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
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