The torque of force `barF= (2 hat i -3 hat j + 4 hat k)` newton acting at the point =` (3 hat i + 2 hat j+ 3 hat k)` metre about origin is `tau` (in N-m). Find x. component of `tau` .

Answer

To find the x-component of the torque \(\tau\) produced by the force \(\bar{F} = (2 \hat{i} - 3 \hat{j} + 4 \hat{k})\) acting at the point \(\bar{R} = (3 \hat{i} + 2 \hat{j} + 3 \hat{k})\) about the origin, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the vectors**:
- Force vector: \(\bar{F} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}\)
- Position vector: \(\bar{R} = 3 \hat{i} + 2 \hat{j} + 3 \hat{k}\)

2. **Use the torque formula**:
- The torque \(\tau\) is given by the cross product of the position vector \(\bar{R}\) and the force vector \(\bar{F}\):
\[ \tau = \bar{R} \times \bar{F} \]

3. **Set up the determinant for the cross product**:
- We can express the cross product using the determinant of a matrix:
\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} \]

4. **Calculate the determinant**:
- Expanding the determinant:
\[ \tau = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} \]

5. **Calculate each of the 2x2 determinants**:
- For \(\hat{i}\):
\[ \begin{vmatrix} 2 & 3 \\ -3 & 4 \end{vmatrix} = (2)(4) - (3)(-3) = 8 + 9 = 17 \]
- For \(-\hat{j}\):
\[ \begin{vmatrix} 3 & 3 \\ 2 & 4 \end{vmatrix} = (3)(4) - (3)(2) = 12 - 6 = 6 \]
- For \(\hat{k}\):
\[ \begin{vmatrix} 3 & 2 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (2)(2) = -9 - 4 = -13 \]

6. **Combine the results**:
- Thus, the torque vector \(\tau\) is:
\[ \tau = 17 \hat{i} - 6 \hat{j} - 13 \hat{k} \]

7. **Extract the x-component**:
- The x-component of \(\tau\) is simply the coefficient of \(\hat{i}\):
\[ \text{x-component of } \tau = 17 \]

### Final Answer:
The x-component of the torque \(\tau\) is \(17 \, \text{N-m}\).