If vectors A → = cos ⁡ ω t i ^ + s i n ω t j ^ and B → = c o s ω t 2 i ^ + sin ⁡ ω t 2 j ^ are functions of…

If vectors A=cosωi^+sinωt j^ and B=cosωti^+sinωtj^ are functions of time, then the value of t at which they are orthogonal to each other is:
  1. t=π2ω
  2. t=πω
  3. t=0
  4. t=π4ω

Solution

A=cos wti^+sinwtj^
B=coswt2i^+sinwt2j^
For these vectors to be orthogonal, A·B=0
A·B=0=cos wt.coswt2+sin wt.sin wt2
=coswt-wt2=coswt2=0
So, as we know, \(\cos (90)=0\).
So, wt2=π2 t=πw

Asked in: NEET 2015 (Phase 2)

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