How is the direction of torque determined?

$A$ wheel of radius $R$ with an axle of radius $R/2$ is as shown in the figure. It is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure: a force $F$ at an angle of $45^{\circ}$ to the tangent of the outer rim,a force $F$ tangential to the inner axle,and a force $2F$ tangential to the outer rim. The magnitude of the net torque acting on the system is nearly ............. $FR$.

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

If force $\vec{F} = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ acts on a particle having position vector $\vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$,then the torque about the origin will be:

In rotational motion,which quantity is analogous to force in linear motion?

$A$ 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: