Q.

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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a

t=πω

b

t=0

c

t=π4ω

d

t=π2ω

answer is A.

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Detailed Solution

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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