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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)$

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