Let P( , , ) be the image of the point Q(3, , 4) in the line x-1 1 = y 2 = z-2 -1 . If + + = 5 , then the value of is equal to

Step-by-step solution

Let the foot of the perpendicular from Q(3, , 4) to the given line be A . Any point on the line can be written as A(t+1, 2t, -t+2) . The direction ratios of the line are (1, 2, -1) . The vector QA is given by (t-2, 2t- , -t-2) . Since QA is perpendicular to the line, their dot product is zero: 1(t-2) + 2(2t- ) - 1(-t-2) = 0 t - 2 + 4t - 2 + t + 2 = 0 6t - 2 = 0 t = 3 Since A is the midpoint of P and Q , we have P = 2A - Q . Thus, the coordinates of P( , , ) are: = 2(t+1) - 3 = 2t - 1 = 2(2t) - = 4t - = 2(-t+2) - 4