$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

$A$ slab is subjected to two forces $\vec F_1$ and $\vec F_2$ of the same magnitude $F$ as shown in the figure. Force $\vec F_2$ is in the $XY$-plane,while force $\vec F_1$ acts along the $z$-axis at the point $(2\hat i + 3\hat j)$. The moment of these forces about point $O$ is:

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 torque of a force $\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $\vec{\tau}$. If the force acts on a particle whose position vector is $\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,then the value of $\vec{\tau}$ will be:

The coordinates of a particle with respect to origin in a given reference frame is $(1, 1, 1) \ m$. If a force of $\vec{F} = \hat{i} - \hat{j} + \hat{k} \ N$ acts on the particle,then the magnitude of the torque (with respect to origin) in the $z$-direction is . . . . . . .

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is: