If the shortest distance between the lines L1 : r = (2 + ) i + (1 - 3 ) j + (3 + 4 ) k, R L2 : r = 2(1 + ) i + 3(1 + ) j + (5 + ) k, R is m n, where (m, n) = 1, then the value of m + n equals
- 384
- 387
- 390
- 377
Answer
(2) 387
Solution
L1: a1 = 2 i + j + 3 k, b1 = i - 3 j + 4 k
L2: a2 = 2 i + 3 j + 5 k, b2 = 2 i + 3 j + k
b1 b2 = vmatrix i & j & k \ 1 & -3 & 4 \ 2 & 3 & 1 vmatrix = -15 i + 7 j + 9 k
| b1 b2| = 225 + 49 + 81 = 355, a2 - a1 = 2 j + 2 k
d = |( a2 - a1) ( b1 b2) || b1 b2| = 14 + 18 355 = 32 355
Since 355 = 5 71, (32, 355) = 1, so m = 32, n = 355
m + n = 387
Answer: (2)