a: \(\overrightarrow{a}=\left(2;-3\right);\overrightarrow{b}=\left(4;8\right);\overrightarrow{c}=\left(-7;3\right)\)
Tọa độ của \(\overrightarrow{a}+\overrightarrow{b}\) là:
\(\left\{{}\begin{matrix}x=2+4=6\\y=-3+8=5\end{matrix}\right.\)
Tọa độ của \(3\overrightarrow{a}-2\overrightarrow{b}+5\overrightarrow{c}\) là:
\(\left\{{}\begin{matrix}x=3\cdot2-2\cdot4+5\left(-7\right)=-37\\y=3\cdot\left(-3\right)-2\cdot8+5\cdot3=-10\end{matrix}\right.\)
b: \(\overrightarrow{2a}+\overrightarrow{u}-\overrightarrow{c}=\overrightarrow{0}\)
=>\(\overrightarrow{u}=-2\overrightarrow{a}+\overrightarrow{c}\)
Tọa độ của vecto u là:
\(\left\{{}\begin{matrix}x=-2\cdot2+\left(-7\right)=-11\\y=-2\cdot\left(-3\right)+3=6+3=9\end{matrix}\right.\)
c: Đặt \(\overrightarrow{c}=x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}\)
=>\(\left\{{}\begin{matrix}-7=2x+4y\\3=-3x+8y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+8y=-14\\-3x+8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=-17\\-3x+8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{17}{7}\\8y=3x+3=\dfrac{-51}{7}+3=-\dfrac{30}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{17}{7}\\y=-\dfrac{30}{7\cdot8}=-\dfrac{30}{56}=-\dfrac{15}{28}\end{matrix}\right.\)
Vậy: \(\overrightarrow{c}=-\dfrac{17}{7}\cdot\overrightarrow{a}+\dfrac{-15}{28}\cdot\overrightarrow{b}\)
d: \(\left|\overrightarrow{a}\right|=\sqrt{2^2+\left(-3\right)^2}=\sqrt{13}\)
\(\overrightarrow{b}-\overrightarrow{c}=\left(11;5\right)\)
=>\(\left|\overrightarrow{b}-\overrightarrow{c}\right|=\sqrt{11^2+5^2}=\sqrt{146}\)
e: \(\overrightarrow{a}\cdot\overrightarrow{b}=2\cdot4+\left(-3\right)\cdot8=-24+8=-16\)
\(\overrightarrow{a}-\overrightarrow{b}=\left(-2;-11\right);\overrightarrow{b}+\overrightarrow{c}=\left(-3;11\right)\)
\(\left(\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{b}+\overrightarrow{c}\right)=\left(-2\right)\cdot\left(-3\right)+\left(-11\right)\cdot11=-121+6=-115\)
f: \(cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\dfrac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{a}\right|\cdot\left|\overrightarrow{b}\right|}=\dfrac{-16}{\sqrt{2^2+\left(-3\right)^2}\cdot\sqrt{4^2+8^2}}\)
\(=\dfrac{-16}{\sqrt{13}\cdot4\sqrt{5}}=-\dfrac{4}{\sqrt{65}}\)
