Let a force $\vec{F} = -F\hat{k}$ act on the origin of a Cartesian frame of reference. The moment of force about a point $(1, -1)$ will be

$A$ force $F_1 = A \hat{j}$ is applied to a point whose radius vector is $r_1 = a \hat{i}$, while a force $F_2 = B \hat{i}$ is applied to the point whose radius vector is $r_2 = b \hat{j}$. Both the radius vectors are determined relative to the origin of the coordinate axes $O$. The moment of the force relative to $O$ is

If force $\vec{F} = -3 \hat{i} + \hat{j} + 5 \hat{k}$ acts at a position vector $\vec{r} = 7 \hat{i} + 3 \hat{j} + \hat{k}$,then the torque $\vec{\tau}$ acting at that point is:

$A$ force $\vec{F}=(2 \hat{i}+3 \hat{j}-5 \hat{k}) \, N$ acts at a point $\vec{r}_1=(2 \hat{i}+4 \hat{j}+7 \hat{k}) \, m$. The torque of the force about the point $\vec{r}_2=(\hat{i}+2 \hat{j}+3 \hat{k}) \, m$ is ............. $N m$.

Let $\vec{F}$ be a force acting on a particle having position vector $\vec{r}$,and $\vec{\tau}$ be the torque of this force about the origin. Then:

The position vectors of two particles of a rigid body are $(3, 0, 0) \ m$ and $(0, 3, 0) \ m$. Forces of $(0, 1, 0) \ N$ and $(0, -1, 0) \ N$ act on these particles respectively. The torque of the couple is ....... $Nm$.