Given plane Π : f(x,y,z) = 4x+3y-z = -1 Need to find point P on Π that is closest to the origin O=(0,0,0).
Solution: First step: check if O is on the plane Π : f(0,0,0)=0 ≠ -1 => O is not on Π Next: We know that the required point must lie on the normal vector <4,3,-1> passing through the origin, i.e. P=(0,0,0)+k<4,3,-1> = (4k,3k,-k) For P to lie on plane Π , it must satisfy 4(4k)+3(3k)-(-k)=-1 Solving for k k=-1/26 => Point P is (4k,3k,-k) = (-4/26, -3/26, 1/26) = (-2/13, -3/26, 1/26) because P is on the normal vector originating from the origin, and it satisfies the equation of plane Π Answer: P(-2/13, -3/26, 1/26) is the point on Π closest to the origin.
