Let P be the point of intersection of the lines x - 21 = y - 45 = z - 21 and x - 32 = y - 23 = z - 32. Then, the shortest distance of P from the line 4x = 2y = z is
- 3 147
- 5 147
- 147
- 6 147
Answer
(1) 3 147
Solution
Points: ( + 2, 5 + 4, + 2) and (2 + 3, 3 + 2, 2 + 3)
+ 2 = 2 + 3, 5 + 4 = 3 + 2 ⇒ = -1, = -1 ⇒ P(1, -1, 1)
4x = 2y = z ⇒ x1 = y2 = z4 (through O, d.r. 1, 2, 4)
Foot Q(k, 2k, 4k): PQ = (k - 1, 2k + 1, 4k - 1) (1, 2, 4)
(k - 1) + 2(2k + 1) + 4(4k - 1) = 0 21k = 3 k = 17
PQ2 = ( 67 )2 + ( 97 )2 + ( 37 )2 = 12649 ⇒ PQ = 3 147
Answer: (1)