Let c and d be vectors such that | c + d| = 29 and c (2 i + 3 j + 4 k) = (2 i + 3 j + 4 k) d. If 1, 2 ( 1 > 2) are the possible values of ( c + d) (-7 i + 2 j + 3 k), then the equation K2x2 + (K2 - 5K + 1)xy + (3K + 22 )y2 - 8x + 12y + 2 = 0 represents a circle, for K equal to :
- 4
- -1
- 2
- 1
Answer
(4) 1
Solution
Let u = 2 i + 3 j + 4 k, | u| = 29
c u = u d = - d u
( c + d) u = 0
c + d = u
| c + d| = 29 | | 29 = 29 = 1
( c + d) (-7 i + 2 j + 3 k) = (-14 + 6 + 12) = 4 = 4
1 = 4, 2 = -4
Equation: K2x2 + (K2 - 5K + 4)xy + (3K - 2)y2 - 8x + 12y - 4 = 0
For a circle, coefficient of xy = 0: K2 - 5K + 4 = 0 K = 1 or 4
and coefficient of x2 = coefficient of y2: K2 = 3K - 2 K = 1 or 2
K = 1
Check: x2 + y2 - 8x + 12y - 4 = 0, radius = 16 + 36 + 4 = 56 (real circle).
Answer: (4)