Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63.

Answer 1

The equation of the plane is 2x + 6y – 3z = 63.

Dividing each term by \(\sqrt{{2^2 + 6^2 + (-3)^2}}\) = \(\sqrt{49}\) = 7, we get

This is the normal form of the equation of plane. 

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

and length of perpendicular from origin to the plane is p = 9. 

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