Q.Find the torque of a force $7\hat{i} + 3\hat{j} - 5\hat{k}$ about the origin. The force acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}$.

The key idea is that torque, a rotational analogue of force, is calculated as the cross product of the position vector and the force vector.

The position vector is given as $\vec{r} = \hat{i} - \hat{j} + \hat{k}$, and the force vector is $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$.

We compute the cross product:

$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}$ …

When a force acts on an object, it can cause linear motion (translation) or rotational motion. Torque is the rotational equivalent of force; it's the physical quantity that causes an object to rotate about an axis.

The effectiveness of a force in causing rotation depends on three factors:

1. Magnitude of the force: A larger force generally produces a larger torque.
2. Distance from the pivot (axis of rotation): The further the force is applied from the pivot, the greater its rotational effect. This distance is represented by the position vector $\vec{r}$ from the pivot to the point of application of the force.
3. Angle between the force and the position vector: Only the component of the force perpendicular to the position vector contributes to rotation. If the force acts directly towards or away from the pivot, it causes no rotation.

The cross product (or vector product) is the mathematical operation that naturally combines these three factors. For two vectors $\vec{A}$ and $\vec{B}$, their cross product $\vec{A} \times \vec{B}$ results in a new vector whose magnitude is $|\vec{A}||\vec{B}|\sin\theta$ (where $\theta$ is the angle between them) and whose direction is perpendicular to both $\vec{A}$ and $\vec{B}$ (given by the right-hand rule). This perfectly aligns with the definition of torque.

Let's apply this concept to the given problem.

1. Identify the given vectors.

   The force vector is $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$.

   The position vector of the particle, relative to the origin (which is our pivot point for "about the origin"), is $\vec{r} = \hat{i} - \hat{j} + \hat{k}$.

2. Set up the cross product.

   We need to calculate $\vec{\tau} = \vec{r} \times \vec{F}$. The cross product of two 3D vectors can be conveniently calculated using a determinant:

Substituting the components of $\vec{r}$ and $\vec{F}$:

3. Calculate the components of the torque vector. We expand the determinant along the first row:
   • $\hat{i}$ component: …

Concept: Torque as the Cross Product of Position and Force

Step 1: Identify the vectors

Position vector $\vec r = \hat i - \hat j + \hat k$; force vector $\vec F = 7\hat i + 3\hat j - 5\hat k$. Torque about the origin: $\vec\tau = \vec r\times\vec F$.

Step 2: Set up the determinant

Step 3: Expand along the first row …