If `vec F` is the force acting in a particle having position vector `vec r` and `vec tau` be the torque of this force about the origin, then

Let vecF be a force acting on a particle having positon vector vecr. Let vecr be the torque of this force about the origin then

If is the force acting on a particle having position vector and be the torque of this force about the origin, then-

A force vec(F) = (2 hat(i) + 3 hat(j) + 4 hat(k)) N is applied to a point having position vector vec(r) = (3 hat(i) + 2 hat(j) + hat(k)) m. Find the torque due to the force about the axis passing through origin.

A particle of mass m moves in the xy-plane with velocity of vec v = v_x hat i+ v_y hat j . When its position vector is vec r = x vec i + y vec j , the angular momentum of the particle about the origin is.

If vec a,vec b,vec c are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin,then

A slab is subjected to two forces vec F_(1) and vec F_(2) of same magnitude F as shown in the figure . Force vec F_(2) is in XY-plane while force F_(1) acts along z-axis at the point (2 vec I + 3 vec j) .The moment of these forces about point O will be :

If vec a,vec b,vec c are the position vectors of the vertices of an equilateral triangle whose orthocentre is t the origin,then write the value of vec a+vec b+vec *