If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line x1= y-11= z-22 is 4, then the sum of all possible values of a is equal to :
- 11
- 9
- 6
- 4
Answer
(3) 6
Solution
Let L(t, 1 + t, 2 + 2t) be the foot of the perpendicular from A(1, 2, 7) on the line. Then AL = (t - 1, t - 1, 2t - 5) is perpendicular to (1, 1, 2), so
(t - 1) + (t - 1) + 2(2t - 5) = 0 6t = 12 t = 2
which gives L(2, 3, 6).
L is the mid-point of A and its image A', we get A' = (2 2 - 1, 2 3 - 2, 2 6 - 7) = (3, 4, 5).
The distance of (a, 2, 5) from A' is 4, so
(a - 3)2 + (2 - 4)2 + (5 - 5)2 = 16
(a - 3)2 = 12 a = 3 2 3
Sum of all possible values of a = 6
Answer: (3)