If the distance of the point P (1,-2,1) from the plane x+2 y -2 z = , where >0 is 5 units, then the foot of the perpendicular from P to the plane is

Step-by-step solution

Given point P(1,-2,1) and plane x+2y-2z= with distance 5 , determine the value of using the distance formula d = |Ax₀ + By₀ + Cz₀ + D| A^2+B^2+C^2 . Substituting P(1,-2,1) into x+2y-2z- =0 gives 5 = |1 - 4 - 2 - | 9 = |-5- | 3 . Since > 0 , |-5- | = 5 + , so 15 = 5 + and = 10 . The normal vector (1,2,-2) from the plane x+2y-2z=10 gives the direction for the perpendicular from P . The line through P is parameterized as (1+ , -2+2 , 1-2 ) . Substituting into the plane equation: (1+ ) + 2(-2+2 ) - 2(1-2 ) = 10 simplif