A force `F=2hat(i)+hat(j)-hat(k)` acts at point A whose position vector is `2hat(i)-hat(j)`. Find the moment of force F about the origin.

Answer

To find the moment of the force \( \mathbf{F} \) about the origin, we will follow these steps:

### Step 1: Identify the vectors
The force vector \( \mathbf{F} \) is given as:
\[ \mathbf{F} = 2\hat{i} + \hat{j} - \hat{k} \]
The position vector \( \mathbf{r} \) of point A is given as:
\[ \mathbf{r} = 2\hat{i} - \hat{j} \]

### Step 2: Use the formula for the moment of force
The moment of force \( \mathbf{M} \) about the origin is given by the cross product:
\[ \mathbf{M} = \mathbf{r} \times \mathbf{F} \]

### Step 3: Set up the determinant for the cross product
To compute the cross product, we will set up a determinant using the unit vectors \( \hat{i}, \hat{j}, \hat{k} \):
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 2 & 1 & -1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Now we will calculate the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} -1 & 0 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -1 & 0 \\ 1 & -1 \end{vmatrix} = (-1)(-1) - (0)(1) = 1 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & 0 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (0)(2) = -2 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} = (2)(1) - (-1)(2) = 2 + 2 = 4 \]

### Step 5: Combine the results
Now substituting back into the expression for \( \mathbf{M} \):
\[ \mathbf{M} = 1\hat{i} - (-2)\hat{j} + 4\hat{k} = \hat{i} + 2\hat{j} + 4\hat{k} \]

### Step 6: Write the final answer
Thus, the moment of force \( \mathbf{F} \) about the origin is:
\[ \mathbf{M} = \hat{i} + 2\hat{j} + 4\hat{k} \]