If the shortest distance between the lines x - 3 = y - 2-1 = z - 11 and x + 2-3 = y + 52 = z - 44 is 44 30, then the largest possible value of | | is equal to .
Answer
43
Solution
Skew lines: d = | ( a2 - a1 ) ( b1 b2 ) | | b1 b2 |
a1 = ( , 2, 1), b1 = (3, -1, 1); a2 = (-2, -5, 4), b2 = (-3, 2, 4)
b1 b2 = -6 i - 15 j + 3 k, | b1 b2 | = 270 = 3 30
a2 - a1 = (-2 - ) i - 7 j + 3 k
( a2 - a1 ) ( b1 b2 ) = 12 + 6 + 105 + 9
= 6 + 126
|6 + 126|3 30 = 44 30 | + 21| = 22
= 1 or = -43 largest | | = 43
Answer: 43