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If vector A=cosωti^+sinωtj^  and  B=cosωt2i^+sinωt2j^  are functions of time,  then the value of t at which they are orthogonal to each other is

 

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By Expert Faculty of Sri Chaitanya
a
t=πω
b
t=0
c
t=π4ω
d
t=π2ω
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detailed solution

Correct option is A

Two vectors A→ and B→ are orthogonal to each other, if  their scalar product is zero i.e. A→·B→=0 .  Here, A→=cosωti^+sinωtj^  and   B→=cosωt2i^+sinωt2j^∴A→·B→=(cosωti^+sinωtj^)·cosωt2i^+sinωt2j^=cosωtcosωt2+sinωtsinωt2(∵i^·i^=j^·j^=1 and i^·j^=j^·i^=0)=cosωt-ωt2(∵cos(A-B)=cosAcosB+sinAsinB) But A→·B→=0 (as A→ and B→ are orthogonal to each other) ∴  cosωt-ωt2=0cosωt-ωt2=cosπ2 or ωt-ωt2=π2ωt2=π2 or t=πω


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