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Find the length of the perpendicular frome the origin to the plane$\overrightarrow{r} . (\overrightarrow{3i}+\overrightarrow{4j}+\overrightarrow{12k})=26$.

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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 $ax+by+cz=q$
The equation $ \overrightarrow r.(3\overrightarrow i+4\overrightarrow j+12\overrightarrow k)=26$ is of the form $ \overrightarrow r.\overrightarrow n=q$
$ \therefore $ the perpendicular distance from the origin to the plane is $ \large\frac{q}{|\overrightarrow n|}=\large\frac{26}{\sqrt{9+16+144}}=\large\frac{26}{\sqrt{169}}=\large\frac{26}{13}=2$ units.
answered Jun 14, 2013 by thanvigandhi_1

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