Question

A force of \( (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{N} \) acts on a particle whose position vector with respect to the origin of the coordinate system is \( (6\hat{i} + b\hat{j} + 12\hat{k}) \, \text{m} \). If the angular momentum of the body is constant, the value of \( b \) is:

• A. \( 6 \)
• B. \( 9 \)
• C. \( 12 \)
• D. \( 3 \)

Hint:

When solving for unknowns in vector problems, ensure to break down each component of the cross product and equate them to zero to maintain consistency, especially when dealing with angular momentum and force relationships.

Answer 1

The Correct Option is B

The angular momentum \( \vec{L} \) of a particle is given by the cross product of its position vector \( \vec{r} \) and its momentum \( \vec{p} \): \[ \vec{L} = \vec{r} \times \vec{p} \] where \( \vec{p} = m\vec{v} \) and \( \vec{v} \) is the velocity of the particle. The force acting on the particle is \( \vec{F} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{N} \), and this force is responsible for changing the angular momentum. Since the angular momentum is constant, the rate of change of angular momentum must be zero. Therefore, we have: \[ \vec{r} \times \vec{F} = 0 \] Substitute \( \vec{r} = (6\hat{i} + b\hat{j} + 12\hat{k}) \, \text{m} \) and \( \vec{F} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \, \text{N} \), and compute the cross product: \[ (6\hat{i} + b\hat{j} + 12\hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = 0 \] Performing the cross product yields: \[ \hat{i}(b \cdot 4 - 12 \cdot 3) - \hat{j}(6 \cdot 4 - 12 \cdot 2) + \hat{k}(6 \cdot 3 - b \cdot 2) = 0 \] Simplifying: \[ \hat{i}(4b - 36) - \hat{j}(24 - 24) + \hat{k}(18 - 2b) = 0 \] Thus, we have the following components: \[ \hat{i}: 4b - 36 = 0 \] \[ \hat{j}: 0 = 0 \quad \text{(this term is automatically satisfied)} \] \[ \hat{k}: 18 - 2b = 0 \] Solving for \( b \) from the first equation: \[ 4b = 36 \quad \Rightarrow \quad b = 9 \] Thus, the correct value of \( b \) is \( 9 \). Hence, the correct answer is option (2).

Answer 2

Step 1: Use the condition for constant angular momentum.
For the angular momentum to be constant, the net torque \( \vec{\tau} \) on the particle must be zero:
\[ \vec{\tau} = \vec{r} \times \vec{F} = 0 \]
Step 2: Write the given vectors.
Position vector: \( \vec{r} = 6\hat{i} + b\hat{j} + 12\hat{k} \)
Force vector: \( \vec{F} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)

Step 3: Compute the cross product \( \vec{r} \times \vec{F} \).
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & b & 12 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(b \cdot 4 - 12 \cdot 3) - \hat{j}(6 \cdot 4 - 12 \cdot 2) + \hat{k}(6 \cdot 3 - b \cdot 2) \] \[ = \hat{i}(4b - 36) - \hat{j}(24 - 24) + \hat{k}(18 - 2b) = \hat{i}(4b - 36) + \hat{k}(18 - 2b) \]
Step 4: Set the torque to zero vector.
\[ \hat{i}(4b - 36) + \hat{k}(18 - 2b) = 0 \Rightarrow 4b - 36 = 0 \quad \text{and} \quad 18 - 2b = 0 \] Solving both: - \( 4b = 36 \Rightarrow b = 9 \) - \( 2b = 18 \Rightarrow b = 9 \) ✅ (both consistent)

Final Answer: \( \boxed{9} \)