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