Find the minimum distance from the point 5 2 7 to the plane

Find the minimum distance from the point (5, 2, 7) to the plane 2x + 3y + z = 12. (Give your answer correct to 2 decimal places.) (Hint: To simplify the computations, minimize the square of the distance.)

Solution

minimum distace is length of the normal passing through (5,2,7) so (x-5,y-2,z-7) are in parallel with 2,3,1 so x-5/2 = y-2/3 = z-7/1 = t so (x,y,z) = (2t+5,3t+2,t+7) subtitute in plane equation so 2(2t+5)+3(3t+2)+t+7 = 12 so t = -11/7 get x,y,z and find distance betwween (x,y,z) and (5,2,7) this is the shortest distance