JEE Main202628 January 2026Morning ShiftMathematicsVector AlgebraActual
For three unit vectors a , b , c satisfying | a - b |²+| b - c |²+| c - a |²=9 and |2 a +k b +k c |=3 , the positive value of k is
Options
- A4
- B3
- C5
- D6
Correct answer
C. 5
Step-by-step solution
Expanding | a - b |^2 + | b - c |^2 + | c - a |^2 = 9 with unit vectors gives 6 - 2( a b + b c + c a ) = 9 . So a b + b c + c a = -3/2 . For the symmetric case where vectors are equally spaced (e.g., on a circle at 120° intervals), we have a b = b c = c a = -1/2 . Expanding |2 a + k b + k c |^2 = 9 gives 4 + 2k^2 + 4k( a b + a c ) + 2k^2 b c = 9 . Substituting the symmetric values: 4 + 2k^2(1/2) + 4k(-1) + 2k^2(-1/2) = 9 , which simplifies to k^2 - 4k - 5 = 0 . Factoring: (k-5)(k+1) = 0 , giving k = 5 as the positi