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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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In each of the following cases determine the direction cosines of the normal to the plane and the distance from origin(d) $5y+8=0$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

$\begin{array}{1 1}(0,1,0);\frac{8}{5} units \\ (0,0,1);\frac{8}{5} units \\ (0,-1,0);\frac{8}{5} units \\ (D) (0,-1,0);\frac{8}{5} units. \end{array} $

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Toolbox:
  • The equation of normal to a plane is $lx+my+nz=d$,where $l,m,n$ are the direction cosines and d is the distance of the perpendicular from the origin.
Step 1:
The given equation is $5y+8=0$
This can be written as $0x+5y+0z=-8$
(i.e)$0x-5y+0z=8$-----(1)
Therefore $\sqrt{0^2+(-5)^2+0^2}=5$
Step 2:
Divide on both sides of eq(1) by 5
$\large\frac{0}{5}$$x-\large\frac{5}{5}$$y+\large\frac{0z}{5}=\frac{8}{5}$
On simplifying we get
$\Rightarrow -y=\large\frac{8}{5}$
This equation of the form $lx+my+nz=d$,where $l,m$ and $n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.
Therefore the direction cosines of the normal to the plane are $(0,-1,0)$ and distance from the origin is $\large\frac{8}{5}$units.
answered May 30, 2013 by sreemathi.v
 

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