Let a line L1 pass through the origin and be perpendicular to the lines L2 : r=(3+t) i+(2t-1) j+(2t+4) k and L3 : r=(3+2s) i+(3+2s) j+(2+s) k, t, s R. If (a, b, c), a Z, is the point on L3 at a distance of 17 from the point of intersection of L1 and L2, then (a+b+c)2 is equal to .
Answer
4
Solution
The directions are (1, 2, 2) for L2 and (2, 2, 1) for L3. L1 is perpendicular to both, so its direction is
vmatrix i & j & k\ 1 & 2 & 2\ 2 & 2 & 1 vmatrix = -2 i + 3 j - 2 k
and L1 is (x, y, z) = (-2, 3, -2).
For L1 L2: 3 + t = -2 , 2t - 1 = 3 , 2t + 4 = -2 .
From the first and third, 3 + t = 2t + 4 t = -1, = -1 (the second gives -3 = -3, true), so the point is P(2, -3, 2).
A point of L3 is Q(3 + 2s, 3 + 2s, 2 + s) with PQ2 = 17:
(1 + 2s)2 + (6 + 2s)2 + s2 = 17 9s2 + 28s + 20 = 0 (s + 2)(9s + 10) = 0
s = - 109 gives a = 79 Z, so it is rejected; s = -2 gives Q(-1, -1, 0).
Hence a + b + c = -2 and (a + b + c)2 = 4.
Answer: 4