Let a → = i ^ + j ^ + k ^ , b → = − i ^ − 8 j ^ + 2 k ^ and c → = 4 i ^ + c 2 j ^ + c 3 k ^ be three vectors such that b → × a → = c → × a → . If the angle between the vector c → and the vector 3 i ^ + 4 j ^ + k ^ is θ , then the greatest integer less than or equal to tan 2 θ is:

Step-by-step solution

Given: a → = i ^ + j ^ + k , b → = − i ^ − 8 j ^ + 2 k ^ and c → = 4 i ^ + c 2 j ^ + c 3 k Also, b → × a → = c → × a → ⇒ b → − c → × a → = 0 ⇒ b → − c → = λ a → ⇒ b → = c → + λ a → ⇒ − i ^ - 8 j ^ + 2 k = 4 i ^ + c 2 j ^ + c 3 k + λ i ^ + j ^ + k ⇒ λ + 4 = − 1 , λ + c 2 = − 8 , λ + c 3 = 2 ⇒ λ = − 5 , c 2 = − 3 , c 3 = 7 ⇒ c → = 4 i ^ − 3 j ^ + 7 k Now, finding angle we get, cos θ = 3 i ^ + 4 j ^ + k ^ · 4 i ^ − 3 j ^ + 7 k ^ 3 i ^ + 4 j ^ + k ^ ⋅ 4 i ^ − 3 j ^ + 7 k ^ ⇒ cos θ = 12 − 12 + 7 26 ⋅ 74 ⇒ cos θ = 7 26 ⋅