The moment of the couple formed by the forces ` 5hati +hatj and -5hati - hatk` acting at the points (9 ,-1,2) and ( 3,-2,1) respectively, is

Options

A. `-hati+hatj+5hatk`
B. `hati-hatj-5hatk`
C. `2hati-2hatj-10hatk`
D. `-2hati+2hatj+10hatk`

Answer

To find the moment of the couple formed by the forces \( \mathbf{F_1} = 5\hat{i} + \hat{j} \) and \( \mathbf{F_2} = -5\hat{i} - \hat{k} \) acting at the points \( P_1(9, -1, 2) \) and \( P_2(3, -2, 1) \), we can follow these steps:

### Step 1: Find the position vector \( \mathbf{P} \)
The position vector \( \mathbf{P} \) is given by the difference between the position vectors of the two points:
\[ \mathbf{P} = \mathbf{P_1} - \mathbf{P_2} \]
Where:
\[ \mathbf{P_1} = 9\hat{i} - 1\hat{j} + 2\hat{k} \]
\[ \mathbf{P_2} = 3\hat{i} - 2\hat{j} + 1\hat{k} \]

Calculating \( \mathbf{P} \):
\[ \mathbf{P} = (9 - 3)\hat{i} + (-1 + 2)\hat{j} + (2 - 1)\hat{k} = 6\hat{i} + 1\hat{j} + 1\hat{k} \]

### Step 2: Choose one of the forces
Since the forces are equal in magnitude but opposite in direction, we can choose one of the forces for our calculations. We will use:
\[ \mathbf{F} = 5\hat{i} + \hat{j} \]

### Step 3: Calculate the moment of the couple \( \mathbf{M} \)
The moment of the couple is given by the cross product:
\[ \mathbf{M} = \mathbf{P} \times \mathbf{F} \]

Substituting the values:
\[ \mathbf{M} = (6\hat{i} + 1\hat{j} + 1\hat{k}) \times (5\hat{i} + 1\hat{j} + 0\hat{k}) \]

### Step 4: Set up the determinant for the cross product
We can express the cross product using a determinant:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & 1 \\ 5 & 1 & 0 \end{vmatrix} \]

### Step 5: Calculate the determinant
Calculating the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & 1 \\ 5 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 1 \\ 5 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(1) = -1 \)
2. \( \begin{vmatrix} 6 & 1 \\ 5 & 0 \end{vmatrix} = (6)(0) - (1)(5) = -5 \)
3. \( \begin{vmatrix} 6 & 1 \\ 5 & 1 \end{vmatrix} = (6)(1) - (1)(5) = 1 \)

Substituting back into the equation for \( \mathbf{M} \):
\[ \mathbf{M} = -1\hat{i} - (-5)\hat{j} + 1\hat{k} = -1\hat{i} + 5\hat{j} + 1\hat{k} \]

### Final Result
Thus, the moment of the couple is:
\[ \mathbf{M} = -1\hat{i} + 5\hat{j} + 1\hat{k} \]