What is the torque of a force `overset(rarr)F=(2 hat i -3 hat j +4 hat k)` Newton acting at a point `overset(rarr)r=(3 hat I +2 hat j + 3 hat k)` metre about the origin ?

Options

A. (a)`6hati -6 hatj +12hatk`
B. (b)`17hat i-6 hat j-13 hatk`
C. (c)`-6hat I +6hat j-12 hatk`
D. (d)`-17hati+6 hat j+13 hatk`

Explanation

To find the torque of the force \(\vec{F} = (2 \hat{i} - 3 \hat{j} + 4 \hat{k})\) Newton acting at the point \(\vec{r} = (3 \hat{i} + 2 \hat{j} + 3 \hat{k})\) meter about the origin, we can use the formula for torque:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

### Step 1: Write down the vectors
We have:
- \(\vec{F} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}\)
- \(\vec{r} = 3 \hat{i} + 2 \hat{j} + 3 \hat{k}\)

### Step 2: Set up the determinant for the cross product
The torque can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of the vectors \(\vec{r}\) and \(\vec{F}\):

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we can expand it as follows:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} = (2)(4) - (3)(-3) = 8 + 9 = 17 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} = (3)(4) - (3)(2) = 12 - 6 = 6 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (2)(2) = -9 - 4 = -13 \]

### Step 4: Combine the results
Now substituting back into the expression for torque:

\[ \vec{\tau} = 17 \hat{i} - 6 \hat{j} - 13 \hat{k} \]

### Final Result
Thus, the torque \(\vec{\tau}\) is:

\[ \vec{\tau} = 17 \hat{i} - 6 \hat{j} - 13 \hat{k} \, \text{N m} \]

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