The distance of the point (1,0,-3) from the plane x-y-z=9 measured parallel to the line x-2 2 = y+2 3 = z-6 -6 is

Step-by-step solution

Let the given point be P(1, 0, -3) . The equation of the line passing through P and parallel to the given line x-2 2 = y+2 3 = z-6 -6 is x-1 2 = y-0 3 = z+3 -6 = r Any point on this line is of the form Q(2r+1, 3r, -6r-3) . Since Q lies on the plane x-y-z=9 , substituting the coordinates of Q in the plane equation gives (2r+1) - (3r) - (-6r-3) = 9 2r + 1 - 3r + 6r + 3 = 9 5r + 4 = 9 5r = 5 r = 1 Substituting r=1 , the coordinates of Q are (3, 3, -9) . The required distance is the distance between P and Q : PQ = (3-1