A force `vec(F)=3hat(i)+2hat(j)-4hat(k)` is applied at the point `(1, -1, 2)`. Find the moment of `vec(F)` about the point `(2, -1, 3)`.

Answer

To find the moment of the force \(\vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k}\) about the point \((2, -1, 3)\) when it is applied at the point \((1, -1, 2)\), we will follow these steps:

### Step 1: Define the position vectors
Let point A be the point where the force is applied, which is given as \(A(1, -1, 2)\). The position vector of point A is:
\[ \vec{A} = 1\hat{i} - 1\hat{j} + 2\hat{k} \]

Let point B be the point about which we are calculating the moment, given as \(B(2, -1, 3)\). The position vector of point B is:
\[ \vec{B} = 2\hat{i} - 1\hat{j} + 3\hat{k} \]

### Step 2: Find the vector \(\vec{R}\) from point A to point B
The vector \(\vec{R}\) from point A to point B is given by:
\[ \vec{R} = \vec{B} - \vec{A} \]
Calculating \(\vec{R}\):
\[ \vec{R} = (2\hat{i} - 1\hat{j} + 3\hat{k}) - (1\hat{i} - 1\hat{j} + 2\hat{k}) = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (3 - 2)\hat{k} = 1\hat{i} + 0\hat{j} + 1\hat{k} \]
Thus,
\[ \vec{R} = \hat{i} + \hat{k} \]

### Step 3: Calculate the moment \(\vec{M}\)
The moment of the force \(\vec{F}\) about point B is given by the cross product:
\[ \vec{M} = \vec{R} \times \vec{F} \]
Substituting the vectors:
\[ \vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k} \]
Now we compute the cross product:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 3 & 2 & -4 \end{vmatrix} \]

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

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} = (0)(-4) - (1)(2) = -2 \]
2. For \(-\hat{j}\):
\[ \begin{vmatrix} 1 & 1 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (1)(3) = -4 - 3 = -7 \quad \Rightarrow \quad +7\hat{j} \]
3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & 0 \\ 3 & 2 \end{vmatrix} = (1)(2) - (0)(3) = 2 \]

### Step 5: Combine the results
Now substituting back into the moment vector:
\[ \vec{M} = -2\hat{i} + 7\hat{j} + 2\hat{k} \]

### Final Answer
Thus, the moment of the force \(\vec{F}\) about the point \((2, -1, 3)\) is:
\[ \vec{M} = -2\hat{i} + 7\hat{j} + 2\hat{k} \]