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Find the unit normal vectors to the plane $2x-y+2z=5$

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  • Equation of the plane whose perpendicular distance from the origin is $P$ and $ \overrightarrow n$ is the unit normal vector to the plane from the origin. $ \overrightarrow r.\overrightarrow n=p$ (vector equation) $ lx+my+nz=p$ where $ \overrightarrow n=l\overrightarrow i+m\overrightarrow j+n\overrightarrow k$ If $\overrightarrow n = a\overrightarrow i+b\overrightarrow j+c\overrightarrow k$ is not the unit normal vector, then the equation is of the form $\overrightarrow r.\overrightarrow n=q$ where $ p=\large\frac{q}{|\overrightarrow n|}$ The cartesian equation is $ax+by+cz=q$
Equation of the plane is $ 2x-y+2z=5$
or $ (x\overrightarrow i+y\overrightarrow j+z\overrightarrow k).(2\overrightarrow i-\overrightarrow j+2\overrightarrow k)=5$
The normal vectors to the plane are $ \overrightarrow n= \pm 2\overrightarrow i-\overrightarrow j+2\overrightarrow k$
The unit normal vectors are $ \overrightarrow n = \pm \large\frac{2\overrightarrow i-\overrightarrow j+2\overrightarrow k}{\sqrt{4+1+4}}=\pm \large\frac{2\overrightarrow i-\overrightarrow j+2\overrightarrow k}{3}$

 

answered Jun 14, 2013 by thanvigandhi_1
edited Jun 24, 2013 by thanvigandhi_1
 

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