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If |vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)...

If `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`, then the value of `|vec(A)+vec(B)|` is

A

`(A^(2)+B^(2)+AB)^(1//2)`

B

`(A^(2)+B^(2)+(AB)/(sqrt(3)))^(1//2)`

C

`A+B`

D

`(A^(2)+B^(2)+sqrt(3)AB)^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`….(i)
but `|vec(A)xxvec(B)|=|vec(A)||vec(B)| sin theta= AB sin theta`
and `vec(A).vec(B)=|vec(A)||vec(B)|cos theta= AB cos theta`
Make these substitution in Eq.(i), we get
`AB sin theta= sqrt(3) AB cos theta`
or `tan theta =sqrt(3)implies theta= 60^(@)` The resultant of vector `vec(A)` and `vec(B)` can be given by the law of parallelogram.
`:. |vec(A)+vec(B)|= sqrt(A^(2)+B^(2)+2AB cos 60^(@))`
`=sqrt(A^(2)+B^(2)+2ABxx1/2)`
`=(A^(2)+B^(2)+AB)^(1//2)`
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