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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 $(d)\;5y + 8 = 0 $

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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 coordinates of the foot of the perpendicular are $(ld,md,nd)$.
Step 1:
Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $(x,y,z)$
$5y+8=0$
This can be written as $0x+5y+0z=-8$
$0x-5y+0z=8$-----(1)
Step 2:
The direction ratios are $0,-5$ and $0$
Therefore $\sqrt{0^2+(-5)^2+0}=5$
Divide on both sides of equ(5) by 5
$\large\frac{0}{5}$$x-\large\frac{5}{5}$$y+\large\frac{0}{5}$$z=\large\frac{8}{5}$
$\Rightarrow -y=\large\frac{8}{5}$
Step 3:
The coordinates of the foot of the perpendicular are $(ld,md,nd)$
Therefore the coordinates of the foot of the perpendicular are $\big(0,1.\big(\large\frac{8}{5}\big)$$,0\big)$
$\Rightarrow \big(0,\large\frac{8}{5}$$,0\big)$
answered May 30, 2013 by sreemathi.v
 

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