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Question Number 79111 by mathocean1 last updated on 22/Jan/20

$$\mathrm{Show}\mathrm{that}$$$$\mathrm{E}=\{(x,\mathrm{y},z)\in {\mathbb{R}}^3/x-2y+z=0\}$$$$\mathrm{is}\mathrm{a}\mathrm{subspace}\mathrm{vector}\mathrm{of}\mathrm{which}\mathrm{we}$$$$\mathrm{will}\mathrm{determine}\mathrm{one}\mathrm{base}.$$$$\mathrm{please}\mathrm{help}\mathrm{sirs}...$$

Commented by mathmax by abdo last updated on 22/Jan/20

$$x-2y+z=0\Rightarrow x=2y-z\Rightarrow (x,y,z)=(2y-z,y,z)$$$$=(2y,y,o)+(-z,0,z)=y(2,1,0)+z(-1,0,1)=y\overrightarrow{u}+z\overrightarrow{v}\Rightarrow$$$$EisavectorialplanewithbaseB=(\overrightarrow{u},\overrightarrow{v})$$$$\overrightarrow{u}(2,1,0)and\overrightarrow{v}(-1,0,1)$$

Commented by mathocean1 last updated on 22/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 23/Jan/20

$$youarewelcome$$

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