The resultant moment of three forces `hati + 2hatj -3hatk, 2hati + 3hatj + 4hatk` and `-hati- hatj + hatk` acting on particle at a point P (0,1,2) about the point A (1,-2,0) is

Options

A. `6 sqrt2`
B. `sqrt140`
C. `sqrt21`
D. none

Answer

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

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

### Step 2: Calculate the Resultant Force
To find the resultant force \( \mathbf{F} \), we add the three forces together:
\[ \mathbf{F} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} \]
Calculating each component:
- 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:
\[ \mathbf{F} = 2\hat{i} + 4\hat{j} + 2\hat{k} \]

### Step 3: Determine the Position Vector \( \mathbf{R} \)
The position vector \( \mathbf{R} \) from point A to point P is given by:
\[ \mathbf{R} = \mathbf{P} - \mathbf{A} = (0, 1, 2) - (1, -2, 0) = (0 - 1, 1 - (-2), 2 - 0) = -\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 4: Calculate the Moment \( \mathbf{M} \)
The moment \( \mathbf{M} \) about point A due to the resultant force \( \mathbf{F} \) is given by the cross product:
\[ \mathbf{M} = \mathbf{R} \times \mathbf{F} \]
Using the determinant method for the cross product:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 2 \\ 2 & 4 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{M} = \hat{i} \left( 3 \cdot 2 - 2 \cdot 4 \right) - \hat{j} \left( -1 \cdot 2 - 2 \cdot 2 \right) + \hat{k} \left( -1 \cdot 4 - 3 \cdot 2 \right) \]
\[ = \hat{i} (6 - 8) - \hat{j} (-2 - 4) + \hat{k} (-4 - 6) \]
\[ = -2\hat{i} + 6\hat{j} - 10\hat{k} \]

### Step 5: Find the Magnitude of the Moment
The magnitude of the moment vector \( \mathbf{M} \) is calculated as:
\[ |\mathbf{M}| = \sqrt{(-2)^2 + 6^2 + (-10)^2} \]
\[ = \sqrt{4 + 36 + 100} = \sqrt{140} \]

### Final Result
Thus, the resultant moment of the forces about point A is:
\[ |\mathbf{M}| = \sqrt{140} \]