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Home  >>  TN XII Math  >>  Vector Algebra
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Find the cartesian form of following planes: $\overrightarrow{r}=(1+s+t) \overrightarrow{i}+(2-s+t)\overrightarrow{j}+(3-2s+2t)\overrightarrow{k}$

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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=(1+s+t)\overrightarrow i+(2-s+t)\overrightarrow j+(3-2s+2t)\overrightarrow k$
Equating the components in the $ x-, y-, z-$ directions.
$ x = 1+s+t\: \: \Rightarrow x-1=s+t$
$ y = 2-s+t\: \: \Rightarrow y-2=-s+t$
$ z = 3-2s+2t\: \: \Rightarrow z-3=-2s+2t$
Step 2
Eliminating $ t, s$ from the above, we have
$ \begin{vmatrix} x-1 & 1 & 1 \\ y-2 & -1 & 1 \\ z-3 & -2 & 2 \end{vmatrix}=0 \Rightarrow (x-1)(-2+2)-(y-2)(2+2)+(z-3)(1+1)=0$
$ \therefore -4y+4+2z-6=0\: or \: 4y-2z=-2$
$ \Rightarrow 2y-z=-1$ is the equation of the plane.
answered Jun 16, 2013 by thanvigandhi_1
 

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