QJEEM25Phc036: JEE Main 2025 Physics - Class 11 JEEM25Phc036 Solution: The coordinates of a particle with respect to origin in a given …

The torque $\vec{\tau}$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$ :

$\vec{\tau} = \vec{r} \times \vec{F}$

Given the position vector $\vec{r} = (1, 1, 1) \, \text{m}$ and the force vector $\vec{F} = \hat{i} - \hat{j} + \hat{k}$ , we need to calculate the cross product:

$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$

Calculating the determinant, we have:

$\vec{\tau} = \hat{i} \left(1 \cdot 1 - 1 \cdot (-1)\right) - \hat{j} \left(1 \cdot 1 - 1 \cdot 1\right) + \hat{k} \left(1 \cdot (-1) - 1 \cdot 1\right)$

This simplifies to:

$\vec{\tau} = \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$

$\vec{\tau} = 2\hat{i} - 0\hat{j} - 2\hat{k}$

The torque vector is $\vec{\tau} = 2\hat{i} - 2\hat{k}$ .

To find the magnitude of the torque in the z-direction, we look at the $\hat{k}$ component:

$\tau_z = -2$

The magnitude of torque in the z-direction is:

$|\tau_z| = 2 \, \text{Nm}$

Thus, the magnitude of the torque in the z-direction is 2 Newton-meters.