The velocity of a particle of mass m is `vec(v) = 5 hat(i) + 4 hat(j) + 6 hat(k)" " "when at" " " vec(r) = - 2 hat(i) + 4 hat (j) + 6 hat(k).` The angular momentum of the particle about the origin is

Options

A. 42 m
B. `m(42hat(j) - 28 hat (k))`
C. `m(42hat(i) - 28 hat (j))`
D. `m(42hat(k) - 28 hat (i))`

Answer

To find the angular momentum of a particle about the origin, we can use the formula:

\[ \vec{L} = \vec{r} \times \vec{p} \]

where \(\vec{L}\) is the angular momentum, \(\vec{r}\) is the position vector, and \(\vec{p}\) is the linear momentum of the particle. The linear momentum \(\vec{p}\) can be expressed as:

\[ \vec{p} = m \vec{v} \]

Given:
- \(\vec{v} = 5 \hat{i} + 4 \hat{j} + 6 \hat{k}\)
- \(\vec{r} = -2 \hat{i} + 4 \hat{j} + 6 \hat{k}\)

### Step 1: Calculate the linear momentum \(\vec{p}\)
Using the formula for linear momentum:

\[ \vec{p} = m \vec{v} = m (5 \hat{i} + 4 \hat{j} + 6 \hat{k}) = 5m \hat{i} + 4m \hat{j} + 6m \hat{k} \]

### Step 2: Calculate the angular momentum \(\vec{L}\)
Now we can calculate the angular momentum using the cross product:

\[ \vec{L} = \vec{r} \times \vec{p} \]

Substituting the values of \(\vec{r}\) and \(\vec{p}\):

\[ \vec{L} = (-2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (5m \hat{i} + 4m \hat{j} + 6m \hat{k}) \]

### Step 3: Set up the determinant for the cross product
We can set up the determinant as follows:

\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 6 \\ 5m & 4m & 6m \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant:

1. For \(\hat{i}\):

\[ \hat{i} \left( 4 \cdot 6m - 6 \cdot 4m \right) = \hat{i} (24m - 24m) = 0 \hat{i} \]

2. For \(\hat{j}\):

\[ -\hat{j} \left( -2 \cdot 6m - 6 \cdot 5m \right) = -\hat{j} ( -12m - 30m) = -\hat{j} (-42m) = 42m \hat{j} \]

3. For \(\hat{k}\):

\[ \hat{k} \left( -2 \cdot 4m - 4 \cdot 5m \right) = \hat{k} (-8m - 20m) = -28m \hat{k} \]

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

\[ \vec{L} = 0 \hat{i} + 42m \hat{j} - 28m \hat{k} \]

Thus, the angular momentum of the particle about the origin is:

\[ \vec{L} = 42m \hat{j} - 28m \hat{k} \]

### Final Answer
The angular momentum of the particle about the origin is:

\[ \vec{L} = 42m \hat{j} - 28m \hat{k} \]