Given `vecF=(4hati-10hatj)and vecr=(5hati-3hatj)`. then torque `vectau` is

Options

A. 1. `-62 hatj`
B. 2. `62 hatk`
C. 3. `38 hati`
D. 4. `-38 hatk`

Answer

To find the torque \(\vec{\tau}\) given the force \(\vec{F} = 4 \hat{i} - 10 \hat{j}\) and the position vector \(\vec{r} = 5 \hat{i} - 3 \hat{j}\), we will use the formula for torque:

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

### Step 1: Write the vectors
We have:
\[ \vec{r} = 5 \hat{i} - 3 \hat{j} \]
\[ \vec{F} = 4 \hat{i} - 10 \hat{j} \]

### Step 2: Set up the cross product
The cross product of two vectors \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) and \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\) can be calculated using the determinant of a matrix:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

In our case, since both vectors lie in the xy-plane, we can treat the k-component as zero:

\[ \vec{r} = 5 \hat{i} - 3 \hat{j} + 0 \hat{k} \]
\[ \vec{F} = 4 \hat{i} - 10 \hat{j} + 0 \hat{k} \]

### Step 3: Calculate the determinant
Now we can set up the determinant:

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -3 & 0 \\ 4 & -10 & 0 \end{vmatrix} \]

### Step 4: Expand the determinant
Calculating the determinant, we have:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 0 \\ 4 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & -3 \\ 4 & -10 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} = (-3)(0) - (0)(-10) = 0 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 5 & 0 \\ 4 & 0 \end{vmatrix} = (5)(0) - (0)(4) = 0 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 5 & -3 \\ 4 & -10 \end{vmatrix} = (5)(-10) - (-3)(4) = -50 + 12 = -38 \]

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

\[ \vec{\tau} = 0 \hat{i} - 0 \hat{j} - 38 \hat{k} = -38 \hat{k} \]

### Final Answer
Thus, the torque \(\vec{\tau}\) is:

\[ \vec{\tau} = -38 \hat{k} \]

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