A force `(2hati+3hatj-hatk) ` newtons acts on a particle having position vector `(hati-hatj+2hatk)` . Find the torque of the force about origin.

Answer

To find the torque of the force about the origin, we can follow these steps:

### Step 1: Identify the position vector and force vector
Given:
- Position vector \( \mathbf{r} = \hat{i} - \hat{j} + 2\hat{k} \) (in newtons)
- Force vector \( \mathbf{F} = 2\hat{i} + 3\hat{j} - \hat{k} \) (in newtons)

### Step 2: Use the formula for torque
The torque \( \mathbf{\tau} \) about the origin is given by the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \):
\[ \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \]

### Step 3: Set up the determinant for the cross product
We can calculate the cross product using the determinant method. We set up the following determinant:
\[ \mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix} \]

### Step 4: Calculate the determinant
To calculate the determinant, we expand it:
\[ \mathbf{\tau} = \hat{i} \begin{vmatrix} -1 & 2 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -1 & 2 \\ 3 & -1 \end{vmatrix} = (-1)(-1) - (2)(3) = 1 - 6 = -5 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5 \]

Putting it all together:
\[ \mathbf{\tau} = -5\hat{i} + 5\hat{j} + 5\hat{k} \]

### Step 5: Write the final torque vector
Thus, the torque vector is:
\[ \mathbf{\tau} = -5\hat{i} + 5\hat{j} + 5\hat{k} \]

### Step 6: Calculate the magnitude of the torque vector
The magnitude of the torque vector is given by:
\[ |\mathbf{\tau}| = \sqrt{(-5)^2 + (5)^2 + (5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} \]

### Step 7: Final answer
The torque of the force about the origin is:
\[ |\mathbf{\tau}| = \sqrt{75} \text{ Newton meter} \]

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