Let a = 2 i + j + k, b = - i + k, c = j - k, where and are integers and = -6. Let the values of the ordered pair ( , ), for which the area of the parallelogram of diagonals a + b and b + c is 212, be ( 1, 1) and ( 2, 2). Then 12 + 12 - 2 2 is equal to
- 21
- 24
- 19
- 17
Answer
(3) 19
Solution
a + b = i + j + 2 k and b + c = - i + j
Area of a parallelogram with diagonals d1, d2 is 12| d1 d2|, and here
( a + b) ( b + c) = -2 i - 2 j + ( + ) k
12 4 2 + 4 + ( + )2 = 212
2 + 2 + 5 2 = 17 2 + 5 2 = 29 (since = -6)
As , are integers, 2 = 4 and 2 = 9; with = -6 this gives ( , ) = (3, -2) or (-3, 2).
In both cases 2 + 2 = 13 and = -6, so
12 + 12 - 2 2 = 13 + 6 = 19
Answer: (3)