What is the torque of the force `vecF=(2hati+3hatj+4hatk)N` acting at the point `vecr=(2hati+3hatj+4hatk)m` about the origin? (Note: Tortue, `vectau=vecrxxvecF`)

Explanation

To find the torque of the force \(\vec{F} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{N}\) acting at the point \(\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{m}\) about the origin, we will use the formula for torque:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

### Step 1: Write down the vectors
We have:
- \(\vec{r} = 2\hat{i} + 3\hat{j} + 4\hat{k}\)
- \(\vec{F} = 2\hat{i} + 3\hat{j} + 4\hat{k}\)

### Step 2: Set up the cross product using the determinant method
The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{r}\) and \(\vec{F}\):

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 2 & 3 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can expand it as follows:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 3 & 4 \\ 3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} \]

### Step 4: Evaluate the 2x2 determinants
Each of the 2x2 determinants evaluates to zero because the rows are identical:

1. \(\begin{vmatrix} 3 & 4 \\ 3 & 4 \end{vmatrix} = 3 \cdot 4 - 4 \cdot 3 = 0\)

2. \(\begin{vmatrix} 2 & 4 \\ 2 & 4 \end{vmatrix} = 2 \cdot 4 - 4 \cdot 2 = 0\)

3. \(\begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 2 = 0\)

### Step 5: Combine the results
Putting it all together, we find:

\[ \vec{\tau} = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = \vec{0} \]

### Final Result
The torque \(\vec{\tau}\) is the null vector:

\[ \vec{\tau} = \vec{0} \]