The torque of force $\vec F = -2\hat i + 2\hat j + 3\hat k$ acting on a point $\vec r = \hat i - 2\hat j + \hat k$ about the origin will be:

$A$ force $\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$ is applied at the intersection point of the plane $x = 2$ and the $x$-axis. The magnitude of the torque of this force about the point $(2, 3, 4)$ is .......... . (Round off to the nearest integer)

What is the torque of a force $\vec{F} = 3\hat{i} + 7\hat{j} + 4\hat{k}$ about the origin,if the force acts on a particle whose position vector is $\vec{r} = 2\hat{i} + 2\hat{j} + 1\hat{k}$?

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:

The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

$A$ uniform cube of side $a$ and mass $m$ rests on a rough horizontal table. $A$ horizontal force $F$ is applied normal to one of the faces at a point that is directly above the centre of the face,at a height $\frac{3a}{4}$ above the base. The minimum value of $F$ for which the cube begins to tilt about the edge is (assume that the cube does not slide):