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a.
\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)
\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)
\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)
b.
\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)
\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)
c.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)
Câu 2:
a: \(\overrightarrow{AB}+\overrightarrow{CD}\)
\(=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{CI}+\overrightarrow{ID}\)
\(=\overrightarrow{IB}+\overrightarrow{ID}=2\overrightarrow{IJ}\)
b: \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{a}-\dfrac{1}{2}\overrightarrow{b}\)
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..............