As shown in figure, suppose a force `vecF` acts on a particle P. The position vector of of P with respect to orgin O is `vecr .` The angle between `vecr` and `vecF` is `theta .` Here, the particle P is not necessarily be a particle of a rigid body.

The vector product of `vecr` and `vecF` is called the torque acting on the particle P, with respect to the point O.

`vectau = vecr xx vecF `

`tau = rFsintheta`

From figure, `rsintheta = OQ` = the perpendicular distance of the line of action of force from O

`tau = (F) ` (perpendicular distance of line of action of force from O)

= Moment of force with respect to point O

Thus, torque is the moment of force with respect to a given reference point. Its dimensional formula is `ML^2T^(-2)` and its unit is N m.
(i) According to the right hand screw rule the direction of torque `(vectau)` is perpendicular to the plane formed by `vecr` and `vecF`
(ii) Since the value of torque `vectau` depends on the reference point, in definition the torque, the reference point must be mention.

The mutual internal force between the particle of a system are equal and opposite, the resultant force and hence the torque produced due to them become zero. Hence, we will not consider the internal force in our discussion.

Suppose for a system of particles the position vector of different particles are `vecr_1, vecr_2, ........................vecr_n, ` and the respective force acting on them are `vecF_1`, `vecF_2..........vecF_n` The resultant torque on the system means the vector sum of the torque acting on every particle of the system.

`vectau = vectau_1 +vectau_2 +.......vectau_n`

`tau =(vecr_1xxvecF_1) + (vecr_2xxvecF_2)+..........(vecr_nxxvecF_n)`

`= sum_(i =1)^n (vecr_ixxvecF_i)`

Suppose a rigid body rotates about a fixed axis OZ, as shown in fig. The net force acting on the particles with position vectors `vecr_1, vecr_2...................vecr_n, ` are `vecF_1, vecF_2, ............vecF_n` respectively.

Considering the force `vecF_n` acting on the particle with position vector `vecr_n` the torque `vectau_n` acting on it is

`vectau_n = vecr_n xx vecF_n`

`= | (hati ,hatj, hatk),(x_n, y_n, z_n),(F_(nx),F_(ny),F_(nz)) |`

From the equation the torque acting on the entire body can be written as a vector sum of the torques acting on all particles, as follows:

`vec tau_n = sum_n(y_n F_(nz) - z_n F_(ny) hati + z_n F_(nx) - x_n F_(ny) hatj) + x_n F_(nz) - y_n F_(nx) hatk)`

For the rotational motion of the rigid body about Z-axis, only the Z- component of the above mentioned torque is responsible. For the rotational motion about Y-axis the Y-component of the torque is responsible. `tau.hatn` component of the torque is responsible for the rotational motion about axis.

To produce the rotational motion of the rigid body external force are required to be applied, but not on all particle of it. As for example, we do not apply force on all the particle of a door to open it or shunt.