1. **Problem statement:** Given a force vector $\vec{F} = l \hat{i} + m \hat{j}$ acting at point $A(4,2)$, the moment vector about the origin is $-22 \hat{k}$ and about point $B(13,1)$ is $22 \hat{k}$. Find $l + m$. 2. **Formula for moment vector:** The moment $\vec{M}$ of a force $\vec{F}$ about a point with position vector $\vec{r}$ is given by: $$\vec{M} = \vec{r} \times \vec{F}$$ where $\times$ denotes the cross product. 3. **Calculate moment about origin:** Position vector of point $A$ relative to origin is $\vec{r}_A = 4 \hat{i} + 2 \hat{j}$. Force vector is $\vec{F} = l \hat{i} + m \hat{j}$. The moment about origin is: $$\vec{M}_O = \vec{r}_A \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & 0 \\ l & m & 0 \end{vmatrix} = (4 \times m - 2 \times l) \hat{k} = (4m - 2l) \hat{k}$$ Given $\vec{M}_O = -22 \hat{k}$, so: $$4m - 2l = -22$$ 4. **Calculate moment about point B:** Position vector of point $B$ is $\vec{r}_B = 13 \hat{i} + 1 \hat{j}$. Position vector of $A$ relative to $B$ is: $$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (4 - 13) \hat{i} + (2 - 1) \hat{j} = -9 \hat{i} + 1 \hat{j}$$ Moment about point $B$ is: $$\vec{M}_B = \vec{r}_{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -9 & 1 & 0 \\ l & m & 0 \end{vmatrix} = (-9m - 1 \times l) \hat{k} = (-9m - l) \hat{k}$$ Given $\vec{M}_B = 22 \hat{k}$, so: $$-9m - l = 22$$ 5. **Solve the system of equations:** From step 3: $$4m - 2l = -22$$ From step 4: $$-9m - l = 22$$ Rewrite second equation: $$-9m - l = 22 \implies l = -9m - 22$$ Substitute $l$ into first equation: $$4m - 2(-9m - 22) = -22$$ $$4m + 18m + 44 = -22$$ $$22m + 44 = -22$$ $$22m = -22 - 44 = -66$$ $$m = \frac{-66}{22} = -3$$ Find $l$: $$l = -9(-3) - 22 = 27 - 22 = 5$$ 6. **Find $l + m$:** $$l + m = 5 + (-3) = 2$$ **Final answer:** $\boxed{2}$ which corresponds to option b).