If the distances of the point (1, 2, a) from the line x - 11 = y2 = z - 11 along the lines L1 : x - 13 = y - 24 = z - ab and L2 : x - 11 = y - 24 = z - ac are equal, then a + b + c is equal to
- 4
- 6
- 7
- 5
Answer
(3) 7
Solution
Let P(1, 2, a). A point on L1 is (1 + 3t, 2 + 4t, a + bt), and it lies on the given line:
3t1 = 2 + 4t2 = a + bt - 11
3t = 1 + 2t t = 1 and a + b - 1 = 3 a + b = 4
So the meeting point is Q1(4, 6, 4) and PQ12 = 9 + 16 + b2 = 25 + b2.
A point on L2 is (1 + s, 2 + 4s, a + cs), so
s = 2 + 4s2 = 1 + 2s s = -1 and a - c - 1 = -1 a = c
So the meeting point is Q2(0, -2, 0) and PQ22 = 1 + 16 + c2 = 17 + c2.
PQ1 = PQ2 c2 - b2 = 8, and with c = a = 4 - b:
(4 - b)2 - b2 = 8
16 - 8b = 8 b = 1, a = 3, c = 3
a + b + c = 7
Answer: (3)