Find the moment (torque) about the point `hat(i)+2hat(j)+3hat(k)` of a force represented by `hat(i)+hat(j)+hat(k)` acting through the point `-2hat(i)+3hat(j)+hat(k)`.

Answer

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

### Step 1: Identify the Force and Points
- The force vector \( \mathbf{F} = \hat{i} + \hat{j} + \hat{k} \).
- The point of application of the force \( \mathbf{P} = -2\hat{i} + 3\hat{j} + \hat{k} \).
- The point about which we want to find the moment \( \mathbf{O} = \hat{i} + 2\hat{j} + 3\hat{k} \).

### Step 2: Calculate the Position Vector
- The position vector from point \( \mathbf{O} \) to point \( \mathbf{P} \) is given by:
\[ \mathbf{r} = \mathbf{P} - \mathbf{O} \]
Substituting the values:
\[ \mathbf{r} = (-2\hat{i} + 3\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) \]
\[ \mathbf{r} = (-2 - 1)\hat{i} + (3 - 2)\hat{j} + (1 - 3)\hat{k} \]
\[ \mathbf{r} = -3\hat{i} + 1\hat{j} - 2\hat{k} \]

### Step 3: Calculate the Moment (Torque)
- The moment (torque) \( \mathbf{M} \) about point \( \mathbf{O} \) is given by the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \):
\[ \mathbf{M} = \mathbf{r} \times \mathbf{F} \]
Substituting the vectors:
\[ \mathbf{M} = (-3\hat{i} + 1\hat{j} - 2\hat{k}) \times (\hat{i} + \hat{j} + \hat{k}) \]

### Step 4: Compute the Cross Product
Using the determinant method to compute the cross product:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 1 & -2 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & -2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & 1 \\ 1 & 1 \end{vmatrix} \]

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

Putting it all together:
\[ \mathbf{M} = 3\hat{i} + 1\hat{j} - 4\hat{k} \]

### Final Answer
The moment (torque) about the point \( \hat{i} + 2\hat{j} + 3\hat{k} \) is:
\[ \mathbf{M} = 3\hat{i} + 1\hat{j} - 4\hat{k} \]

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