JEE Main202628 January 2026Morning ShiftMathematicsVector AlgebraActual
Let P Q R be a triangle such that P Q =-2 i - j +2 k and PR =a i +b j -4 k , a, b Z . Let S be the point on QR, which is equidistant from the lines PQ and PR. If | PR |=9 and PS = i -7 j +2 k , then the value of 3 a-4 b is _ _ _ _
Correct answer
0
Step-by-step solution
Given: PQ = -2 i - j + 2 k , PR = a i + b j - 4 k ( a, b Z ), PS = i - 7 j + 2 k , | PR | = 9 From | PR | = 9 : a^2 + b^2 + 16 = 81 a^2 + b^2 = 65 ...(1) Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So: = PQ PS | PQ || PS | = -2+7+4 3 3 6 = 9 9 6 = 1 6 Similarly: 1 6 = PS PR | PS || PR | = a - 7b - 8 3 6 9 a - 7b = 35 ...(2) From (1) and (2): a = 7, b = -4 3a - 4b = 21 + 16 = 37 (NTA Answer) However, QS = 3 i - 6 j and SR = 6 i + 3 j - 6 k are not parallel, so Q, S, R are not colline