A force `vecF=alphahati+3hatj+6hatk` is acting at a point `vecr=2hati-6hatj-12hatk`. The value of `alpha` for which angular momentum about origin is conserved is:

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of \(\alpha\) for which the angular momentum about the origin is conserved. This occurs when the torque is zero.

### Step-by-step Solution:

1. **Understanding Torque**:
The torque \(\vec{\tau}\) is given by the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\):
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
For angular momentum to be conserved, we need \(\vec{\tau} = 0\).

2. **Given Vectors**:
The force vector is given as:
\[ \vec{F} = \alpha \hat{i} + 3 \hat{j} + 6 \hat{k} \]
The position vector is given as:
\[ \vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k} \]

3. **Calculating the Cross Product**:
We need to compute \(\vec{r} \times \vec{F}\):
\[ \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix} \]

4. **Expanding the Determinant**:
Using the determinant formula, we can calculate:
\[ \vec{r} \times \vec{F} = \hat{i} \begin{vmatrix} -6 & -12 \\ 3 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -12 \\ \alpha & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ \alpha & 3 \end{vmatrix} \]

- For \(\hat{i}\):
\[ \begin{vmatrix} -6 & -12 \\ 3 & 6 \end{vmatrix} = (-6)(6) - (-12)(3) = -36 + 36 = 0 \]

- For \(\hat{j}\):
\[ \begin{vmatrix} 2 & -12 \\ \alpha & 6 \end{vmatrix} = (2)(6) - (-12)(\alpha) = 12 + 12\alpha \]

- For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -6 \\ \alpha & 3 \end{vmatrix} = (2)(3) - (-6)(\alpha) = 6 + 6\alpha \]

5. **Putting it Together**:
Therefore, the torque vector becomes:
\[ \vec{\tau} = 0 \hat{i} - (12 + 12\alpha) \hat{j} + (6 + 6\alpha) \hat{k} \]
This simplifies to:
\[ \vec{\tau} = - (12 + 12\alpha) \hat{j} + (6 + 6\alpha) \hat{k} \]

6. **Setting Torque to Zero**:
For the torque to be zero, both components must be zero:
\[ - (12 + 12\alpha) = 0 \quad \text{(1)} \]
\[ 6 + 6\alpha = 0 \quad \text{(2)} \]

7. **Solving the Equations**:
From equation (1):
\[ 12 + 12\alpha = 0 \implies \alpha = -1 \]

From equation (2):
\[ 6 + 6\alpha = 0 \implies \alpha = -1 \]

8. **Conclusion**:
The value of \(\alpha\) for which the angular momentum about the origin is conserved is:
\[ \alpha = -1 \]