Two unlike forces of equal magnitudes `3hat(i)+hat(k)` and `-3hat(i)-hat(k)` acting at the points `hat(i)+2hat(j)-hat(k)` and `2hat(i)-hat(j)+3hat(k)` respectively. Find the moment of the coupie formed by the forces.

Answer

To find the moment of the couple formed by the two forces, we will follow these steps:

### Step 1: Identify the Forces and Points
We have two forces:
- **Force F1**: \( \vec{F_1} = 3\hat{i} + \hat{k} \) acting at point \( A = \hat{i} + 2\hat{j} - \hat{k} \)
- **Force F2**: \( \vec{F_2} = -3\hat{i} - \hat{k} \) acting at point \( B = 2\hat{i} - \hat{j} + 3\hat{k} \)

### Step 2: Calculate the Position Vector from A to B
The position vector \( \vec{r} \) from point A to point B is given by:
\[ \vec{r} = \vec{B} - \vec{A} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) \]
Calculating this:
\[ \vec{r} = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (3 + 1)\hat{k} = \hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 3: Calculate the Moment of the Couple
The moment of the couple \( \vec{M} \) is given by the cross product of the position vector \( \vec{r} \) and one of the forces. We can use either force; here we will use \( \vec{F_1} \):
\[ \vec{M} = \vec{r} \times \vec{F_1} \]
Substituting the values:
\[ \vec{M} = (\hat{i} - 3\hat{j} + 4\hat{k}) \times (3\hat{i} + \hat{k}) \]

### Step 4: Compute the Cross Product
Using the determinant method for the cross product:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 3 & 0 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{M} = \hat{i} \begin{vmatrix} -3 & 4 \\ 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 3 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -3 & 4 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (0)(4) = -3 \)
2. \( \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} = (1)(1) - (4)(3) = 1 - 12 = -11 \)
3. \( \begin{vmatrix} 1 & -3 \\ 3 & 0 \end{vmatrix} = (1)(0) - (-3)(3) = 0 + 9 = 9 \)

Putting it all together:
\[ \vec{M} = -3\hat{i} + 11\hat{j} + 9\hat{k} \]

### Step 5: Final Result
Thus, the moment of the couple is:
\[ \vec{M} = -3\hat{i} + 11\hat{j} + 9\hat{k} \]

### Step 6: Magnitude of the Moment
To find the magnitude of the moment:
\[ |\vec{M}| = \sqrt{(-3)^2 + (11)^2 + (9)^2} = \sqrt{9 + 121 + 81} = \sqrt{211} \]