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 |
whereiandf 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)
wheretheta 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)
Giventheta=60^@
=>| vecv_f - vecv_i |=4xx2sin(60/2)=4