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 K ² x²+ ( K ²-5 ~K + ₁ ) x y+ (3 ~K + ₂ 2 ) y²-8 x+12 y+ ₂=0 represents a circle, for K equal to :

Step-by-step solution

From c a = a d where a = 2 i + 3 j + 4 k , we get ( c + d ) a = 0 . So c + d = t a . Given | c + d | = 29 : |t| 29 = 29 , so t = 1 . ( c + d ) (-7 i + 2 j + 3 k ) = (-14 + 6 + 12) = 4 . ₁ = 4 , ₂ = -4 . For circle: Coeff of x^2 = Coeff of y^2 : K^2 = 3K - 2 K = 1, 2 . Coeff of xy = 0 : K^2 - 5K + 4 = 0 K = 1, 4 . Common value: K = 1 .