Question
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.
Solution
The equation of plane is
2x – 3y + 4z – 6 = 0 ...(1)
Direction ratios of the normal to the plane (1) are 2, –3, 4.
Dividing each by 
we get the direction cosines of the normal as 
∴ dividing (1) throughout by 
we get, 
This is of the form l x + m y + n z = p, where p is the distance of the plane from the origin.
distance of plane from the origin = 
.
Let (x1, y1, z1) be the coordinates of the foot of perpendicular drawn from origin 0(0, 0, 0) to the plane (1).
∴ direction ratios of OP are .x1– 0, y, –0, –0 i.e. x1, y1, z1
∴ Direction cosines of OP are
Since direction cosines and direction ratios of a line are proportional
Since P(x1, y1, z1) lies on plane (1)
Tips: -
If d is the distance from the origin and l, m, n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd).
