Find the coordinates of the foot of the perpendicular drawn from the origin to the plane i) `2x+3y+4z-12=0`, ii) `5y+8=0`.

Answer

i) `(24/29, 36/29, 48/29)`, ii) `(0,-8/5,0)`

Explanation

Given, equation is `2x+3y+4z=12`…………..(i)
D.r.'s of normal to the plane are 2,3,4.
Equation of the line through O(0,0,0) and perpendicular to plane (i) is
`x/2=y/3=z/4=lambda` (say).
A general point on (ii), is `N(2lambda,3lambda,4lambda)`.
If this point lies on plane (i), then
`(2 xx 2lambda)+(3 xx 3lambda)+(4 xx 4lambda)=12 rArr 29N = 12 rArr lambda=12/29`.
Hence,the required point is `(24/29,36/29,48/29)`.
ii) The given plane is `0x-5y+0z=8`..................(i)
D.r.'s of normal to the plane are 0,-5,0.
Equation of the line through O(0,0,0) and perpendicular to plane (i) are
`x/0=y/-5=z/0=lambda` (say)..............(ii)
A general point on this line (i), we have
`-5 xx (-5lambda)=8 rArr 25lambda=8 rArr lambda=8/25`.
Hence, the required point is `N(0,-8/5,0)`.