A vector of magnitude 4 unit rotates through an angle of #60degree#.Then magnitude of its change is?

options are

  1. #2sqrt3 unit#
  2. 6 unit
  3. 8 unit
  4. 4unit

1 Answer
A08
Aug 2, 2018

4.

Explanation:

The magnitude of change in vector is

#| vecv_f - vecv_i |#
where #iandf# subscripts show initial and final vectors.

We know that magnitude of a vector is #=# square root of the dot product of vector with itself.
As the magnitude #4# does not change we have

#| vecv_f - vecv_i |=sqrt( 4^2 + 4^2 - 2xx4xx4costheta)#
where #theta# is the angle between the vectors.
#=>| vecv_f - vecv_i |=4sqrt( 2(1 - costheta))#

Converting into half angle formula we get

#| vecv_f - vecv_i |=4xx2sin(theta/2)#
Given #theta=60^@#
#=>| vecv_f - vecv_i |=4xx2sin(60/2)=4#

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