Question

Solve the following :

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.

Answer 1

The equation of the plane is 2x + 3y + 6z = 49.
Dividing each term by

`sqrt(2^2 + 3^2 + (-6)^2)`

= `sqrt(49)`

= 7,
we get

`(2)/(7)x + (3)/(7)y - (6)/(7)z = (49)/(7)` = 7

This is the normal form of the equation of plane.

∴ the direction cosines of the perpendicular drawn from the origin to the plane are

l = `(2)/(7), m = (3)/(7), n = (6)/(7)`

and length of perpendicular from origin to the plane is p = 7.

∴ the coordinates of the foot of the perpendicular from the origin to the plane are
`(lp, mp, np) "i.e." (2, 3, 6)`