IF the force represented by ` 3hatj +2hatk` is acting through the point ` 5hati +4hatj -3hatk` , then its moment about th point (1,3,1) is

Options

A. ` 14 hati -8 hatj +12hatk`
B. ` -14hati +8hatj -12hatk`
C. ` -6hati -hatj +9hatk`
D. ` 6hati +hatj -9hatk`

Explanation

To find the moment of the force represented by the vector \( \vec{F} = 3\hat{j} + 2\hat{k} \) about the point \( (1, 3, 1) \), we will follow these steps:

### Step 1: Identify the points and vectors
The force vector is given as:
\[ \vec{F} = 3\hat{j} + 2\hat{k} \]
The point through which the force is acting is:
\[ \vec{P} = 5\hat{i} + 4\hat{j} - 3\hat{k} \]
The point about which we need to find the moment is:
\[ \vec{Q} = 1\hat{i} + 3\hat{j} + 1\hat{k} \]

### Step 2: Calculate the position vector \( \vec{r} \)
The position vector \( \vec{r} \) from point \( \vec{Q} \) to point \( \vec{P} \) is calculated as:
\[ \vec{r} = \vec{P} - \vec{Q} = (5\hat{i} + 4\hat{j} - 3\hat{k}) - (1\hat{i} + 3\hat{j} + 1\hat{k}) \]
Calculating the components:
\[ \vec{r} = (5 - 1)\hat{i} + (4 - 3)\hat{j} + (-3 - 1)\hat{k} = 4\hat{i} + 1\hat{j} - 4\hat{k} \]

### Step 3: Calculate the moment \( \vec{M} \)
The moment \( \vec{M} \) about point \( \vec{Q} \) is given by the cross product:
\[ \vec{M} = \vec{r} \times \vec{F} \]
Substituting the vectors:
\[ \vec{M} = (4\hat{i} + 1\hat{j} - 4\hat{k}) \times (0\hat{i} + 3\hat{j} + 2\hat{k}) \]

### Step 4: Set up the determinant for the cross product
We can set up the determinant as follows:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & -4 \\ 0 & 3 & 2 \end{vmatrix} \]

### Step 5: Calculate the determinant
Calculating the determinant:
\[ \vec{M} = \hat{i} \begin{vmatrix} 1 & -4 \\ 3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & -4 \\ 0 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 1 \\ 0 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -4 \\ 3 & 2 \end{vmatrix} = (1)(2) - (-4)(3) = 2 + 12 = 14 \]
2. For \( -\hat{j} \):
\[ \begin{vmatrix} 4 & -4 \\ 0 & 2 \end{vmatrix} = (4)(2) - (-4)(0) = 8 \]
3. For \( \hat{k} \):
\[ \begin{vmatrix} 4 & 1 \\ 0 & 3 \end{vmatrix} = (4)(3) - (1)(0) = 12 \]

Putting it all together:
\[ \vec{M} = 14\hat{i} - 8\hat{j} + 12\hat{k} \]

### Final Result
Thus, the moment about the point \( (1, 3, 1) \) is:
\[ \vec{M} = 14\hat{i} - 8\hat{j} + 12\hat{k} \]

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