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Find the cartesian form of following planes:$ \overrightarrow{r}=(s-2t) \overrightarrow{i}+(3-t)\overrightarrow{j}+(2s+t)\overrightarrow{k}$

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Step 1
Let $ \overrightarrow r=x\overrightarrow i+y\overrightarrow j+z\overrightarrow k$
We have $ x\overrightarrow i+y\overrightarrow j+z\overrightarrow k=(s-2t)\overrightarrow i+(3-t)\overrightarrow j+(2s+t)\overrightarrow k$
Equating the components in the $ x, y, z$ directions
$y=3-t \Rightarrow y-3=-t$
Step 2
Eliminating $ t, s$
$\begin{vmatrix} x & 1 & -2 \\ y-3 & 0 & -1 \\ z & 2 & 1 \end{vmatrix} = 0 \Rightarrow x(0+2)-(y-3)(1+4)+z(-1-0)=0$
i.e., $2x-5y+15-z=0 \: or \: 2x-5y-z=-15$ is the cartesian equation of the plane.
answered Jun 16, 2013 by thanvigandhi_1

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