What is the moment about the point `hat(i) +2hat(j)-hat(k)` of a force represented by ` 3 hat(i) + hat(k)` acting through the point `2hat(i) - hat(j)+3hat(k)` ?

Options

A. `-3hat(i)+11hat(j)+9hat(k)`
B. `3hat(i)+2hat(j)+9hat(k)`
C. `3hat(i)+4hat(j)+9hat(k)`
D. `hat(i)+hat(j)+hat(k)`

Explanation

To find the moment about the point \( \hat{i} + 2\hat{j} - \hat{k} \) of a force represented by \( 3\hat{i} + \hat{k} \) acting through the point \( 2\hat{i} - \hat{j} + 3\hat{k} \), we will follow these steps:

### Step 1: Identify the position vectors
Let:
- \( \mathbf{r_1} = 2\hat{i} - \hat{j} + 3\hat{k} \) (point where the force is applied)
- \( \mathbf{r_2} = \hat{i} + 2\hat{j} - \hat{k} \) (point about which we are calculating the moment)
- \( \mathbf{F} = 3\hat{i} + \hat{k} \) (force vector)

### Step 2: Calculate the position vector \( \mathbf{r} \)
The position vector \( \mathbf{r} \) from point \( \mathbf{r_2} \) to point \( \mathbf{r_1} \) is given by:
\[ \mathbf{r} = \mathbf{r_1} - \mathbf{r_2} \]
Substituting the values:
\[ \mathbf{r} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) \]

### Step 3: Simplify the position vector
Now, simplify the expression:
\[ \mathbf{r} = (2\hat{i} - \hat{i}) + (-\hat{j} - 2\hat{j}) + (3\hat{k} + \hat{k}) \]
\[ \mathbf{r} = \hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 4: Calculate the moment \( \tau \)
The moment \( \tau \) about point \( \mathbf{r_2} \) is given by the cross product:
\[ \tau = \mathbf{r} \times \mathbf{F} \]
Substituting the values:
\[ \tau = (\hat{i} - 3\hat{j} + 4\hat{k}) \times (3\hat{i} + \hat{k}) \]

### Step 5: Set up the determinant for the cross product
We can calculate the cross product using the determinant:
\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 3 & 0 & 1 \end{vmatrix} \]

### Step 6: Calculate the determinant
Calculating the determinant:
\[ \tau = \hat{i} \begin{vmatrix} -3 & 4 \\ 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 3 & 0 \end{vmatrix} \]

Calculating each of these:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -3 & 4 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (0)(4) = -3 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 4 \\ 3 & 1 \end{vmatrix} = (1)(1) - (3)(4) = 1 - 12 = -11 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & -3 \\ 3 & 0 \end{vmatrix} = (1)(0) - (3)(-3) = 0 + 9 = 9 \]

### Step 7: Combine the results
Putting it all together:
\[ \tau = -3\hat{i} + 11\hat{j} + 9\hat{k} \]

### Final Result
Thus, the moment about the point \( \hat{i} + 2\hat{j} - \hat{k} \) is:
\[ \tau = -3\hat{i} + 11\hat{j} + 9\hat{k} \]