A vector of magnitude 4 unit rotates through an angle of #60degree#.Then magnitude of its change is?
options are
#2sqrt3 unit#
- 6 unit
- 8 unit
- 4unit
options are
#2sqrt3 unit# - 6 unit
- 8 unit
- 4unit
1 Answer
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
As the magnitude
#| 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#