If 2a^+b→=a^×b→ (where a^ is a unit vector), then |3b→+4a^|=
We have 2a^+b^=a^×b^--1
Taking dot product with a^×b^, we get
(a^×b^)⋅(a^×b^)=0⇒ b→=λa^
So, from (1), λ=−2⇒b→=−2a^
∴|3b→+4a^|=|3(−2a^)+4a^|=|2a^|=2