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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $(c)\; x + y + z = 1 $

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  • The equation of the form $lx+my+nz=d$ where $l,m,n$ are the direction cosines of the normal to the plane and $d$ is the distance of the normal from the origin.
  • The coordinate of the foot of the perpendicular are $(ld,md,nd)$.
Step 1:
The given equation of the plane is $x+y+z=1$------(1)
Hence the direction cosines of the normal are $(1,1,1)$
Therefore $\sqrt{1^2+1^2+1^2}=\sqrt 3$
Divide both sides of the equation by $\sqrt 3$
$\large\frac{1}{\sqrt 3}$$x+\large\frac{1}{\sqrt 3}$$y+\large\frac{1}{\sqrt 3}$$z=\large\frac{1}{\sqrt 3}$
Step 2:
Therefore the coordinates of the foot of the perpendicular are $(ld,md,nd)$
(i.e)$\big(\large\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}\big)$
$\Rightarrow (\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$
Hence the coordinates are $(\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$
answered May 30, 2013 by sreemathi.v
 

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