Find the vector moment of the forces :
`hat(i)+2hat(j)-3hat(k), 2hat(i)+3hat(j)+4hat(k)` and `-hat(i)-hat(j)+hat(k)`
acting on a particle at a point P (0, 1, 2) about the point `A(1, -2, 0)`.

Answer

To find the vector moment of the forces acting on a particle at point P(0, 1, 2) about the point A(1, -2, 0), we will follow these steps:

### Step 1: Identify the Forces
The forces acting on the particle are:
1. \( \mathbf{F_1} = \hat{i} + 2\hat{j} - 3\hat{k} \)
2. \( \mathbf{F_2} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
3. \( \mathbf{F_3} = -\hat{i} - \hat{j} + \hat{k} \)

### Step 2: Calculate the Resultant Force
To find the resultant force \( \mathbf{F_R} \), we sum all the forces:
\[ \mathbf{F_R} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} \]
Calculating the components:
- For \( \hat{i} \): \( 1 + 2 - 1 = 2 \)
- For \( \hat{j} \): \( 2 + 3 - 1 = 4 \)
- For \( \hat{k} \): \( -3 + 4 + 1 = 2 \)

Thus, the resultant force is:
\[ \mathbf{F_R} = 2\hat{i} + 4\hat{j} + 2\hat{k} \]

### Step 3: Determine the Position Vectors
Next, we need the position vectors of points A and P:
- Position vector of A: \( \mathbf{A} = \hat{i} - 2\hat{j} + 0\hat{k} \)
- Position vector of P: \( \mathbf{P} = 0\hat{i} + 1\hat{j} + 2\hat{k} \)

### Step 4: Calculate the Position Vector from A to P
The position vector \( \mathbf{r} \) from A to P is given by:
\[ \mathbf{r} = \mathbf{P} - \mathbf{A} \]
Calculating the components:
- For \( \hat{i} \): \( 0 - 1 = -1 \)
- For \( \hat{j} \): \( 1 - (-2) = 3 \)
- For \( \hat{k} \): \( 2 - 0 = 2 \)

Thus, the position vector is:
\[ \mathbf{r} = -\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 5: Calculate the Moment of the Force
The moment \( \mathbf{M} \) of the force about point A is given by the cross product:
\[ \mathbf{M} = \mathbf{r} \times \mathbf{F_R} \]
Substituting the vectors:
\[ \mathbf{M} = (-\hat{i} + 3\hat{j} + 2\hat{k}) \times (2\hat{i} + 4\hat{j} + 2\hat{k}) \]

### Step 6: Compute the Cross Product
We can compute the cross product using the determinant:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 2 \\ 2 & 4 & 2 \end{vmatrix} \]
Calculating the determinant:
- For \( \hat{i} \): \( (3 \cdot 2 - 2 \cdot 4) = 6 - 8 = -2 \)
- For \( \hat{j} \): \( -(-1 \cdot 2 - 2 \cdot 2) = 2 + 4 = 6 \)
- For \( \hat{k} \): \( -(-1 \cdot 4 - 3 \cdot 2) = 4 + 6 = 10 \)

Thus, the moment vector is:
\[ \mathbf{M} = -2\hat{i} + 6\hat{j} - 10\hat{k} \]

### Step 7: Final Result
The vector moment of the forces about point A is:
\[ \mathbf{M} = -2\hat{i} + 6\hat{j} - 10\hat{k} \]