The moment of a force represented by `F=hat(i)+2hat(j)+3hat(k)` about the point `2hat(i)-hat(j)+hat(k)` is equal to

Options

A. `5hat(i)-5hat(j)+5hat(k)`
B. `5hat(i)+5hat(j)-6hat(k)`
C. `-5hat(i)-5hat(j)+5hat(k)`
D. `-5hat(i)-5hat(j)+2hat(k)`

Answer

To find the moment of the force represented by the vector \( \mathbf{F} = \hat{i} + 2\hat{j} + 3\hat{k} \) about the point \( \mathbf{r_0} = 2\hat{i} - \hat{j} + \hat{k} \), we will use the formula for the moment of a force, which is given by:

\[ \mathbf{M} = \mathbf{r} \times \mathbf{F} \]

where \( \mathbf{r} \) is the position vector from the point about which we are calculating the moment to the point of application of the force.

### Step 1: Determine the position vector \( \mathbf{r} \)

The position vector \( \mathbf{r} \) is calculated as:

\[ \mathbf{r} = \mathbf{r_{F}} - \mathbf{r_{0}} \]

Assuming the force \( \mathbf{F} \) is applied at the origin (0, 0, 0), we have:

\[ \mathbf{r_{F}} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \hat{0} \]

Thus,

\[ \mathbf{r} = \hat{0} - (2\hat{i} - \hat{j} + \hat{k}) = -2\hat{i} + \hat{j} - \hat{k} \]

### Step 2: Set up the cross product

Now we will calculate the moment \( \mathbf{M} \) using the cross product \( \mathbf{M} = \mathbf{r} \times \mathbf{F} \):

\[ \mathbf{M} = (-2\hat{i} + \hat{j} - \hat{k}) \times (\hat{i} + 2\hat{j} + 3\hat{k}) \]

### Step 3: Write the cross product in determinant form

We can express this cross product using the determinant of a matrix:

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

### Step 4: Calculate the determinant

Now, we will expand the determinant:

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

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} -2 & -1 \\ 1 & 3 \end{vmatrix} = (-2)(3) - (-1)(1) = -6 + 1 = -5 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} -2 & 1 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (1)(1) = -4 - 1 = -5 \]

### Step 5: Combine the results

Now substituting back into the expression for \( \mathbf{M} \):

\[ \mathbf{M} = 5\hat{i} + 5\hat{j} - 5\hat{k} \]

Thus, we can write:

\[ \mathbf{M} = -5\hat{i} - 5\hat{j} + 5\hat{k} \]

### Final Answer

The moment of the force about the point is:

\[ \mathbf{M} = -5\hat{i} - 5\hat{j} + 5\hat{k} \]