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:

Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$.

$A$ 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 ............. $N m$.

$A$ force $F = 2.0\,N$ acts on a particle $P$ in the $xz-$ plane. The force $F$ is parallel to the $x-$ axis. The particle $P$ (as shown in the figure) is at a distance $3\,m$ from the origin,and the line joining $P$ with the origin makes an angle of $30^\circ$ with the $x-$ axis. The magnitude of the torque on $P$ with respect to the origin $O$ (in $N-m$) is:

$A$ couple produces:

$A$ force $\vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ is applied at a point with position vector $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point with position vector $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is: