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\(\overrightarrow{u}+\overrightarrow{v}=\left(1+2;2-3\right)=\left(3;-1\right)\)
\(\overrightarrow{u}-\overrightarrow{v}=\left(1-2;2+3\right)=\left(-1;5\right)\)
\(2\overrightarrow{u}=\left(2;4\right)\)
\(3\overrightarrow{v}=\left(6;-9\right)\)
\(2\overrightarrow{u}+3\overrightarrow{v}=\left(2+6;4-9\right)=\left(8;-5\right)\)
Lời giải:
\(\overrightarrow{a}-\overrightarrow{u}+2\overrightarrow{v}=\overrightarrow{0}\)
\(\Leftrightarrow (3-x;2)-(1;-2)+2(3;-1)=(0;0)\)
\(\Rightarrow \left\{\begin{matrix} 3-x-1+2.3=0\\ 2-(-2)+2(-1)=0\end{matrix}\right.\) (vô lý)
a/ \(\overrightarrow{u}=2\left(2;3\right)+3\left(4;1\right)-\left(1;1\right)\)
\(\Leftrightarrow\overrightarrow{u}=\left(4+12-1;6+3-1\right)=\left(15;8\right)\)
b/ \(\left(1;1\right)=x\left(2;3\right)+y\left(4;1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}2x+4y=1\\3x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{10}\\y=\frac{1}{10}\end{matrix}\right.\Rightarrow\overrightarrow{c}=\frac{3}{10}\overrightarrow{a}+\frac{1}{10}\overrightarrow{b}\)
Lời giải:
a) Gọi vecto \(\overrightarrow{u}(m,n)\)
\(\left\{\begin{matrix} \overrightarrow{u}\perp \overrightarrow{a}\\ \overrightarrow{u}.\overrightarrow{b}=-4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3m+6n=0\\ -2m+5n=-4\end{matrix}\right.\)
\(\Rightarrow m=\frac{8}{9}; n=\frac{-4}{9}\)
Vậy \(\overrightarrow{u}(\frac{8}{9}; \frac{-4}{9})\)
b) Gọi vecto \(\overrightarrow{v}(m,n)\)
\(\left\{\begin{matrix} \overrightarrow{v}\perp \overrightarrow{a}\\ |\overrightarrow{v}|=\sqrt{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3m+6n=0\\ m^2+n^2=2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} m=-2n\\ m^2+n^2=2\end{matrix}\right.\Rightarrow (-2n)^2+n^2=2\)
\(\Rightarrow n=\pm \sqrt{\frac{2}{5}}\)
\(\Rightarrow m=\mp 2\sqrt{\frac{2}{5}}\) (tương ứng)
Vậy..............
vecto a=-2*vecto u+vecto v
=>xa=-2*2+3=-1 và ya=-2*3+(-2)=-8