Let P be the foot of the perpendicular drawn from the point Q( , 4, 1) to the line x-1 1 = y-3 -1 = z 2 . If the point P lies on the plane x + y + z = 6 , then the value of is

Step-by-step solution

Let the coordinates of a general point on the line x-1 1 = y-3 -1 = z 2 = t be (t+1, -t+3, 2t) . Since P is the foot of the perpendicular, it lies on the line, so P (t+1, -t+3, 2t) for some value of t . It is given that P lies on the plane x + y + z = 6 . Substituting the coordinates of P into the plane's equation: (t+1) + (-t+3) + (2t) = 6 2t + 4 = 6 2t = 2 t = 1 Thus, the coordinates of P are (1+1, -1+3, 2(1)) = (2, 2, 2) . The direction ratios of the vector QP are (2- , 2-4, 2-1) = (2- , -2, 1) . The direction r