Let a→ and b→ be two vectors such that |a→|=1,|b→|=4 and a→⋅b→=2. If c→=(2a→×b→)−3b→ then find the angle between

|a→|=1,|b→|=4,a→⋅b→=2c→=(2a→×b→)−3b→ or c→+3b→=2a→×b→∵ a→⋅b→=2⇒ |a→|⋅|b→|cos⁡θ=2or  cos⁡θ=2|a→|⋅|b→|=24=12or   θ=π3or  θ=π3⇒ |c→+3b→|2=|2a→×b→|2or  |c→|2+9|b→|2+2c→⋅3b→=4|a→|2|b→|2sin2⁡θor  |c→|2+144+6b→⋅c→=48or  |c→|2+96+6(b→⋅c→)=0------ior  c→=2a→×b→−3b→⇒b→⋅c→=0−3×16=−48Putting value of b b→c→ in Eq. (i), we have|c→|2+96−6×48=0=48×4=192Again, putting the value of c→ in Eq. (i), we have 192+96+6|b→|⋅|c→|cos⁡α=0or  6×4×83cos⁡α=−288or   cos⁡α=−2886×4×83=−323=−32or   α=5π6