Three forces `hati+2hatj-3hatk,2hati+3hatj+4hatk and hati-hatj+hatk` acting on a particle at the point (0,1,2) the magnitude of the moment of the forces about the point (1,-2,0) is

Options

A. `2sqrt(35)`
B. `6sqrt(10)`
C. `4sqrt(7)`
D. none of these

Answer

To find the magnitude of the moment of the forces about the point (1, -2, 0), we will follow these steps:

### Step 1: Identify the Forces
The forces acting on the particle 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 Net Force
To find the net force \( \mathbf{F_{net}} \), we sum the individual forces:
\[ \mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} + \mathbf{F_3} \]
Calculating the components:
- \( \hat{i} \) components: \( 1 + 2 + 1 = 4 \)
- \( \hat{j} \) components: \( 2 + 3 - 1 = 4 \)
- \( \hat{k} \) components: \( -3 + 4 + 1 = 2 \)

Thus,
\[ \mathbf{F_{net}} = 4\hat{i} + 4\hat{j} + 2\hat{k} \]

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

### Step 4: Calculate the Moment of Force
The moment of force \( \mathbf{\tau} \) is given by the cross product:
\[ \mathbf{\tau} = \mathbf{AP} \times \mathbf{F_{net}} \]
We can compute this using the determinant:
\[ \mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 2 \\ 4 & 4 & 2 \end{vmatrix} \]

### Step 5: Expand the Determinant
Calculating the determinant:
\[ \mathbf{\tau} = \hat{i} \begin{vmatrix} 3 & 2 \\ 4 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 2 \\ 4 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 3 \\ 4 & 4 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 3 & 2 \\ 4 & 2 \end{vmatrix} = (3)(2) - (2)(4) = 6 - 8 = -2 \)
2. \( \begin{vmatrix} -1 & 2 \\ 4 & 2 \end{vmatrix} = (-1)(2) - (2)(4) = -2 - 8 = -10 \)
3. \( \begin{vmatrix} -1 & 3 \\ 4 & 4 \end{vmatrix} = (-1)(4) - (3)(4) = -4 - 12 = -16 \)

Thus,
\[ \mathbf{\tau} = -2\hat{i} + 10\hat{j} - 16\hat{k} \]

### Step 6: Calculate the Magnitude of the Moment
The magnitude of \( \mathbf{\tau} \) is given by:
\[ |\mathbf{\tau}| = \sqrt{(-2)^2 + (10)^2 + (-16)^2} = \sqrt{4 + 100 + 256} = \sqrt{360} \]
This simplifies to:
\[ |\mathbf{\tau}| = 6\sqrt{10} \]

### Final Answer
The magnitude of the moment of the forces about the point (1, -2, 0) is \( 6\sqrt{10} \).

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