Torque measured turning effect as $\tau = rF\sin\theta$. The The Cross Product packages that size together with a direction, which you need as soon as forces don't all lie in one plane.
Key idea
The torque about a point O of a force $\vec{F}$ applied at position $\vec{r}$ (measured from O to where the force acts) is $$\vec{\tau} = \vec{r}\times\vec{F}, \qquad |\vec{\tau}| = rF\sin\theta.$$ Its direction, by the right-hand rule, is along the axis the force tends to turn the object about. In a plane, $\tau_z = r_xF_y - r_yF_x$: positive is counterclockwise seen from $+z$.
Why it works
Torque should grow with the force and with the lever arm, and only the part of the force perpendicular to $\vec{r}$ should turn anything; pushing straight toward the pivot does nothing. The cross product has exactly these properties: its size is $rF\sin\theta$, zero when $\vec{F}$ is along $\vec{r}$. And its direction answers "turning about which axis, and which way?" in one vector, which lets torques from different directions add as vectors.
Worked example: A force in the plane
A force $\vec{F} = \langle 3, 4, 0 \rangle\ \text{N}$ acts at $\vec{r} = \langle 2, -1, 0 \rangle\ \text{m}$. Find the torque about the origin.
- $\tau_z = r_xF_y - r_yF_x = (2)(4) - (-1)(3) = 11\ \text{N·m}$.
- The $x$- and $y$-components are zero, because both vectors lie in the $xy$-plane.
Answer: $\vec{\tau} = \langle 0, 0, 11 \rangle\ \text{N·m}$.
Worked example: Three dimensions
A force $\vec{F} = \langle 0, 3, 4 \rangle\ \text{N}$ acts at $(1, 2, 0)\ \text{m}$. Find the torque about the origin.
- $$\vec{\tau} = \langle r_yF_z - r_zF_y,\ r_zF_x - r_xF_z,\ r_xF_y - r_yF_x \rangle$$
- $$= \langle 2(4) - 0(3),\ 0(0) - 1(4),\ 1(3) - 2(0) \rangle = \langle 8, -4, 3 \rangle\ \text{N·m}$$
- Check: $\vec{\tau}\cdot\vec{r} = 8 - 8 + 0 = 0$ and $\vec{\tau}\cdot\vec{F} = 0 - 12 + 12 = 0$. The torque is perpendicular to both, as a cross product must be.
Answer: $\langle 8, -4, 3 \rangle\ \text{N·m}$.
Because torques are vectors, several of them add component by component, and an object is in rotational equilibrium only when all three components of the total are zero. In Static Equilibrium Problems you'll pick the point about which to take torques so that unknown forces drop out.
Common mistake
Two slips: measuring $\vec{r}$ from the origin when the torque is wanted about another point, and reversing the order. About a point Q, use $\vec{r} = \vec{r}_{\text{force}} - \vec{r}_Q$; the same force has different torques about different points. And $\vec{F}\times\vec{r} = -\vec{r}\times\vec{F}$, so the wrong order flips every sign. Check with the right-hand rule: curl your fingers the way the force would turn the object; your thumb must point along $\vec{\tau}$.