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Find the vector and cartesian equations of a plane which is at a distance of $18$ units from the origin and which is normal to the vector $\overrightarrow{2i}+\overrightarrow{7j}+\overrightarrow{8k}$

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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 $ax+by+cz=q$
Step 1
Here $p=18$ units, $\overrightarrow n=2i+7\overrightarrow j+8\overrightarrow k$
$ \therefore \overrightarrow n=\large\frac{2\overrightarrow i+7\overrightarrow j+8\overrightarrow k}{\sqrt{4+49+64}} = \large\frac{2\overrightarrow i+7\overrightarrow j+8\overrightarrow k}{\sqrt{117}} = \large\frac{2\overrightarrow i+7\overrightarrow j+8\overrightarrow k}{3\sqrt{13}}$
Step 2
The vector equation of the plane is $ \overrightarrow r.\overrightarrow n=p$
$(x\overrightarrow i+y\overrightarrow j+z\overrightarrow k).\large\frac{(2\overrightarrow i+7\overrightarrow j+8\overrightarrow k)}{3\sqrt{13}}=18$
or $ (x\overrightarrow i+y\overrightarrow j+z\overrightarrow k).(2\overrightarrow i+7\overrightarrow j+8\overrightarrow k)=54\sqrt{13}$
The cartesian equation is $2x+7y+8z=54\sqrt{13}$
answered Jun 14, 2013 by thanvigandhi_1

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