Let the shortest distance between the lines x-33 = y- -1 = z-31 and x+3-3 = y+72 = z- 4 be 3 30. Then the positive value of 5 + is
- 42
- 40
- 48
- 46
Answer
(4) 46
Solution
Line 1: point P(3, , 3), direction b1 = 3 i - j + k
Line 2: point Q(-3, -7, ), direction b2 = -3 i + 2 j + 4 k
b1 b2 = vmatrix i & j & k \ 3 & -1 & 1 \ -3 & 2 & 4 vmatrix = -6 i - 15 j + 3 k
| b1 b2| = 36 + 225 + 9 = 270 = 3 30
PQ = -6 i - (7 + ) j + ( - 3) k
PQ ( b1 b2) = 36 + 15(7 + ) + 3( - 3)
= 132 + 15 + 3
Shortest distance:
|132 + 15 + 3 |3 30 = 3 30
⇒ |132 + 15 + 3 | = 270
⇒ 15 + 3 = 138 or -402
⇒ 5 + = 46 or -134
∴ Positive value of 5 + = 46
Answer: (4)