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vecF = a hat i + 3hat j+ 6 hat k and vec...

`vecF = a hat i + 3hat j+ 6 hat k` and `vec r = 2hat i-6hat j -12 hat k`. The value of `a` for which the angular momentum is conserved is

A

-1

B

2

C

zero

D

1

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To find the value of \( a \) for which the angular momentum is conserved, we need to analyze the given vectors and apply the concept of torque. ### Step-by-Step Solution: 1. **Identify the vectors**: We have the force vector \( \vec{F} = a \hat{i} + 3 \hat{j} + 6 \hat{k} \) and the position vector \( \vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k} \). 2. **Torque Calculation**: The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \): \[ \vec{\tau} = \vec{r} \times \vec{F} \] 3. **Set up the cross product**: \[ \vec{r} = \begin{pmatrix} 2 \\ -6 \\ -12 \end{pmatrix}, \quad \vec{F} = \begin{pmatrix} a \\ 3 \\ 6 \end{pmatrix} \] The cross product can be calculated using the determinant of a matrix: \[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ a & 3 & 6 \end{vmatrix} \] 4. **Calculate the determinant**: \[ \vec{\tau} = \hat{i} \begin{vmatrix} -6 & -12 \\ 3 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -12 \\ a & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ a & 3 \end{vmatrix} \] - For \( \hat{i} \): \[ = -6 \cdot 6 - (-12) \cdot 3 = -36 + 36 = 0 \] - For \( \hat{j} \): \[ = 2 \cdot 6 - (-12) \cdot a = 12 + 12a \] - For \( \hat{k} \): \[ = 2 \cdot 3 - (-6) \cdot a = 6 + 6a \] 5. **Combine the results**: \[ \vec{\tau} = 0 \hat{i} - (12 + 12a) \hat{j} + (6 + 6a) \hat{k} \] Therefore, the torque vector is: \[ \vec{\tau} = 0 \hat{i} - (12 + 12a) \hat{j} + (6 + 6a) \hat{k} \] 6. **Condition for Angular Momentum Conservation**: For angular momentum to be conserved, the torque must be zero: \[ 12 + 12a = 0 \quad \text{and} \quad 6 + 6a = 0 \] 7. **Solve for \( a \)**: - From \( 12 + 12a = 0 \): \[ 12a = -12 \implies a = -1 \] - From \( 6 + 6a = 0 \): \[ 6a = -6 \implies a = -1 \] Thus, the value of \( a \) for which the angular momentum is conserved is: \[ \boxed{-1} \]
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