The sum of all values of , for which the shortest distance between the lines x+1 = y-2-1 = z-4- and x = y-12 = z-12 is 2, is
- -6
- -8
- 8
- 6
Answer
(1) -6
Solution
Line 1 passes through A(–1, 2, 4) with b1 = i - j - k, and line 2 through B(0, 1, 1) with b2 = i + 2 j + 2 k.
b1 b2 = vmatrix i & j & k \ & -1 & - \ & 2 & 2 vmatrix = 0 i - 3 2 j + 3 k
With AB = i - j - 3 k, we get AB ( b1 b2) = 3 2 - 9 and | b1 b2| = 3| | 2 + 1, so
Shortest distance = |3 2 - 9 |3| | 2 + 1 = | - 3| 2 + 1 = 2
( - 3)2 = 2( 2 + 1)
2 + 6 - 7 = 0
= 1 or -7
∴ Sum of all values of = 1 + (-7) = -6
Answer: (1)