If \( \vec{F}=2 \hat{\imath}+3 \hat{\jmath} \) acts at point \( C \) and \( \overrightarrow{A B}=4 \hat{\imath}+6 \hat{\jmath} \) and \( \overrightarrow{M_{A}}=\left(m^{2}+4\right) \hat{k}, \overrightarrow{M_{B}}= \) \( (4 \mathrm{~m}) \hat{k} \), then \( m=\cdots \ldots \) \( \begin{array}{llll}\text { (a) } 3 & \text { (b) } 2 & \text { (c) } 4 & \text { (d) } 1\end{array} \)

To solve for \( m \), let's analyze the given information step by step. **Given:** - Force vector: \( \vec{F} = 2 \hat{\imath} + 3 \hat{\jmath} \) - Vector from point \( A \) to \( B \): \( \overrightarrow{AB} = 4 \hat{\imath} + 6 \hat{\jmath} \) - Moment about point \( A \): \( \overrightarrow{M_A} = (m^2 + 4) \hat{k} \) - Moment about point \( B \): \( \overrightarrow{M_B} = 4m \hat{k} \) **Relationship Between Moments:** The relationship between the moments about two points \( A \) and \( B \) due to a force \( \vec{F} \) is given by: \[ \overrightarrow{M_A} = \overrightarrow{M_B} + \overrightarrow{AB} \times \vec{F} \] **Calculating the Cross Product:** \[ \overrightarrow{AB} \times \vec{F} = (4 \hat{\imath} + 6 \hat{\jmath}) \times (2 \hat{\imath} + 3 \hat{\jmath}) \] \[ = 4\hat{\imath} \times 2\hat{\imath} + 4\hat{\imath} \times 3\hat{\jmath} + 6\hat{\jmath} \times 2\hat{\imath} + 6\hat{\jmath} \times 3\hat{\jmath} \] \[ = 0 + 12\hat{k} - 12\hat{k} + 0 = 0 \] **Simplifying the Equation:** Since \( \overrightarrow{AB} \times \vec{F} = 0 \), \[ \overrightarrow{M_A} = \overrightarrow{M_B} \] \[ m^2 + 4 = 4m \] \[ m^2 - 4m + 4 = 0 \] \[ (m - 2)^2 = 0 \] \[ m = 2 \] **Answer:** **Option (b) 2**