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\(\left|\overrightarrow{a}+\overrightarrow{b}\right|^2=\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
\(=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2+2\overrightarrow{a}.\overrightarrow{b}\)
\(=5^2+12^2+2.5.12.cos\left(\overrightarrow{a},\overrightarrow{b}\right)\)
\(=169+120cos\left(\overrightarrow{a},\overrightarrow{b}\right)=13^2\)
Suy ra: \(cos\left(\overrightarrow{a};\overrightarrow{b}\right)=0\).
\(\overrightarrow{a}\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left(\overrightarrow{a}\right)^2+\overrightarrow{a}.\overrightarrow{b}=5^2+5.12.0=25\).
Mặt khác \(\overrightarrow{a}\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left|\overrightarrow{a}\right|.\left|\overrightarrow{a}+\overrightarrow{b}\right|.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\)
\(=5.13.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\).
Vì vậy \(25=5.13.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\).
\(cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)=\dfrac{5}{13}\).
Vậy góc giữa hai véc tơ \(\overrightarrow{a}\) và \(\overrightarrow{a}+\overrightarrow{b}\) là \(\alpha\) sao cho \(cos\alpha=\dfrac{5}{13}\).
b) \(\left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|\) khi vectơ a và vectơ b cùng hướng
a) \(\overrightarrow{a}+\overrightarrow{b}=\left(2;-2\right)+\left(1;4\right)=\left(3;2\right)\).
\(\overrightarrow{a}-\overrightarrow{b}=\left(2;-2\right)-\left(1;4\right)=\left(1;-6\right)\).
\(2\overrightarrow{a}+3\overrightarrow{b}=2\left(2;-2\right)+3\left(1;4\right)=\left(4;-4\right)+\left(3;12\right)\)\(=\left(7;8\right)\).
c) Gọi x và y là hai số thực để:
\(\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}=x\left(2;-2\right)+y\left(1;4\right)=\left(2x+y;-2x+4y\right)\)
Từ đó suy ra: \(\left\{{}\begin{matrix}2x+y=5\\-2x+4y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\).
Vậy \(\overrightarrow{c}=2\overrightarrow{a}+1\overrightarrow{b}\).
\(\left(a+2b\right)^2=28\Leftrightarrow a^2+4b^2+4ab=28\)
\(\Rightarrow ab=\frac{28-4^2-4.3^2}{4}=-6\)
\(\Rightarrow cos\left(a;b\right)=-\frac{6}{4.3}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)
\(A^2=\left|3a+5b\right|^2=9a^2+25b^2+30ab=9.1+25.1+30.3=124\)
\(\Rightarrow A=2\sqrt{31}\)
\(u.v=0\Leftrightarrow\left(2a+3b\right)\left(-15a+14b\right)=0\)
\(\Leftrightarrow-30a^2+42b^2-17ab=0\)
\(\Leftrightarrow ab=\frac{-30.4^2+42.3^2}{17}=-6\)
\(\Rightarrow cos\left(a;b\right)=\frac{ab}{\left|a\right|\left|b\right|}=-\frac{6}{12}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)
\(\left|\overrightarrow{a}+\overrightarrow{a}\right|=\left|2\overrightarrow{a}\right|=2\left|\overrightarrow{a}\right|=2.50=100\)
Akai Haruma