matrix}4=2x+\left(-1\right)y\\-6=0x+\dfrac{1}{2}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-12\\2x-\left(-12\right)=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-12\\2x=4+\left(-12\right)=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-4\\y=-12\end{matrix}\right.\)
vậy: \(\overrightarrow{c}=-4\cdot\overrightarrow{a}-12\cdot\overrightarrow{b}\)
a,
\(\overrightarrow{d}=2.\left(2,0\right)-3.\left(-1,\dfrac{1}{2}\right)+5\left(4,-6\right)=\left(27,-\dfrac{63}{2}\right)\)
b.
Đặt \(\overrightarrow{c}=x.\overrightarrow{a}+y.\overrightarrow{b}\)
\(\Rightarrow\left\{{}\begin{matrix}4=2.x+\left(-1\right).y\\-6=0.x+\dfrac{1}{2}.y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-12\end{matrix}\right.\)
Vậy \(\overrightarrow{c}=-4.\overrightarrow{a}-12.\overrightarrow{b}\)
