Question

[Sách Giáo Khoa]

Cho \(\overrightarrow{a}=\left(2;1\right);\overrightarrow{b}=\left(3;-4\right);\overrightarrow{c=}\left(-7;2\right)\)

a) Tìm tọa độ của vectơ \(\overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}-4\overrightarrow{c}\)

b) Tìm tọa độ vectơ \(\overrightarrow{x}\) sao cho :  \(\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\)

c) Tìm các số k và h sao cho : \(\overrightarrow{c}=k\overrightarrow{a}+h\overrightarrow{b}\)

Answer 1

[Bùi Thị Vân]

a) \(\overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}-4\overrightarrow{c}=3\left(2;1\right)+2\left(3;-4\right)-4\left(-7;2\right)\)
\(=\left(6;3\right)+\left(6;-8\right)-\left(-28;8\right)\)
\(=\left(6+6+28;3-8-8\right)=\left(40;-13\right)\).
b) \(\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Leftrightarrow\overrightarrow{x}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Leftrightarrow\overrightarrow{x}=\left(3;-4\right)-\left(-7;2\right)-\left(2;1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(3+7-2;-4-2-1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(8;-7\right)\).
c) Có \(\overrightarrow{c}\left(-7;2\right)=k\overrightarrow{a}+h\overrightarrow{b}=k\left(2;1\right)+h\left(3;-4\right)\)
\(=\left(2k+3h;k-4h\right)\).
Từ đó suy ra: \(\left\{{}\begin{matrix}2k+3h=-7\\k-4h=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}k=-2\\h=-1\end{matrix}\right.\).