Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line x - 11 = y2 = z - 11. Then the distance of Q from the line x - 93 = y - 92 = z - 5-2 is
- 8
- 7
- 6
- 5
Answer
(2) 7
Solution
Any point of the first line: F(1 + t, 2t, 1 + t); direction (1, 2, 1)
PF = (t - 2, 2t - 2, t), and PF (1, 2, 1)
(t - 2) + 2(2t - 2) + t = 0 6t - 6 = 0 t = 1
foot F(2, 2, 2)
Q = 2F - P = (1, 2, 3)
Second line passes through B(9, 9, 5) with direction d = (3, 2, -2)
BQ = (-8, -7, -2)
BQ d = vmatrix i & j & k \ -8 & -7 & -2 \ 3 & 2 & -2 vmatrix = 18 i - 22 j + 5 k
Distance = | BQ d|| d| = 324 + 484 + 25 17 = 833 17 = 49 = 7
Answer: (2)