The image of a point P(3,5,3) in the line x 1 = y-1 2 = z-2 3 is P^ (a, b, c) . Then a+b+c=

Step-by-step solution

Let the line be L: x 1 = y-1 2 = z-2 3 = k . Any point Q on the line is given by Q(k, 2k+1, 3k+2) . The direction vector of the line is v = i + 2 j + 3 k . The vector PQ = (k-3) i + (2k+1-5) j + (3k+2-3) k = (k-3) i + (2k-4) j + (3k-1) k . Since PQ is perpendicular to the line, PQ v = 0 . (k-3)(1) + (2k-4)(2) + (3k-1)(3) = 0 . k - 3 + 4k - 8 + 9k - 3 = 0 . 14k - 14 = 0 k = 1 . The foot of the perpendicular Q is (1, 2(1)+1, 3(1)+2) = (1, 3, 5) . Since Q is the midpoint of PP^ , where P^ = (a, b, c) , we have 3+a 2 =