At a given instant of time the position vector of a particle moving in a circle with a velocity `3hat i - 4 hat j + 5 hat k` is `hat i + 9 hat j - 8 hat k`. Its angular velocity at that time is:

Options

A. `((13 hati-29hatj-31hatk))/(sqrt(146))`
B. `((13 hati-29 hatj-31hatk))/146`
C. `((13 hati+29 hatj-31hatk))/(sqrt(146))`
D. `((13 hati+29hatj+31hatk))/146`

Answer

B

Explanation

`vec(omega)=(vec(r)xxvec(v))/(r^(2))`
`3hati-4hatj+5hatk=(xhati+yhatj+zhatk)xx(hati+9hatj-8hatk)`