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Home  >>  TN XII Math  >>  Vector Algebra
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The foot of the perpendicular drawn from the origin to the plane is $(8 , -4 , 3 )$ find the equation of the plane.

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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
$P(8, -4, 3)$ is the foot of the perpendicular from the origin to the plane
$ \therefore \overrightarrow {OP} = \overrightarrow n=8\overrightarrow i-4\overrightarrow j+3\overrightarrow k$
and $p=|8\overrightarrow i-4\overrightarrow j+3\overrightarrow k|=\sqrt{64+16+9}=\sqrt{89}$
$ \hat n=\large\frac{8\overrightarrow i-4\overrightarrow j+3\overrightarrow k}{\sqrt{89}}$
Step 2
The vector equations of the plane is $ \overrightarrow r.\overrightarrow n=p$
$(x\overrightarrow i+y\overrightarrow j+z\overrightarrow k).\large\frac{(8\overrightarrow i-4\overrightarrow j+3\overrightarrow k)}{\sqrt{89}}=\sqrt{89}$
i.e., $(x\overrightarrow i+y\overrightarrow j+z\overrightarrow k).\large\frac{(8\overrightarrow i-4\overrightarrow j+3\overrightarrow k)}{\sqrt{89}}=\sqrt{89}$
Step 3
The cartesian equation is $ 8x-4y+3z=89$

 

answered Jun 14, 2013 by thanvigandhi_1
edited Jun 25, 2013 by thanvigandhi_1
 

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