Question

The torque of a force \( \mathbf{F} = 4\hat{i} - 2\hat{j} + 3\hat{k} \) about the origin is \( \mathbf{r} \). The force acts on a particle having position vector \( \hat{i} + 2\hat{j} - \hat{k} \). Then the torque \( \mathbf{r} \) is:

• A. \( 4\hat{i} - \hat{j} - 10\hat{k} \)
• B. \( 8\hat{i} - 7\hat{j} + 6\hat{k} \)
• C. \( 4\hat{i} - 7\hat{j} - 10\hat{k} \)
• D. \( 8\hat{i} + \hat{j} - 10\hat{k} \)

Hint:

When calculating torque using the cross product, remember to use the determinant of the matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the position and force vectors.

Answer 1

The Correct Option is A

The torque \( \mathbf{r} \) is given by the cross product of the position vector \( \mathbf{r} \) and the force \( \mathbf{F} \): \[ \mathbf{r} = \mathbf{r} \times \mathbf{F} \] The position vector \( \mathbf{r} = \hat{i} + 2\hat{j} - \hat{k} \) and the force \( \mathbf{F} = 4\hat{i} - 2\hat{j} + 3\hat{k} \). We can calculate the cross product using the determinant formula: \[ \mathbf{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -1
4 & -2 & 3 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{r} = \hat{i} \begin{vmatrix} 2 & -1
-2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1
4 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2
4 & -2 \end{vmatrix} \] \[ \mathbf{r} = \hat{i} (2 \times 3 - (-1) \times (-2)) - \hat{j} (1 \times 3 - (-1) \times 4) + \hat{k} (1 \times (-2) - 2 \times 4) \] \[ \mathbf{r} = \hat{i} (6 - 2) - \hat{j} (3 + 4) + \hat{k} (-2 - 8) \] \[ \mathbf{r} = \hat{i} (4) - \hat{j} (7) + \hat{k} (-10) \] Thus, the torque is: \[ \mathbf{r} = 4\hat{i} - \hat{j} - 10\hat{k} \]