If the equation of the line passing through the point (0, - 12, 0 ) and perpendicular to the lines r = ( i + a j + b k ) and r = ( i - j - 6 k ) + (-b i + a j + 5 k ) is x-1-2 = y+4d = z-c-4, then a + b + c + d is equal to :
- 13
- 14
- 12
- 10
Answer
(2) 14
Solution
The point (0, - 12, 0 ) lies on x-1-2 = y+4d = z-c-4:
0-1-2 = 12 - 12+4d = 12 d = 7 and -c-4 = 12 c = 2
Direction ratios of the line: (-2, 7, -4)
Perpendicular to (1, a, b): -2 + 7a - 4b = 0 ...(i)
Perpendicular to (-b, a, 5): 2b + 7a - 20 = 0 ...(ii)
(ii) − (i): 6b - 18 = 0 b = 3, then 7a = 14 a = 2
a + b + c + d = 2 + 3 + 2 + 7 = 14
Answer: (2)