Let P be the foot of the perpendicular from the point Q(2, -2, 6) to the line x-1 2 = y+1 1 = z-2 2 . If P' is the image of P in the plane 2x + y + 2z = 5 , then the square of the distance of P' from the origin is

Step-by-step solution

Let a general point on the line x-1 2 = y+1 1 = z-2 2 = t be (2t+1, t-1, 2t+2) . Let this point be P . The direction ratios of the line segment QP are: (2t+1 - 2, t-1 - (-2), 2t+2 - 6) = (2t-1, t+1, 2t-4) Since QP is perpendicular to the given line, the dot product of their direction ratios is zero: 2(2t-1) + 1(t+1) + 2(2t-4) = 0 4t - 2 + t + 1 + 4t - 8 = 0 9t - 9 = 0 t = 1 Substituting t = 1 , the coordinates of P are (3, 0, 4) . Now, we find the image P'(x, y, z) of the point P(3, 0, 4) in the plane 2x + y + 2z -