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If Vectors vec(A)= cos omega hat(i)+ sin...

If Vectors `vec(A)= cos omega hat(i)+ sin omega hat(j)` and `vec(B)=(cos)(omegat)/(2)hat(i)+(sin)(omegat)/(2)hat(j)`are functions of time. Then the value of `t` at which they are orthogonal to each other is

A

`t=0`

B

`t= (pi)/(4omega)`

C

`t = (pi)/(2omega)`

D

`t= (pi)/(omega)`

Text Solution

Verified by Experts

The correct Answer is:
4
`vecA *vecB=0`
`cos omega t cos"" (omegat)/(2) + sin omega t sin""(omegat)/(2) =0`
`cos(omegat- (omegat)/(2)) =0 rArr cos""(omegat)/(2) =0`
`rArr (omegat)/(2) = (pi)/(2) rArr t= (pi)/(omega)`
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