If r → = i ^ + 2 j ^ + 3 k ^ + λ i ^ - j ^ + k ^ and r → = i ^ + 2 j ^ + 3 k ^ + μ i ^ + j ^ - k ^ are two lines, then find the equation of acute angle bisector of two lines.

Given data r → = i ^ + 2 j ^ + 3 k ^ + λ i ^ - j ^ - k ^ . . . line 1 and r → = i ^ + 2 j ^ + 3 k ^ + μ i ^ + j ^ - k ^ . . . line 2 Position vector of P 1 , 2 , 3 and let a → = i ^ - j ^ + k ^ , b → = i ^ + j ^ - k ^ We know that: a → · b → = a → b → cosθ So, cosθ = a → · b → a → b → cosθ = i ^ - j ^ + k ^ i ^ + j ^ - k ^ 1 + 1 + 1 1 + 1 + 1 cosθ = 1 - 1 - 1 9 = - 1 3 Since 1 is acute angle because cosθ is - ve So direction of acute angle bisector. = i ^ - j ^ + k ^ 3 · i ^ + j ^ - k ^ 3 = - 2 j ^ + 2 k ^ 3 Acute angle bisector = i ^ + 2 j ^ + 3 k ^ + δ - 2 j ^ + 2 k ^ 3 Let δ is any constant So r → = i ^ + 2 j ^ + 3 k ^ + t - j ^ + k ^ let t = - 2 δ 3 So acute angle bisector r → = i ^ + 2 j ^ + 3 k ^ + t - j ^ + k ^

Show that equation of angle bisectors of line r → = a → + λ b → and r → = b → + μ a → are r → = ( a → + b → ) + γ ( | b → | a → ± | a → | b → )

∙ Let L 1 ≡ r 1 → = a → + λ b → & L 2 ≡ r 2 → = b → + μ b → be two lines and their point of intersection be P . Solving L 1 & L 2 ⇒ a → + λ b → = b → + μ a → ⇒ a → 1 - μ + b → λ - 1 = 0 Since a → & b → are non-collinear vectors. λ - 1 = 0 , 1 - μ = 0 ⇒ λ = μ = 1 So, P ≡ a → + b → u → = b → , u ^ = b → b → ν → = a → , ν ^ = a → a → So, equation to angle bisectors is given by r → = a → + b → + t u ^ ± ν ^ ⇒ r → = a → + b → + μ b → b → ± a → a → ⇒ r → = a → + b → + ν b → a → ± a → b → Where ν = - μ a → b → Hence proved