Let the image of the point P(1, 6, a) in the line L: x 1 = y-1 2 = z-a+1 b , b > 0 , be ( a 3 , 0, a+c ) . If S( , , ) , > 0 , is the point on L such that the distance of S from the foot of perpendicular from the point P on L is 2 14 , then + + is equal to:

Step-by-step solution

Let the foot of the perpendicular from P(1, 6, a) to the line L be M . Since the image of P is P' ( a 3 , 0, a+c ) , M is the midpoint of P and P' . M = ( 1 + a 3 2 , 6 + 0 2 , a + a + c 2 ) = ( a+3 6 , 3, a + c 2 ) Since M lies on the line L: x 1 = y-1 2 = z-a+1 b , we substitute its coordinates into the equation of the line: a+3 6 1 = 3-1 2 = a + c 2 - a + 1 b a+3 6 = 1 = c 2 + 1 b From a+3 6 = 1 , we get a = 3 . From c 2 + 1 b = 1 , we get 2b = c + 2 c = 2b - 2 . The vector PP' is perpendicular to the direction