The foot of the perpendicular from the point P(1,6,3) to the line x 1 = y-1 2 = 2-z -3 is

Step-by-step solution

The given line is x 1 = y-1 2 = z-2 3 = . Any point on the line is given by Q( , 2 + 1, 3 + 2) . The vector PQ = ( - 1, 2 + 1 - 6, 3 + 2 - 3) = ( - 1, 2 - 5, 3 - 1) . The direction vector of the line is v = (1, 2, 3) . Since PQ is perpendicular to the line, PQ v = 0 . 1( - 1) + 2(2 - 5) + 3(3 - 1) = 0 . - 1 + 4 - 10 + 9 - 3 = 0 . 14 - 14 = 0 = 1 . Substituting = 1 into the coordinates of Q , we get Q(1, 2(1) + 1, 3(1) + 2) = (1, 3, 5) . Answer: (1,3,5)