Cho 3 vecto \(\overrightarrow{u}\left(1;2;3\right),\overrightarrow{v}\left(2;2-1\right),\overrightarrow{w}\left(4;0;-4\right)\). Tìm tọa độ của vecto \(\overrightarrow{x}\), biết
a, \(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}\)
b,\(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}+2\overrightarrow{w}\)
c, \(\overrightarrow{x}=2\overrightarrow{u}+4\overrightarrow{v}-\overrightarrow{w}\)
d,\(2\overrightarrow{x}-3\overrightarrow{u}=\overrightarrow{w}\)
e, \(2\overrightarrow{u}+\overrightarrow{v}-\overrightarrow{w}+3\overrightarrow{x}=\overrightarrow{0}\)
Lời giải:
a)
\(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}=(1-2, 2-2,3-(-1))=(-1,0,4)\)
b)
\(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}+2\overrightarrow{w}=(1-2+2.4,2-2+2.0; 3-(-1)+2(-4))\)
\(=(7, 0, -4)\)
c)
\(\overrightarrow{x}=2\overrightarrow{u}+4\overrightarrow{v}-\overrightarrow{w}=(2.1+4.2-4, 2.2+4.2-0, 2.3+4.(-1)-(-4))\)
\(=(6,12,6)\)
d)
\(2\overrightarrow{x}=3\overrightarrow{u}+\overrightarrow{w}=3(1,2,3)+(4,0,-4)=(3.1+4, 3.2+0,3.3+(-4))\)
\(=(7,6,5)\Rightarrow \overrightarrow{x}=(\frac{7}{2}, 3, \frac{5}{2})\)
e)
\(3\overrightarrow{x}=-2\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}=-2(1,2,3)-(2,2,-1)+(4,0,-4)\)
\(=(-2,-4,-6)-(2,2,-1)+(4,0,-4)=(-2-2+4,-4-2+0,-6-(-1)+(-4))\)
\(=(0,-6,-9)\Rightarrow \overrightarrow{x}=(0,-2,-3)\)
