Question

Find the distance from the point (4, 1, −1) to the plane x − 2y + 2z + 3 = 0

Answer 1

I believe the answer is -17????

Answer 2

Distance from point (4,1,-1) to plane x-2y+2z+3 = 0 is 1. The formula for the distance from a point to a plane is d = abs(Ax + By + Cz + D) / sqrt(A^2 + B^2 + C^2) where d = distance (x,y,z) = coordinates of point A,B,C,D = coefficients of equation for the plane For this problem, the values for A,B,C,and D are A=1, B=-2, C=2, and D=3. So substituting the known values into the equation, we get: d = abs(1*x - 2*y + 2*z + 3) / sqrt(1^2 + (-2)^2 + 2^2) d = abs(1*4 - 2*1 + 2*(-1) + 3) / sqrt(1 + 4 + 4) d = abs(4 - 2 - 2 + 3) / sqrt(9) d = abs(3) / 3 d = 1