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Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $(b)\; 3y + 4z – 6 = 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 is $(ld,md,nd)$.
Step 1:
The given equation of the plane is $3y+4z-6=0$
This can be written as $0x+3y+4z=6$------(1)
Hence the direction ratios of the normal are $0,3$ and $4$
Therefore $\sqrt{0^2+3^2+4^2}=5$
Step 2:
Dividing on both sides of equ(1) by 5 we get
Therefore the coordinates of the foot of the perpendicular are $(ld,md,nd)$
(i.e) $\big(0,\large\frac{3}{5}$$\times\large\frac{6}{5},\large\frac{4}{5}$$\times \large\frac{6}{5}\big)$
$\Rightarrow \big(0,\large\frac{18}{25},\frac{24}{25}\big)$
Hence the coordinates are $\big(0,\large\frac{18}{25},\frac{24}{25}\big)$
answered May 31, 2013 by sreemathi.v

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