The linear velocity perpendicular to radius vector of a particle moving with angular velocity `omega=2hatK` at position vector `r=2hati+2hatj` is

The linear velocity of a rotating body is given by v =omega xx r , where omega is the angular velocity and r is the radius vector. The angular velocity of a body omega=hati-2hatj+2hatk and their radius vector r=4hatj-3hatk, |v| is -

The linear velocity of rotating body is given by vecv = vecomega xx vecr where vecomega is the angular velocity and vecr is the radius vector. The angular velocity of body vecomega = hati - 2hatj + 2hatk and this radius vector vecr = 4hatj - 3hatk , then |vecv| is

The linear velocity of a particle on a rotating body is given by vec v = vec omega xx vec r" where "vec omega is the angular velocity and vec r is the radius vector. What is the value of |v| if vec omega = hati -2hatj+2hatk and vec r =4hatj-3hatk ?

The radius vector vec r=2i+5k and the angular velocity of a particle is vec w=3i-4k Then the linear velocity of the particle is

The radius vector vec r=2i+5k and the angular velocity of a particle is vec w=3i-4k .Then the linear velocity of the particle is

A unit vector perpendicular to the plane passing through the points whose position vectors are hati-hatj+2hatk, 2hati-hatk and 2hati+hatk is

A unit vector perpendicular to the plane passing through the points whose position vectors are hati-hatj+2hatk, 2hati-hatk and 2hati+hatk is

The radius vector vecr =2i+5k and the angular velocity of a particle is w = 3i - 4k. Then the linear velocity of the particle is

The foot of the perpendicular drawn from a point with position vector hati+4hatk on the joining the points having position vectors as -11hatj+3hatk an 2hati-3hatj+hatk has the position vector