In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.(a) `2x + 3y + 4z -12 = 0` (b) `3y + 4z - 6 = 0`(c) `x + y + z = 1` (d) `5y + 8 = 0`

Explanation

i)Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1)

The given equation is,`2x+3y+4z-12=0`

so,`2x+3y+4z=12 ` ...(a)

The directional ratios of normal are` 2,3,4`

so,`√(2^2+3^2+4^2 )=√29`

dividing both sides of a by `√29` we get,

`(2x)/(√29)+(3y)/(√29)+(4z)/(√29)=12/(√29)`

This equation is in the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
So,
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