Find the moment of the couple formed by the forces `5hat(i)+hat(k)` and `-5hat(i)-hat(k)` acting at the points `(9, -1, 2)` and `(3, -2,1)` respectively.

Answer

To find the moment of the couple formed by the forces \( \mathbf{F_1} = 5\hat{i} + \hat{k} \) and \( \mathbf{F_2} = -5\hat{i} - \hat{k} \) acting at the points \( A(9, -1, 2) \) and \( B(3, -2, 1) \) respectively, we can follow these steps:

### Step 1: Identify the position vectors
The position vector of point A is:
\[ \mathbf{A} = 9\hat{i} - 1\hat{j} + 2\hat{k} \]
The position vector of point B is:
\[ \mathbf{B} = 3\hat{i} - 2\hat{j} + 1\hat{k} \]

### Step 2: Calculate the vector \( \mathbf{AB} \)
The vector \( \mathbf{AB} \) is given by:
\[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (3 - 9)\hat{i} + (-2 + 1)\hat{j} + (1 - 2)\hat{k} \]
Calculating this gives:
\[ \mathbf{AB} = -6\hat{i} + 1\hat{j} - 1\hat{k} = -6\hat{i} + \hat{j} - \hat{k} \]

### Step 3: Calculate the moment due to force \( \mathbf{F_1} \)
The moment \( \mathbf{M_1} \) due to force \( \mathbf{F_1} \) at point A is given by:
\[ \mathbf{M_1} = \mathbf{AB} \times \mathbf{F_1} \]
Substituting the values:
\[ \mathbf{M_1} = (-6\hat{i} + \hat{j} - \hat{k}) \times (5\hat{i} + \hat{k}) \]

### Step 4: Calculate the cross product
Using the determinant method for the cross product:
\[ \mathbf{M_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 1 & -1 \\ 5 & 0 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \mathbf{M_1} = \hat{i} \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -6 & -1 \\ 5 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -6 & 1 \\ 5 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i}(1 \cdot 1 - 0 \cdot -1) - \hat{j}(-6 \cdot 1 - 5 \cdot -1) + \hat{k}(-6 \cdot 0 - 5 \cdot 1) \]
\[ = \hat{i}(1) - \hat{j}(-6 + 5) + \hat{k}(0 - 5) \]
\[ = \hat{i} + \hat{j} - 5\hat{k} \]

### Step 5: Calculate the moment due to force \( \mathbf{F_2} \)
The moment \( \mathbf{M_2} \) due to force \( \mathbf{F_2} \) at point B is given by:
\[ \mathbf{M_2} = \mathbf{BA} \times \mathbf{F_2} \]
Where \( \mathbf{BA} = \mathbf{A} - \mathbf{B} = (9 - 3)\hat{i} + (-1 + 2)\hat{j} + (2 - 1)\hat{k} = 6\hat{i} + 1\hat{j} + 1\hat{k} \).

Calculating \( \mathbf{M_2} \):
\[ \mathbf{M_2} = (6\hat{i} + \hat{j} + \hat{k}) \times (-5\hat{i} - \hat{k}) \]
Using the determinant method:
\[ \mathbf{M_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & 1 \\ -5 & 0 & -1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & 1 \\ -5 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 1 \\ -5 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i}(-1) - \hat{j}(-6 + 5) + \hat{k}(0 - (-5)) \]
\[ = -\hat{i} + \hat{j} + 5\hat{k} \]

### Step 6: Find the total moment of the couple
The total moment \( \mathbf{M} \) is given by:
\[ \mathbf{M} = \mathbf{M_1} + \mathbf{M_2} \]
Substituting the values:
\[ \mathbf{M} = (\hat{i} + \hat{j} - 5\hat{k}) + (-\hat{i} + \hat{j} + 5\hat{k}) \]
\[ = (1 - 1)\hat{i} + (1 + 1)\hat{j} + (-5 + 5)\hat{k} \]
\[ = 0\hat{i} + 2\hat{j} + 0\hat{k} = 2\hat{j} \]

### Final Answer
The moment of the couple is:
\[ \mathbf{M} = 2\hat{j} \]