The torque due to the force $(2\hat{i}+\hat{j}+2\hat{k})$ about the origin, acting on a particle whose position vector is $(\hat{i}+\hat{j}+\hat{k})$ , would be

Given Data:

• Force vector: $$\mathbf{F} = (2\hat{i} + \hat{j} + 2\hat{k})$$
• Position vector: $$\mathbf{r} = (\hat{i} + \hat{j} + \hat{k})$$

Step 1: Formula for Torque

Torque ($\boldsymbol{\tau}$) is given by:

$$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$$

where $\times$ represents the cross product.

Step 2: Compute the Cross Product

Using the determinant method:

$$\boldsymbol{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$

Expanding along the first row:

$$\boldsymbol{\tau} = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$$

Step 3: Compute the 2×2 Determinants

For $\hat{i}$:

1. $$\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1 \times 2) - (1 \times 1) = 2 - 1 = 1$$

For $\hat{j}$:

1. $$\begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = (1 \times 2) - (1 \times 2) = 2 - 2 = 0$$

For $\hat{k}$:

1. $$\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = (1 \times 1) - (1 \times 2) = 1 - 2 = -1$$

Step 4: Final Torque Vector

$$\boldsymbol{\tau} = (1)\hat{i} - (0)\hat{j} + (-1)\hat{k}$$

 $$\boldsymbol{\tau} = \hat{i} - \hat{k}$$

Final Answer:

$$\boldsymbol{\tau} = \hat{i} - \hat{k}$$